We have: \(log{ x }^{ y }=log\quad { e }^{ x-y }\)
\(ylogx=(x-y)\)
\(y(1+log\quad x)=x\)
\(y=\frac { x }{ 1+log\quad x } \)
\(\frac { dy }{ dx } =\frac { \left( 1+log\quad x \right) .1-x\left( 0+\frac { 1 }{ x } \right) }{ { \left( 1+logx \right) }^{ 2 } } \)
\(=\frac { 1+log\quad x-1 }{ { \left( 1+logx \right) }^{ 2 } } =\frac { log\quad x }{ { \left( 1+logx \right) }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { log\quad x }{ { \left( log\quad e+log\quad x \right) }^{ 2 } } \)

We have: \({ x }^{ y }={ e }^{ x-y }\)
Taking logs., \(ylogx=(x-y)\)
\(y(1+log\quad x)=x\)
\(y=\frac { x }{ 1+log\quad x } \)
\(\frac { dy }{ dx } =\frac { \left( 1+log\quad x \right) .1-x\left( 0+\frac { 1 }{ x } \right) }{ { \left( 1+logx \right) }^{ 2 } } \)
\(=\frac { 1+log\quad x-1 }{ { \left( 1+logx \right) }^{ 2 } } =\frac { log\quad x }{ { \left( 1+logx \right) }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { log\quad x }{ { \left( log\quad e+log\quad x \right) }^{ 2 } } \)

Let y=\({ sin }^{ -1 }\left( \frac { { 2 }^{ x+1 }.{ 3 }^{ x } }{ 1+{ \left( 36 \right) }^{ x } } \right) \)
\(={ sin }^{ -1 }\left( \frac { { 2.2 }^{ x }.{ 3 }^{ x } }{ 1+{ \left( 6 \right) }^{ 2x } } \right) ={ sin }^{ -1 }\left( \frac { { 2.6 }^{ x } }{ 1+{ \left( 6 \right) }^{ 2x } } \right) \)
Put \({ 6 }^{ x }=tan\theta \) so that \(\theta ={ tan }^{ -1 }\left( { 6 }^{ x } \right) \)
Then \(y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
\(={ sin }^{ -1 }\left( sin2\theta \right) =2\theta =2{ tan }^{ -1 }\left( { 6 }^{ x } \right) \)
\(\frac { dy }{ dx } =2.\frac { 1 }{ 1+{ \left( { 6 }^{ x } \right) }^{ 2 } } .\frac { d }{ dx } \left( { 6 }^{ x } \right) \)
\(=\frac { { 2.6 }^{ x }log6 }{ 1+{ \left( 6 \right) }^{ 2x } } =\frac { { 2 }^{ x+1 }.{ 3 }^{ x } }{ 1+{ \left( 36 \right) }^{ x } } \)

We have: \(xy={ e }^{ x-y }\) ....(1)
Diff.w.r.t.x, \(x.\frac { dy }{ dx } +y.1={ e }^{ x-y }.\frac { d }{ dx } (x-y)\)
\(x.\frac { dy }{ dx } +y={ e }^{ x-y }\left[ 1-\frac { dy }{ dx } \right] \)
\(\left( x+{ e }^{ x-y } \right) \frac { dy }{ dx } ={ e }^{ x-y }-y\)
\(\frac { dy }{ dx } =\frac { { e }^{ x-y }-y }{ x+{ e }^{ x-y } }\)
\(=\frac { xy-x }{ x+xy } \)
\(=\frac { y(x-1) }{ x(y+1) } \)
which is true.

Let y = \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) \)
Put \(x=tan\theta \)
\(y={ tan }^{ -1 }\frac { \sqrt { 1+{ tan }^{ 2 }\theta } -1 }{ tan\theta } \)
\(={ tan }^{ -1 }\left( \frac { sec\theta -1 }{ tan\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 1-cos\theta }{ sin\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ sin }\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { { sin }\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } \right) ={ tan }^{ -1 }\left( { tan }\frac { \theta }{ 2 } \right) \)
\(=\frac { \theta }{ 2 } =\frac { 1 }{ 2 } { tan }^{ -1 }x\)
Hence \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \frac { 1 }{ 1+{ x }^{ 2 } } =\frac { 1 }{ 2(1+{ x }^{ 2 }) } \)

We have: \(x={ e }^{ x/y },\)
Taking logs., \(log\quad x=\frac { x }{ y } \Rightarrow ylogx=x\)
Diff.w.r.t.x, \(y.\frac { 1 }{ x } +logx.\frac { dy }{ dx } =1\)
\( \frac { dy }{ dx } =\frac { 1-y/x }{ log\quad x } =\frac { x-y }{ xlogx } \)

We have: \({ tan }^{ -1 }\left( { x }^{ 2 }+{ y }^{ 2 } \right) =a\)
\({ x }^{ 2 }+{ y }^{ 2 }=tan\quad a\)
Diff.w.r.t.x, 2x+2y
\(\frac { dy }{ dx } =0\Rightarrow \frac { dy }{ dx } =-\frac { x }{ y } \)

We have: y = tan x + sec x
Dif.w.r.t.x, \(\frac { dy }{ dx } ={ sec }^{ 2 }x+secxtanx\)
\(=\frac { 1 }{ { cos }^{ 2 }x } +\frac { sinx }{ { cos }^{ 2 }x } =\frac { 1+sin\quad x }{ { cos }^{ 2 }x } \)
\(=\frac { 1+sin\quad x }{ 1-{ sin }^{ 2 }x } =\frac { 1+sin\quad x }{ \left( 1+sin\quad x \right) \left( 1-sin\quad x \right) } \)
\(=\frac { 1 }{ 1-sin\quad x } \)
Again diff.w.r.t. x,
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
which is true.

The given function is redefined as below:
\(f\left( x \right) =\begin{cases} { x }^{ 2 },\quad if\quad x\ge 0 \\ { -x }^{ 2 },\quad if\quad x<0 \end{cases}\)
\(Lf'(0)=\lim _{ h\rightarrow 0 }{ \frac { f(0-h)-f(0) }{ -h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -{ h }^{ 2 }-0 }{ -h } } =\lim _{ h\rightarrow 0 }{ (h) } =0\)
\(Rf'(0)=\lim _{ h\rightarrow 0 }{ \frac { { h }^{ 2 }-0 }{ h } } =\lim _{ h\rightarrow 0 }{ (h) } =0\)
\(Thus\ Lf'(0)=Rf'(0)\)
\(Hence,\ 'f'\ is\ derivable\ at\ x=0.\)

We have \(sin\quad x=\frac { 2t }{ 1+{ t }^{ 2 } } \quad and\quad tan\quad y=\frac { 2t }{ 1-{ t }^{ 2 } } \)
\(x={ sin }^{ -1 }\left[ \frac { 2t }{ 1+{ t }^{ 2 } } \right] and\quad y={ tan }^{ -1 }\left[ \frac { 2t }{ 1-{ t }^{ 2 } } \right] \)
\( Put\quad t=tan\theta \)
\(x={ sin }^{ -1 }\left[ \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right] \)
\(y={ tan }^{ -1 }\left[ \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right]\)
\(x={ sin }^{ -1 }\left( sin\quad 2\theta \right) \)
\(y={ tan }^{ -1 }\left( tan\quad 2\theta \right) \)
\( x=2\theta \)
\(y=2\theta \)
\(x=2{ tan }^{ -1 }t\)
\(and\quad y=2{ tan }^{ -1 }t\)
\(\frac { dx }{ dt } ={ \frac { 2 }{ 1+{ t }^{ 2 } } },\frac { dy }{ dt } =\frac { 2 }{ 1+{ t }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { 2/\left( 1+{ t }^{ 2 } \right) }{ 2/\left( 1+{ t }^{ 2 } \right) } =1\)

We have: \( f\left( x \right) =\frac { 1 }{ x-1 } \)
\(f[f(x)]=\frac { 1 }{ f\left( x \right) -1 } =\frac { 1 }{ \frac { 1 }{ x-1 } -1 } =\frac { x-1 }{ 1-x+1 } \)
\(=\frac { x-1 }{ 2-x } \)
Clearly, points of discontinuity are x = 1 and x = 2.

Here cos y = x cos(a + y)
\(Diff.\quad w.r.t.\quad y,\quad we\quad get:\)
\(\frac { dx }{ dy } =\frac { cos\quad y\left( -sin\left( \alpha +y \right) .1 \right) }{ { cos }^{ 2 }\left( a+y \right) } \)
\(=\frac { sin\left( a+y \right) cosy-cos\left( a+y \right) siny }{ { cos }^{ 2 }\left( a+y \right) } \)
\(=\frac { sin\left( \alpha +y-y \right) }{ { cos }^{ 2 }\left( a+y \right) } =\frac { sin\alpha }{ { cos }^{ 2 }\left( a+y \right) } \)
\(Hence,\frac { dy }{ dx } =\frac { 1 }{ dx/dy } =\frac { { cos }^{ 2 }\left( a+y \right) }{ sin\quad \alpha } \)

\(Let\quad y={ x }^{ x }+{ x }^{ a }+{ a }^{ x }+{ a }^{ a }\)
\(\frac { dy }{ dx } =\frac { d }{ dx } \left( { x }^{ x } \right) +\frac { d }{ dx } \left( { x }^{ a } \right) +\frac { d }{ dx } \left( { a }^{ x } \right) +\frac { d }{ dx } \left( { a }^{ a } \right) \)
\( { x }^{ x }(1+log\quad x)+a\quad { x }^{ a-1 }+{ a }^{ x }\quad log\quad a+0\)
\( { x }^{ x }(1+log\quad x)+a\quad { x }^{ a-1 }+{ a }^{ x }\quad log\quad a\)

\(Let\ y={ sin }^{ -1 }(x\sqrt { x } )\)
\( \frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-{ \left( x\sqrt { x } \right) }^{ 2 } } } .\frac { d }{ dx } (x\sqrt { x } )\)
\(=\frac { 1 }{ \sqrt { 1-{ x }^{ 3 } } } \frac { d }{ dx } \left( { x }^{ 3/2 } \right) \)
\(=\frac { 1 }{ \sqrt { 1-{ x }^{ 3 } } } \frac { 3 }{ 2 } { x }^{ \frac { 3 }{ 2 } -1 }\)
\(=\frac { 3 }{ 2 } \sqrt { \frac { x }{ 1-{ x }^{ 3 } } } \)

\(Let\ y={ \left( 5x \right) }^{ 3\ cos\ 2x }\)
\(Taking\ logs.,\ log\ y=log\ { \left( 5x \right) }^{ 3\ cos\ 2x }\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ y } .\frac { dy }{ dx } \)
\(=3\left[ cos\quad 2x.\frac { 1 }{ 5x } (5)+log(5x).(-sin\quad 2x)2 \right] \)
\(\Rightarrow \frac { dy }{ dx } \)
\(=3y\left[ \frac { cos\quad 2x }{ x } -2sin\quad 2x\quad log\quad 5x \right]\)
\( Hence\quad \frac { dy }{ dx } ={ \left( 5x \right) }^{ 3\quad cos\quad 2x }\left[ 3\frac { cos\quad 2x }{ x } -6sin\quad 2x\quad log\quad 5x \right] \)

\(Let\quad y={ sin }^{ 3 }x+{ cos }^{ 6 }x\)
\(\frac { dy }{ dx } ={ 3sin }^{ 2 }x\frac { d }{ dx } \left( sin\quad x \right) +6{ cos }^{ 5 }x\frac { d }{ dx } (cos\quad x)\)
\(={ 3sin }^{ 2 }x(cos\quad x) +6{ cos }^{ 5 }x(-sin\quad x)\)
\(=3\quad sin\quad x.cos\quad x(sin\quad x-2{ cos }^{ 4 }x)\)

\(Let\quad y={ \left( { 3x }^{ 2 }-9x+5 \right) }^{ 9 }\)
\(\frac { dy }{ dx } =9{ \left( { 3x }^{ 2 }-9x+5 \right) }^{ 8 }\frac { d }{ dx } \left( { 3x }^{ 2 }-9x+5 \right) \)
\(=9{ \left( { 3x }^{ 2 }-9x+5 \right) }^{ 8 }(6x-9+0)\)
\(=27(2x-3){ \left( { 3x }^{ 2 }-9x+5 \right) }^{ 8 }\)

\(At\quad x=0:\)
\(\lim _{ x\rightarrow { 0 }^{ - } }{ \lambda ({ x }^{ 2 }-2x) } \)
\(=\lambda (0-0)=0\)
\(\lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ (4x+1) } =4(0)+1=1\)
\(Thus\quad \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) \neq } \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } for\quad no\quad value\quad o\quad \lambda \)
\(Hence,\quad 'f'\quad is\quad continuous\quad at\quad x=0\quad for\quad no\quad longer\quad value\quad of\quad \lambda \)
\((ii)At\quad x=1:\)
\(\lim _{ x\rightarrow 1 }{ f\left( x \right) } =\lim _{ x\rightarrow 1 }{ (4x+1) } =4(1)+1=5\)
\(And\quad f(1)=4(1)+1=5.\)
\(Thus\quad \lim _{ x\rightarrow 1 }{ f\left( x \right) } =f(1)\quad for\quad any\quad value\quad of\quad \lambda \)
\( Hence,\quad 'f'\quad is\quad continuous\quad or\quad any\quad value\quad of\quad \lambda \)

\(At\quad x=1\)
\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ - } }{ \left( { x }^{ 10 }-1 \right) }\)
\( \lim _{ h\rightarrow 0 }{ ({ \left( 1-h \right) }^{ 10 }-1) } \)
\(={ \left( 1-0 \right) }^{ 10 }-1=1-1=0\)
\(\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ \left( { x }^{ 2 } \right) } =\lim _{ h\rightarrow 0 }{ { \left( 1+h \right) }^{ 2 } }\)
\( ={ \left( 1+0 \right) }^{ 2 }=1\)
\(Thus\quad \lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } \neq \lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } \)
\('f'\quad is\quad discontinuous\quad at\quad x=1\)
\(At\quad x=c<1:\)
\( \lim _{ x\rightarrow c }{ f\left( x \right) } =\lim _{ x\rightarrow c }{ { x }^{ 10 } } -1={ c }^{ 10 }-1=f(c)\)
\('f'\quad is\quad continuous\quad at\quad x=c<1\)
\(At\quad x=c>1:\)
\(\lim _{ x\rightarrow c }{ f\left( x \right) } =\lim _{ x\rightarrow c }{ { x }^{ 2 } } ={ c }^{ 2 }=f(c)\)
\('f'\quad is\quad continuous\quad at\quad x=c>1\)
\(Hence,\quad the\quad point\quad of\quad discontinuity\quad is\quad x=-1.\)

\((i)\lim _{ x\rightarrow 0 }{ f\left( x \right) = } \lim _{ x\rightarrow 0 }{ (5x-3) } =0-3=-3\)
\(f(0)=-3\)
\(Thus\ \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f(0)\)
f is continuous at x = 0
\((ii)\lim _{ x\rightarrow -3 }{ f\left( x \right) } =\lim _{ x\rightarrow -3 }{ (5x-3) }\)
\( =\lim _{ h\rightarrow 0 }{ \left[ 5(-3+h)-3 \right] } \)
\(=\lim _{ h\rightarrow 0 }{ (-18+5h) } \)
\(=-18+5(0)=-18\)
\(Thus\ \lim _{ x\rightarrow -3 }{ f\left( x \right) =f(-3) } \)
f is continuous at x = -3
\((iii)\lim _{ x\rightarrow 5 }{ f\left( x \right) = } \lim _{ x\rightarrow 5 }{ (5x-3) } \)
\(=\lim _{ h\rightarrow 0 }{ \left[ 5(5+h)-3 \right] } \)
\(=\lim _{ h\rightarrow 0 }{ (22+5h)=22+0=22 } \)
\(f(5)=5(5)-3=25-3=22\)
\(Thus\ \lim _{ x\rightarrow 5 }{ f\left( x \right) } =f(5)\)
f is continuous at x = 5

Let f(x) = cos ^–1 (sin x). Observe that this function is defined for all real numbers. We may rewrite this function as
f(x) = cos ^–1 (sin x)
\(={ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 2 } -x \right) \right] \)
\(=\frac { \pi }{ 2 } -x\)
\(Hence,\ f(x)=0-1=-1\)
\((ii)\ Let\ f\left( x \right) ={ tan }^{ -1 }\left( \frac { sin\quad x }{ 1+cosx } \right) \)
Observe that this function is defined for all real numbers, where cos x ≠ – 1; i.e., at all odd multiplies of π. We may rewrite this function as
\(={ tan }^{ -1 }\left( \frac { 2sin\quad x/2\quad cos\quad x/2 }{ 2{ cos }^{ 2 }x/2 } \right) \)
\(=\tan ^{-1}\left[\tan \left(\frac{x}{2}\right)\right]=\frac{x}{2}\)
Observe that we could cancel \(\cos \left(\frac{x}{2}\right)\) in both numerator and denominator as it is not equal to zero. Thus \(f^{\prime}(x)=\frac{1}{2}\)
\((iii)Let\ f\left( x \right) ={ sin }^{ -1 }\left( \frac { { 2 }^{ x+1 } }{ 1+{ 4 }^{ x } } \right) \)
To find the domain of this function we need to find all x such that \(-1 \leq \frac{2^{x+1}}{1+4^x} \leq 1\). Since the quantity in the middle is always positive, we need to find all x such that \(\frac{2^{x+1}}{1+4^x} \leq 1, \text { i.e.}\) all x such that \(2^{x+1} \leq 1+4^x\). We may rewrite this as \(2 \leq \frac{1}{2^x}+2^x\) which is true for all x. Hence the function is defined at every real number. By putting 2^x = tan θ, this function may be rewritten as
\( f(x)= \sin ^{-1}\left[\frac{2^{x+1}}{1+4^x}\right] \)
\(= \sin ^{-1}\left[\frac{2^x \cdot 2}{1+\left(2^x\right)^2}\right] \)
\( =\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^2 \theta}\right] \)
\(= \sin ^{-1}[\sin 2 \theta] \\ = 2 \theta=2 \tan { }^{-1}\left(2^x\right) \)
\(f^{\prime}(x)= 2 \cdot \frac{1}{1+\left(2^x\right)^2} \cdot \frac{d}{d x}\left(2^x\right) \)
\(= \frac{2}{1+4^x} \cdot\left(2^x\right) \log 2 \)
\(= \frac{2^{x+1} \log 2}{1+4^x}\)

\((i)\quad { e }^{ -x }\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } ={ e }^{ -x }.\frac { d }{ dx } (-x)={ e }^{ -x }\quad (-1)=-{ e }^{ -x }\)
\((ii)\quad Let\quad y=sin(log\quad x)\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } =cos(log\quad x).\frac { d }{ dx } (log\quad x)\)
\(=cos(log\quad x).\frac { 1 }{ x } =\frac { cos(log\quad x) }{ x } ,x>0\)
\((iii)\quad Let\quad y={ cos }^{ -1 }({ e }^{ x })\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } =\frac { -1 }{ \sqrt { 1-{ \left( { e }^{ x } \right) }^{ 2 } } } .\frac { d }{ dx } ({ e }^{ x })\)
\(=-\frac { 1 }{ \sqrt { 1-{ e }^{ 2x } } } .{ e }^{ x }=\frac { -{ e }^{ x } }{ \sqrt { 1-{ e }^{ 2x } } } \)
\((iv)\quad Let\quad y={ e }^{ cos\quad x }\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } ={ e }^{ cos\quad x }\frac { d }{ dx } (cos\quad x)\)
\(={ e }^{ cos\quad x }(-sin\quad x)=-(sin\quad x){ e }^{ cos\quad x }\)

Clearly the function is defined at every real number. By inspection, it seems prudent to partition the domain of definition of f into three disjoint subsets of the real line.
Let \( \mathrm{D}_1=\{x \in \mathbf{R}: x<0\}, \mathrm{D}_2=\{0\} \text { and } \mathrm{D}_3=\{x \in \mathbf{R}: x>0\} \)
Case 1 At any point in D1 , we have f(x) = x 2 and it is easy to see that it is continuous there
Case 2 At any point in D3 , we have f(x) = x and it is easy to see that it is continuous there
Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f(0) = 0. The left hand limit of f at 0 is
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x^2=0^2=0\)
The right hand limit of f at 0 is
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\)
Thus \(\lim _{x \rightarrow 0} f(x)=0=f(0)\)and hence f is continuous at 0. This means that f is continuous at every point in its domain and hence, f is a continuous function

The function f is defined at all points of the real line
Case 1: Case 1 If c < 1, then f(c) = c + 2. Therefore,
\(\lim _{ x\rightarrow c }{ f\left( x \right) = } \lim _{ x\rightarrow c }{ (x+2) } =c+2\)
Thus, f is continuous at all real numbers lessthan 1.
Case 2: If c > 1, then f(c) = c – 2. Therefore,
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-2)=c-2=f(c)\)
Thus, f is continuous at all points x > 1.
Case 3 If c = 1, then the left hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+2)=1+2=3\)
The right hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-2)=1-2=-1\)
Since the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1.
Hence x = 1 is the only point of discontinuity of f.
The graph of the function is given

\(We\ have:y={ cos }^{ -1 }x\)
\(cos\quad y=x\)
\(Diff.\quad w.r.t.x,\quad -sin\quad y\frac { dy }{ dx } =1\)
\(\frac { dy }{ dx } =-cosec\quad y\)
\( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-(-cosec\quad y\quad cot\quad y).\frac { dy }{ dx } \)
\(=cosec\quad y\quad cot\quad y(-cosec\quad y)\)
\(=-cot\quad y\quad { cosec }^{ 2 }y\)

\(We\quad have\quad y=5\quad cosx-3sinx\)
\( \frac { dy }{ dx } =-5\quad sin\quad x-3\quad cos\quad x\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-5\quad cos\quad x+3\quad sinx\)
\(=-(5\quad cos\quad x-3\quad sin\quad x)\)
\(=-y\)
\(Hence,\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

\(Let\ y=sin\ (log\ x)\)
\(\frac { dy }{ dx } =cos(log\ x).\frac { 1 }{ x } \)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { x.\frac { d }{ dx } (cos(log\ x))-cos(log\ x).1 }{ { x }^{ 2 } } \)
\(=\frac { x.(-sin(log\ x))\frac { 1 }{ x } -cos(log\ x) }{ { x }^{ 2 } } \)
\(=-\frac { sin(log\ x)+cos(log\ x) }{ { x }^{ 2 } } \)

\(Let\ y=log(log\ x)\)
\(\frac { dy }{ dx } =\frac { 1 }{ log\quad x } .\frac { 1 }{ x } =\frac { 1 }{ x\quad log\quad x } \)
\( and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { \frac { d }{ dx } (x\quad log\quad x) }{ { \left( x\quad log\quad x \right) }^{ 2 } } \)
\(=-\frac { x.\frac { 1 }{ x } +log\quad x.1 }{ { \left( x\quad log\quad x \right) }^{ 2 } } \)
\( =-\frac { 1+log\quad x }{ { \left( x\quad log\quad x \right) }^{ 2 } }\)

\(Let\quad y={ tan }^{ -1 }x\)
\(\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } ={ \left( 1+{ x }^{ 2 } \right) }^{ -1 }\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =(-1){ \left( 1+{ x }^{ 2 } \right) }^{ -2 }(0+2x)\)
\(=-\frac { 2x }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } \)

\(Let\ y={ e }^{ 6x }cos\ 3x\)
\(\frac { dy }{ dx } ={ e }^{ 6x }.(-sin3x.3)+cos\ 3x.{ e }^{ 6x }.6\)
\(={ e }^{ 6x }(6\ cos\ 3x-3sin\ 3x)\)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ 6x }\left[ 6(-sin\ 3x\ (3)-3\ cos3x.(3) \right] +(6\ cos\ 3x-3\ sin\ 3x){ e }^{ 6x }.6\)
\(={ e }^{ 6x }(-18\ sin\ 3x-9\ cos\ 3x+36\ cos\ 3x-18\ sin\ 3x)\)
\(={ e }^{ 6x }(-36\q uad sin\ 3x+27\ cos\ 3x)\)
\(=9{ e }^{ 6x }(3\ cos\ 3x-4\ sin\ x)\)

\(Let\ y={ e }^{ x }sin5x\)
\(\frac { dy }{ dx } ={ e }^{ x }.(cos\ 5x.5)+sin\ 5x.{ e }^{ x }\)
\(={ e }^{ x }(sin\ 5x+5\ cos\ 5x)\)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ x }(cos5x.5-5sin5x.5)\)
\(+(sin\ 5x+5\ cos5x){ e }^{ x }\)
\(={ e }^{ x }(5\ cos\ 5x-25\ sin\ 5x +sin\ 5x+5cos\ 5x)\)
\(=2{ e }^{ x }(5\ cos\ 5x-12\ sin\ 5x)\)

\(Let\ y={ x }^{ 3 }log\ x\)
\(\frac { dy }{ dx } ={ x }^{ 3 }.\frac { 1 }{ x } +log\ x.({ 3x }^{ 2 })\)
\(={ x }^{ 2 }+{ 3x }^{ 2 }logx\)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2x+3\left( { x }^{ 2 }.\frac { 1 }{ x } +logx.2x \right) \)
\(=2x+3x+6xlogx\)
\(=x(5+6logx)\)

\(Let\ y=log\ x\)
\(\frac { dy }{ dx } =\frac { 1 }{ x } ={ x }^{ -1 }\)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =(-1){ x }^{ -2 }=-\frac { 1 }{ { x }^{ 2 } } \)

\(Let\ y=x\ cos\ x\)
\(\frac { dy }{ dx } =x(-sin\ x)+cos\ x(1)\)
\(-x\ sin\ x+cos\ x\)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-(x\ cos\ x+sin\ x.1)-sin\)
\( =-x\ cos\ x-2\ sin\ x.\)

\(Let\quad y={ x }^{ 20 }\)
\( \frac { dy }{ dx } =20{ x }^{ 19 }\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =(20)(19){ x }^{ 18 }=380{ x }^{ 18 }\)

\(Let\quad y={ x }^{ 2 }+3x+2\)
\(\frac { dy }{ dx } =2x+3\quad and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\)

\(We\quad have:\)
\(X=\sqrt { { a }^{ { sin }^{ -1 }t } } and\quad y=\sqrt { { a }^{ { cos }^{ -1 }t } } \)
\(i.e.,x={ \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }and\quad y={ \left( { \alpha }^{ { cos }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }\)
\(\frac { dx }{ dt } =\frac { 1 }{ 2 } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ -\frac { 1 }{ 2 } }\frac { d }{ dt } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }\)
\(=\frac { 1 }{ 2 } \frac { 1 }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } { \alpha }^{ { sin }^{ -1 }t }{ log }_{ e }\alpha \frac { d }{ dx } \left( { sin }^{ -1 }t \right) \)
\(=\frac { { log }_{ e }\alpha }{ 2 } .\sqrt { { \alpha }^{ { sin }^{ -1 }t } } .\frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(Similarly\quad \frac { dy }{ dt } =-\frac { { log }_{ e }\alpha }{ 2 } \sqrt { { \alpha }^{ { cos }^{ -1 }t } } \frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { \sqrt { { \alpha }^{ { cos }^{ -1 }t } } }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } =-\frac { y }{ x } .,\quad which\quad is\quad true.\)

\(We\quad have\quad :\quad x=a\quad sec\theta ,\quad y=b\quad tan\theta\)
\( \frac { dx }{ d\theta } =a\quad sec\theta tan\theta \quad and\quad \frac { dy }{ d\theta } =b{ sec }^{ 2 }\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { b{ sec }^{ 2 }\theta }{ \alpha sec\theta tan\theta } \)
\(=\frac { b\quad sec\theta }{ \alpha tan\theta } =\frac { b }{ a } cosec\theta \)

\(We\quad have\quad :\quad x=a(\theta -sin\theta ),\quad y=a(1+cos\theta )\)
\(\frac { dx }{ d\theta } =a(1-cos\theta ),\quad \frac { dy }{ d\theta } =a(0-sin\theta )=-a\quad sin\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } \)
\(=\frac { -a\quad sin\theta }{ a(1-cos\theta ) } =\frac { -sin\theta }{ 1-cos\theta } \)
\(=\frac { -2sin\theta /2cos\theta /2 }{ 2{ sin }^{ 2 }\theta /2 } =-\frac { cos\theta /2 }{ sin\theta /2 } =-cot\frac { \theta }{ 2 } \)

\(We\ have\ :\ x=sin\ t,\ y=cos\ 2t\)
\(\frac { dx }{ dt } =cos\quad t,\quad \frac { dy }{ dt } =-sin\quad 2t.(2)\)
\(=-2\quad sin2t\)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { -2\quad sin2t }{ cos\quad t } =-\frac { -4sintcost }{ cos{ t } } \)
\(=-4\quad sint\)

\(We\quad have\quad :\quad x=4t,\quad y=\frac { 4 }{ t } \)
\(\frac { dx }{ dt } =4,\quad \frac { dy }{ dt } =-\frac { 4 }{ { t }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { -4/{ t }^{ 2 } }{ 4 } =-\frac { 1 }{ { t }^{ 2 } } \)

\(We\ have\ :\ x=a\ cos\theta ,\ y=b\ cos\theta \)
\(\frac { dx }{ d\theta } =-a\quad sin\theta ,\quad \frac { dy }{ d\theta } =-b\quad sin\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { -b\quad sin\theta }{ -a\quad sin\theta } =\frac { b }{ a } \)

\(We\quad have\quad :\quad x={ 2at }^{ 2 },y={ at }^{ 4 }\)
\(\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { { 4at }^{ 3 } }{ 4at } ={ t }^{ 2 }\)

\(We\quad have\quad :\quad xy={ e }^{ (x-y) }\)
\(Taking\quad logs.,log(xy)=log{ e }^{ (x-y) }\)
\(logx+logy=(x-y)log\quad e\)
\(logx+logy=(x-y)\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } \)
\(\Rightarrow \left( \frac { 1 }{ y } +1 \right) \frac { dy }{ dx } =1-\frac { 1 }{ x } \)
\(\Rightarrow \frac { 1+y }{ y } \frac { dy }{ dx } =\frac { x-1 }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)

\(We\quad have\quad { x }^{ y }={ y }^{ x }\)
\(Taking\quad logs.,log{ x }^{ y }=log{ y }^{ x }\Rightarrow y\quad logx=x\quad logy\)
\(Diff.\quad w.r.t.x,y\frac { 1 }{ x } +logx.\frac { dy }{ dx } \)
\(=x.\frac { 1 }{ y } .\frac { dy }{ dx } +logy.1\)
\(\Rightarrow \left( logx-\frac { x }{ y } \right) \frac { dy }{ dx } =logy-\frac { y }{ x } \)
\(\Rightarrow \frac { 1 }{ y } (ylogx-x)\frac { dy }{ dx } =\frac { 1 }{ x } (xlogy-y)\)
\(Hence,\quad \frac { dy }{ dx } =\frac { y(x\quad logy-y) }{ x(y\quad logx-x) } \)

\(Let\quad y={ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(Taking\quad logs.,\quad log\quad y=log{ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(\\ =2log(x+3)+3log(x+4)+4log(x+5)\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ y } \frac { dy }{ dx } \)
\(=\frac { 2 }{ x+3 } (1+0)+\frac { 3 }{ x+4 } (1+0)+\frac { 4 }{ x+5 } (1+0)\)
\(\Rightarrow \frac { dy }{ dx } =y\left[ \frac { 2 }{ x+3 } +\frac { 3 }{ x+4 } +\frac { 4 }{ x+5 } \right] \)
\(Hence,\quad \frac { dy }{ dx } ={ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(=\left[ \frac { 2 }{ x+3 } +\frac { 3 }{ x+4 } +\frac { 4 }{ x+5 } \right] \)

\(Let\quad y={ \left( log\quad x \right) }^{ cos\quad x }\)
\(Taking\quad logs,\)
\(log\quad y=log\quad { \left( log\quad x \right) }^{ cos\quad x }\)
\(=cos\quad x\quad log(log\quad x)\)
\(Diff.\quad w.r.t.\quad x,\quad \frac { 1 }{ y } .\frac { dy }{ dx } \)
\(=cos\quad x.\frac { 1 }{ log\quad x } \frac { 1 }{ x } +log(log\quad x)(-sin\quad x)\)
\(\frac { dy }{ dx } =y\left[ \frac { cos\quad x }{ x\quad log\quad x } -sinxlog(log\quad x) \right] \)
\(={ \left( log\quad x \right) }^{ cos\quad x }\left[ \frac { cos\quad x }{ x\quad log\quad x } -sin\quad xlog(log\quad x) \right] \)

\(Let\ y=cosx.\ cos\ 2x.\ cos\ 3x\)
\(Taking\ logs,\)
\(log\ y=log(os\ x.os\ 2x.\ cos\ 3x)\)
\(=log\ cosx+log\ cos2x+\ log\ cos3x\)
\(Diff.\quad w.r.t.\quad \frac { 1 }{ y } \frac { dy }{ dx } \)
\(=\frac { 1 }{ cos\quad x } (-sin\quad x)+\frac { 1 }{ cos\quad 2x } (-2sin\quad 2x)+\frac { 1 }{ cos\quad 3x } (-sin\quad 3x)\)
\(\frac { dy }{ dx } =y\left[ -tan\quad x-2\quad tan\quad 2x-3\quad tan\quad 3x \right] \)
\(=cos\quad x.os\quad 2x.\quad cos\quad 3x\)
\(\left[ -tan\ x-2\ tan\ 2x-3\ tan3x \right] \)
\(=-cos\ x\ cos\ 2x\ cos\ 3x\)

\(Let\quad y=cos(log\quad x+{ e }^{ x })\)
\(\frac { dy }{ dx } =-sin(log\quad x+{ e }^{ x }).\frac { d }{ dx } (log\quad x+{ e }^{ x })\)
\(=-sin(log\quad x+{ e }^{ x })\left( \frac { 1 }{ x } +{ e }^{ x } \right) \)
\(=-\frac { 1 }{ x } sin(log\quad x+{ e }^{ x })(1+{ xe }^{ x }),x>0\)

\(Let\quad y=\frac { cos\quad x }{ log\quad x } \)
\(\frac { dy }{ dx } =\frac { logx.(-sinx)-cosx.\frac { 1 }{ x } }{ { \left( logx \right) }^{ 2 } } \)
\(=-\frac { sinxlogx+\frac { 1 }{ x } cosx }{ { \left( logx \right) }^{ 2 } } ,\quad x>0\)

\(Let\quad y=log(logx)\)
\(\frac { dy }{ dx } =\frac { 1 }{ log\quad x } .\frac { d }{ dx } (log\quad x)\)
\(=\frac { 1 }{ log\quad x } .\frac { 1 }{ x } =\frac { 1 }{ xlogx } ,x>1\)

\(Let\quad y=\sqrt { { e }^{ \sqrt { x } } } ={ \left( { e }^{ \sqrt { x } } \right) }^{ 1/2 }\)
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } { \left( { e }^{ \sqrt { x } } \right) }^{ -1/2 }\frac { d }{ dx } \left( { e }^{ \sqrt { x } } \right) \)
\(=\frac { 1 }{ 2\sqrt { { e }^{ \sqrt { x } } } } { e }^{ \sqrt { x } }\frac { d }{ dx } \left( { e }^{ { x }^{ 1/2 } } \right) \)
\(=\frac { 1 }{ 2\sqrt { { e }^{ \sqrt { x } } } } { e }^{ \sqrt { x } }\frac { 1 }{ 2 } { x }^{ -1/2 }=\frac { { e }^{ \sqrt { x } } }{ 4\sqrt { x } \sqrt { { e }^{ \sqrt { x } } } } ,x>0\)

\(Let\quad y={ e }^{ x }+{ e }^{ { x }^{ 2 } }+......{ e }^{ { x }^{ 5 } }\)
\(=\frac { dy }{ dx } ={ e }^{ x }+{ e }^{ { x }^{ 2 } }\frac { d }{ dx } \left( { x }^{ 2 } \right) +....+{ e }^{ { x }^{ 5 } }.\frac { d }{ dx } \left( { x }^{ 5 } \right)\)
\( ={ e }^{ x }+{ e }^{ { x }^{ 2 } }.2x+....+{ e }^{ { x }^{ 5 } }5{ x }^{ 4 }\)
\(={ e }^{ x }+{ 2xe }^{ { x }^{ 2 } }.2x+....+5{ x }^{ 4 }{ e }^{ { x }^{ 5 } }\)

\(Let\quad y=log(cos\quad { e }^{ x })\)
\(\frac { dy }{ dx } =\frac { 1 }{ cos{ e }^{ x } } .\frac { d }{ dx } (cos{ e }^{ x })\)
\(\frac { 1 }{ cos{ e }^{ x } } (-sin\quad { e }^{ x }).\frac { d }{ dx } ({ e }^{ x })\)
\(=-tan\quad { e }^{ x }{ e }^{ x }=-{ e }^{ x }tan\quad { e }^{ x }\)

\(Let\quad y=sin({ tan }^{ -1 }{ e }^{ -x })\)
\(\frac { dy }{ dx } =cos({ tan }^{ -1 }{ e }^{ -x })\frac { 1 }{ 1+{ \left( { e }^{ -x } \right) }^{ 2 } } \frac { d }{ dx } ({ e }^{ -x })\)
\(=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) .\frac { 1 }{ 1+{ e }^{ -2x } } { e }^{ -x }\frac { d }{ dx } (-x)\)
\(=cos({ tan }^{ -1 }{ e }^{ -x })\frac { 1 }{ 1+{ e }^{ -2x } } { e }^{ -x }(-1)\)
\(=\frac { -{ e }^{ -x } }{ 1+{ e }^{ -2x } } cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \)

\(Let\quad y={ e }^{ x^{ 3 } }\)
\(\frac { dy }{ dx } ={ e }^{ x^{ 3 } }\frac { d }{ dx } ({ x }^{ 3 })\)
\(={ e }^{ x^{ 3 } }.{ 3 }x^{ 2 }={ 3 }x^{ 2 }{ e }^{ x^{ 3 } }\)

\(Let\quad y={ e }^{ { sin }^{ -1 }x }\)
\(\frac { dy }{ dx } ={ e }^{ { sin }^{ -1 }x }\frac { d }{ dx } ({ sin }^{ -1 }x)\)
\(={ e }^{ { sin }^{ -1 }x }\left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\(=\frac { { e }^{ { sin }^{ -1 }x } }{ \sqrt { 1-{ x }^{ 2 } } } ,x\epsilon (-1,1)\)

\(Let\quad y=\frac { { e }^{ x } }{ sin\quad x } \)
\( \frac { dy }{ dx } =\frac { sinx\frac { d }{ dx } \left( { e }^{ x } \right) -{ e }^{ x }\frac { d }{ dx } (sin\quad x) }{ { sin }^{ 2 }x } \)
\(=\frac { sinx\quad { e }^{ x }-{ e }^{ x }.cos\quad x }{ { sin }^{ 2 }x } \)
\(=\frac { { e }^{ x }(sin\quad x-cos\quad x) }{ { sin }^{ 2 }x } \)

\(Here\quad y={ sec }^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) \)
\(={ sec }^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) \)
\(={ sec }^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) \left( \frac { 1 }{ cos2\theta } \right)\)
\( { sec }^{ -1 }(sec\quad 2\theta )=2\theta =2{ cos }^{ -1 }x\)
\(Hence\quad \frac { dy }{ dx } =-\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } } \)

\(Here\quad y={ sin }^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) \)
\(={ sin }^{ -1 }\left( 2sin\theta \sqrt { { sin }^{ 2 }\theta } \right) \)
\(={ sin }^{ -1 }(2sin\theta \sqrt { { sin }^{ 2 }\theta } )\)
\(={ sin }^{ -1 }\left( 2sin\theta cos\theta \right) \)
\(={ sin }^{ -1 }(sin2\theta )\)
\(=2\theta =2{ sin }^{ -1 }x\)
\(Hence\quad \frac { dy }{ dx } =0-\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } } \)

\(Here\quad y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \)
\(={ cos }^{ -1 }\left( \frac { 2{ tan }\theta }{ 1+{ tan }^{ 2 }\theta } \right)\)
\(={ cos }^{ -1 }(sin2\theta )\)
\(={ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 2 } -\theta \right) \right] \)
\(=\frac { \pi }{ 2 } -2\theta =\frac { \pi }{ 2 } -2{ tan }^{ -1 }x\)
\(Hence\quad \frac { dy }{ dx } =0-\frac { 2 }{ 1+{ x }^{ 2 } } =-\frac { 2 }{ 1+{ x }^{ 2 } } \)

\(Here\quad y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \)
\(={ sin }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
\(={ sin }^{ -1 }(cos2\theta )\)
\(={ sin }^{ -1 }\left[ sin\left( \frac { \pi }{ 2 } -\theta \right) \right] \)
\(=\frac { \pi }{ 2 } -2\theta =\frac { \pi }{ 2 } -2{ tan }^{ -1 }x\)
\(Hence\quad \frac { dy }{ dx } =0-2.\frac { 1 }{ 1+{ x }^{ 2 } } =-\frac { 2 }{ 1+{ x }^{ 2 } } \)

\(Here\quad y={ cos }^{ -1 }\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(={ cos }^{ -1 }\frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \)
\(={ cos }^{ -1 }(cos2\theta )=2\theta =2{ tan }^{ -1 }x\)
\(Hence,\quad \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)

\(Here\quad y={ tan }^{ -1 }\frac { 3x-{ x }^{ 3 } }{ 1-3{ x }^{ 2 } } \)
\(={ tan }^{ -1 }\frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \)
\(={ tan }^{ -1 }(tan3\theta )=3\theta =3{ tan }^{ -1 }x\)
\(Hence,\quad \frac { dy }{ dx } =\frac { 3 }{ 1+{ x }^{ 2 } } \)

\(Here\quad y={ sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } \)
\(={ sin }^{ -1 }\left( \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
\(={ sin }^{ -1 }(sin2\theta )=2\theta =2{ tan }^{ -1 }x\)
\(Hence,\quad \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)

\(we\ have\ :\ { sin }^{ 2 }x+cos^{ 2 }y=1\)
\(Diff\quad w.r.t.x,\)
\(2sin\ x\ \frac { d }{ dx } (sin\ x)+2cos\ y\frac { d }{ dx } (cos\ y)=0\)
\(\Rightarrow 2\ sin\ xcos\ x+2\ cos\ y(-sin\ y)\frac { dy }{ dx } =0\)
\(Hence,\ \frac { dy }{ dx } =\frac { 2\ sinxcosx }{ 2\ cosysiny } =\frac { sin2x }{ sin2y } \)

\(we\ have\ :\ { sin }^{ 2 }y+cosxy=\pi \)
\(Diff\ w.r.t.x,\)
\(\Rightarrow 2sin\ y\ cos\ y\frac { dy }{ dx } -sin\ xy\left( x.\frac { dy }{ dx } +y.1 \right) =0\)
\(\Rightarrow (2\ siny\ cos\ y-x\ sinxy)\frac { dy }{ dx } =y\ sin\ xy\)
\(Hence,\ \frac { dy }{ dx } =\frac { y\ sin\ xy }{ sin\ 2y-x\ sin\ xy } \)

\(we\ have\ :\ { x }^{ 3 }+{ x }^{ 2 }y+{ xy }^{ 2 }+{ y }^{ 3 }=81\)
\(Diff\ w.r.t.x,\ { 3x }^{ 2 }+\left( { x }^{ 2 }\frac { dy }{ dx } +2xy \right) \)
\(+\left( x.2y\frac { dy }{ dx } +{ y }^{ 2 }.1 \right) +3{ y }^{ 2 }\frac { dy }{ dx } =0\)
\(\Rightarrow ({ x }^{ 2 }+{ 2xy }+{ 3y }^{ 2 })\frac { dy }{ dx } =-({ 3x }^{ 2 }+{ 2xy }+{ y }^{ 2 })\)
\(Hence,\ \frac { dy }{ dx } =-\frac { { 3x }^{ 2 }+{ 2xy }+{ y }^{ 2 } }{ { x }^{ 2 }+{ 2xy }+{ 3y }^{ 2 } } \)

\(we\ have\quad :\ { x }^{ 2 }+xy+{ y }^{ 2 }=100\)
\(Diff\ w.r.t.x,\ 2x+\left( x.\frac { dy }{ dx } +y.1 \right) +2y\frac { dy }{ dx } =0\)
\(\Rightarrow (x+2y)\frac { dy }{ dx } =\quad -2x-y\)
\(Hence,\ \frac { dy }{ dx } =-\frac { 2x+y }{ x+2y } \)

\(we\ have\ :\ xy+{ y }^{ 2 }=tan\ x+y\)
\(Diff\ w.r.t.x,\ \left( x\frac { dy }{ dx } +y.1 \right) +2y\frac { dy }{ dx } \)
\(={ sec }^{ 2 }x+\frac { dy }{ dx } \)
\(\Rightarrow (x+2y-1)\frac { dy }{ dx } ={ sec }^{ 2 }x-y\)
\(Hence,\ \frac { dy }{ dx } =\frac { { sec }^{ 2 }x-y }{ x+2y-1 } \)

\(We\quad have:\quad ax+{ by }^{ 2 }=cos\quad y\)
\( Diff\quad w.r.t.\quad x,\quad a+2b\quad y\frac { dy }{ dx } =-sin\quad y\quad \frac { dy }{ dx } \)
\((2by+siny)\frac { dy }{ dx } \)
\(Hence, \frac { dy }{ dx } =-\frac { a }{ 2by+sin\quad y }\)

\(We\quad have:\quad 2x+3y=\quad sin\quad y\)
\(Dif.\quad w.r.t.\quad x,\quad 2+3\frac { dy }{ dx } =cosy\frac { dy }{ dx } \)
\(\Rightarrow (cos\quad y-3)\frac { dy }{ dx } =2\)
\(Hence,\ \frac { dy }{ dx } =\frac { 2 }{ cos\quad y-3 } \)

\(Let\quad y=\quad cos\sqrt { x } \)
\(Put\quad \sqrt { x } =t,\quad y=cos\quad t\)
\(\frac { dy }{ dt } =-sin\quad t,\quad \frac { dt }{ dv } =\frac { 1 }{ 2\sqrt { x } } \)
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(=(-sin\quad t).\frac { 1 }{ 2\sqrt { x } } =-\frac { sin\sqrt { x } }{ 2\sqrt { x } } \)

\(Let\quad y=2\sqrt { cot\quad { x }^{ 2 } } \)
Put x² = t, cot t = s.,
\(y=2\sqrt { s } \)
\(\frac { dy }{ ds } =2.\frac { 1 }{ 2\sqrt { s } } =\frac { 1 }{ \sqrt { s } } \)
\(\frac { ds }{ dt } =-{ { { cosec }^{ 2 } }t },\quad \frac { dt }{ dx } =2x\)
\(\frac { dy }{ dx } =\frac { dy }{ ds } .\frac { ds }{ dt } .\frac { dt }{ dx } \)
\(=\frac { 1 }{ \sqrt { s } } (-{ { cosec }^{ 2 } }t).(2x)\)
\(=\frac { -2x }{ \sqrt { cot\quad x } } { { cosec }^{ 2 } }t\)
\(=\frac { -2x }{ \sqrt { { cot\quad }x^{ 2 } } } { cosec }^{ 2 }\quad { x }^{ 2 }\)

\(Let\quad y=sec(tan\quad \sqrt { x } ).\)
\(Put\quad \sqrt { x } =t,\quad s=tan\quad t.\)
\(y=sec\quad s,\quad s=tan\quad t,\quad t=\sqrt { x } \)
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { ds }{ dt } .\frac { dt }{ dx } \)
\(=(sec\quad s\quad tan\quad s)\quad \left( { sec }^{ 2 }\quad t \right) \left( \frac { 1 }{ 2\sqrt { x } } \right) \)
\(=(sec(tan\quad t)tan(tan\quad t)){ sec }^{ 2 }.t\)
\(=\frac { 1 }{ \sqrt { 2x } } sec(tan\sqrt { x } )tan(tan(\sqrt { x } )){ sec }^{ 2 }\sqrt { x } \)

\(Put\quad ax+b=t\)
\(\frac { dy }{ dx } =\frac { dy }{ dt } \frac { dt }{ dx } =cost(a+0)\)
\(=a\quad cos(ax+b)\)

\(Let\quad y=cos(sin\quad x)\)
\(Put\quad sin\quad x=t.\)
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(=(-sin\quad t).(cos\quad x)\)
\(-[sin(sin\quad x)]\quad os\quad x\)

\(Let\quad y=sin({ x }^{ 2 }+5)\)
\(Put\quad { x }^{ 2 }+5=t\)
\(y=sin\quad t,\quad where\quad t={ x }^{ 2 }+5\)
\( \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(=cos\quad t.\quad (2x+0)=2xcos({ x }^{ 2 }+5)\)

\( \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}\right.\left.-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right] \)

\(We\quad have:\quad f(x)=\left| cos\quad x \right| \)
\( { D }_{ f }=R\)
\( Let\quad c\in { D }_{ f }\quad arbitrary\)
\(\lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ \left| cos\quad x \right| } =\left| cos\quad c \right| =f(c)\)
\('f'\quad is\quad continuous\quad at\quad x=c.\)
\( 'f'\quad is\quad continuous\quad on\quad R\)

\(We\quad have:\quad f(x)=cos{ \left( x \right) }^{ 2 }\)
\( { D }_{ f }=R\)
\(Let\quad c\in { D }_{ f },\quad arbitrary.\)
\( \lim _{ x\rightarrow c }{ f(x)=\lim _{ x\rightarrow c }{ cos({ x }^{ 2 }) } } =cos\quad { c }^{ 2 }=f(c)\)
\(\Rightarrow 'f'\quad is\quad continuous\quad at\quad x=c.\)
\(\Rightarrow 'f'\quad is\quad continuous\quad on\quad R.\)

\(=\lim _{ x\rightarrow { 5 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 5 }^{ - } }{ (kx+1) } \)
\(=\lim _{ h\rightarrow 0 }{ (k(5-h)+1) } \)
\(=k(5-0)+1=5k+1\)
\(=\lim _{ x\rightarrow { 5 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 5 }^{ + } }{ (3x-5) } \)
\(=\lim _{ h\rightarrow 0 }{ (3(5+h)-5) } \)
\(3(5+0)-5=15-5=10\)
\(For\quad continuity\quad at\quad x=5,\)
\(5k+1=10=10\Rightarrow 5k+1=10\)
\(k=\frac { 9 }{ 5 } \)

\(\lim _{ x\rightarrow { \pi }^{ - } }{ f(x) } =\lim _{ x\rightarrow { \pi }^{ - } }{ (kx+1) } \)
\(=\lim _{ h\rightarrow 0 }{ (k(\pi -h)+1) } \)
\(=k(\pi -0)+1=k\pi +1.\)
\(=\lim _{ x\rightarrow { \pi }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 2 }^{ + } }{ 3=3 } \)
\(=\lim _{ h\rightarrow 0 }{ cos(\pi +h) } \)
\(=\lim _{ h\rightarrow 0 }{ (-cos } h)=-cos\quad 0=-1\)
\(Also\quad f(\pi )=k\pi +1\)
\(For\quad continuity\quad at\quad x=\pi ,\quad \lim _{ x\rightarrow { \pi }^{ - } }{ f(x) } \)
\(k\pi +1=-1\Rightarrow k=\frac { -2 }{ \pi } \)

\(\lim _{ x\rightarrow { 2 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 2 }^{ - } }{ { \quad k }x^{ 2 } } \)
\(\lim _{ h\rightarrow 0 }{ k{ \left( 2-h \right) }^{ 2 } } \)
\(=k{ \left( 2-0 \right) }^{ 2 }=4k\)
\(\lim _{ x\rightarrow { 2 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 2 }^{ + } }{ \quad 3 } =3\)
\(f(2)=k{ \left( 2 \right) }^{ 2 }=4k\)
\(For\quad continuity\quad at\quad x=2,\)
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 2 }^{ + } }{ \quad f(x) } =f(2)\)
\(4k=3=4k\Rightarrow k=\frac { 3 }{ 4 } \)

Let \(y=x^x -2^{\sin x}\)
Let \(u=x^x, v=2^{\sin x}\)
y=u-v
Differentiating both sides w.r.t. x.
\(\frac{d y}{d x}=\frac{d(u-v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\)

\(At\quad x=0:\)
\(\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { sinx }{ x } } =\lim _{ h\rightarrow 0 }{ \frac { sin(-h) }{ -h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -sin\quad h }{ -h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { sin\quad h }{ h } } =1\)
\( \lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ (x+1) } =0+1=1\)
\( f(0)=0+1=1\)

\(f(x)=\left\{\begin{array}{rll} \frac{k \cos x}{\pi-2 x}, & \text { if } & x \neq \frac{\pi}{2} \\ 3, & \text { if } & x=\frac{\pi}{2} \end{array}\right.\)
f(x) is continuous at \(x=\pi / 2\)
So
\( \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=f\left(\frac{\pi}{2}\right) \)
\(​​​​​​​\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi^*}{2}} \frac{k \cos x}{\pi-2 x}=3 \)
\( \lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}=3 \)
 \(\lim _{h \rightarrow 0} \frac{k \sin h}{2 h}=\lim _{h \rightarrow 0} \frac{-k \sin h}{-2 h}=3\)
\( \frac{k}{2}(1)=3 \Rightarrow k=6\)

\(\lim _{ x\rightarrow 3 }{ f(x) } =\lim _{ x\rightarrow 3 }{ ({ 2x }^{ 2 }-1) } \)
\(=\lim _{ h\rightarrow 0 }{ \left[ 2{ \left( 3+h \right) }^{ 2 }-1 \right] } \)
\(\lim _{ h\rightarrow 0 }{ (2(9+6h+{ h }^{ 2 })-1) } \)
\(\lim _{ h\rightarrow 0 }{ (17+12h+ } 2{ h }^{ 2 })\)
\(=17+0+0=17\)
\(Thus\quad \lim _{ x\rightarrow 3 }{ f(x) } =f(3)\)
'f' is continuous at x = 3.

\(Let\ y={ sin }^{ 2 }x\quad and\quad u={ \quad e }^{ cosx }\)
\( \frac { dy }{ dx } ={ \quad e }^{ cosx }.(-sinx)=-sinx{ \quad e }^{ cosx }\)
\(\frac { dy }{ du } =\frac { { dy }/{ dx } }{ { du }/{ dx } } =\frac { 2sinxcosx }{ -sinx{ \quad e }^{ cosx } } =-\frac { 2cosx }{ { \quad e }^{ cosx } } \)

\(We\ have:\ f(x)={ x }^{ 2 } ....(1)\)
Since f(x) is a polynomial in x,\(\)
it is continuous in [2, 4]
\((II)\quad f'(x)=2x ...(2),\quad which\quad exists\quad in\quad (2,4)\)
Thus, both the conditions of Mean Value Theorem are satisfied.
There will exist atleast one point c in (2, 4) such that:
\(f(c)=\frac { f(b)-f(a) }{ b-a } .....(3)\)
\(But\quad f'(c)=2c,\quad f(a)=f(2)={ 2 }^{ 2 }=4\)
\(and\quad f(b)=f(4)={ 4 }^{ 2 }=16\)
\(From(3),\quad 2c=\frac { 16-4 }{ 4-2 } \Rightarrow 2c=6\Rightarrow c=3\epsilon (2,4)\)
Hence, Mean Value Theorem is applicable and c = 3.

\(We\quad have:\quad y={ x }^{ 3 }+tan\quad x\)
\(\frac { dy }{ dx } ={ 3x }^{ 2 }+{ sec }^{ 2 }x\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =6x+2secx.\frac { d }{ dx } (sec\quad x)\)
\(=6x+2secx.secxtanx\)
\(=6x+2{ sec }^{ 2 }xtanx\)

Let x = a cos³ θ, y = a sin³ θ. Then
\( x^{\frac{2}{3}}+y^{\frac{2}{3}} =\left(a \cos ^3 \theta\right)^{\frac{2}{3}}+\left(a \sin ^3 \theta\right)^{\frac{2}{3}} \)
\( =a^{\frac{2}{3}}\left(\cos ^2 \theta+\left(\sin ^2 \theta\right)=a^{\frac{2}{3}}\right. \)
Hence, \(x=a \cos ^3 \theta, y=a \sin ^3 \theta\) is parametric equation of \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\)
\(Therefore\ \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}=-\tan \theta=-\sqrt[3]{\frac{y}{x}} \)
\(=\frac { -\sqrt [ 3 ]{ \frac { y }{ a } } }{ \sqrt [ 3 ]{ \frac { y }{ x } } } \)
\(=-\sqrt [ 3 ]{ \frac { y }{ x } } \)

We have, \(\frac{d x}{d \theta}=a(1+\cos \theta), \frac{d y}{d \theta}=a(\sin \theta)\)
Therefore \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)}=\tan \frac{\theta}{2}\)

\(We\quad have:\quad x={ at }^{ 2 },\quad y=2at.\)
\(\frac { dx }{ dt } =2at\quad and\quad \frac { dy }{ dt } =2a.\)
\(\frac { dy }{ dx } =\frac { { dy }/{ dt } }{ { dx }/{ dt } } =\frac { 2a }{ 2at } =\frac { 1 }{ t } ,t\neq 0\)

Given that
\(x=a\ cos\theta ,\ y=a\ sin\theta \)
\(\frac { dx }{ d\theta } =-a\ sin\theta \ and\ \frac { dy }{ d\theta } =a\ cos\theta \)
\(Hence,\ \frac { dy }{ dx } =\frac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\frac { a\ cos\theta }{ -a\ sin\theta } =-cot\theta \)

\(Let\quad y={ x }^{ sinx\quad }...(1)\)
\(Taking\quad logs.,\quad log\quad y={ log\quad x }^{ sinx }=sinxlogx.\)
\(Diff.\quad w.r.t.\quad x,\quad \frac { 1 }{ y } .\frac { dy }{ dx } =sinx.\frac { 1 }{ x } +logx.cosx\)
\(\frac { dy }{ dx } =y\left[ \frac { sinx }{ x } +cosxlogx \right] \)
\(={ x }^{ sinx\quad }\left[ \frac { sinx }{ x } +cosxlogx \right] \)

First, observe that the domain of log function is set of all positive real numbers. So the above equation is not true for non-positive real numbers.
The given equation is \(x={ e }^{ logx }\)
Now, let y = e^{log x}
\(Now,let\quad y={ e }^{ logx }\)
\(If\ y>0,\ taking\ logs.,log\ y=log\left( { e }^{ logx } \right) =logx.loge\)
Thus \(y=x.\)
\(Hence,\ x={ e }^{ logx } \) is true only for positive values of x
One of the striking properties of the natural exponential function in differential calculus is that it doesn’t change during the process of differentiation.

We differentiate the relationship directly with respect to x, i.e.,
\(\frac{d y}{d x}+\frac{d}{d x}(\sin y)=\frac{d}{d x}(\cos x)\)
which implies using chain rule
\(\frac{d y}{d x}+\cos y \cdot \frac{d y}{d x}=-\sin x\)
This gives, \( \frac { dy }{ dx } =-\frac { sin\quad x }{ 1+cos\quad y } \)
\(where\ y\neq (2n+1)\pi ,\ n\epsilon I.\)

One way is to solve for y and rewrite the above as
\(y=x-\pi\)
\(\text {But then }\frac{d y}{d x}=1\)
Alternatively, directly differentiating the relationship w.r.t., x, we have
\(\frac{d}{d x}(x-y)=\frac{d \pi}{d x}\)
Recall that \(\frac{d \pi}{d x}\) means to differentiate the constant function taking value p everywhere w.r.t., x. Thus
\(\frac{d}{d x}(x)-\frac{d}{d x}(y)=0\)
which implies that
\(\frac{d y}{d x}=\frac{d x}{d x}=1\)

\(x^y+y^x=1\)
Let \(u=x^y, v=y^x\)
Hence,
u+v=1
Differentiating both sides w.r.t. x.
\(\frac{d(v+u)}{d x}=\frac{d(1)}{d x} \)
\(\frac{d v}{d x}+\frac{d u}{d x}=0\)
(Derivative of constant is 0 )

Let y=tan(2x+3)=tan t, where t=2x+3
\(\frac { dy }{ dt } ={ sec }^{ 2 }\ t\ and\ \)
\(\frac { dt }{ dx } =2(1)+0=2.\)
\(By\quad chain\quad rule,\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(={ sec }^{ 2 }\ t.2\)
\(=2 { sec }^{ 2 }(2x+3)\)

Observe that the given function is a composite of two functions. Indeed, if t = u(x) = x² and v(t) = sin t, then
f(x) = (v o u) (x) = v(u(x)) = v(x 2 ) = sin x²
\(\text {Put } t=u(x)=x^2 \text {. Observe that } \frac{d v}{d t}=\cos t \text { and } \frac{d t}{d x}=2 x\) exist. Hence, by chain rule
\(\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos t \cdot 2 x\)
It is normal practice to express the final result only in terms of x. Thus
\(\frac{d f}{d x}=\cos t \cdot 2 x=2 x \cos x^2\)

Observe that the function is defined for every real number. The function f may be thought of as a composition g o h of the two functions g and h, where g(x) = sin x and h(x) = x². Since both g and h are continuous functions, by Theorem 2, it can be deduced that f is a continuous function

Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x.
Then (h o g) (x) = h(g (x)) = h (1– x + | x |)
= |1– x + | x || = f(x)
we have seen that h is a continuous function.
Hence g being a sum of a polynomial function and the modulus function is continuous.
But then f being a composite of two continuous functions is continuous.

Observe that the function is defined at all real numbers except at 0. Domain of definition of this function is
\(\mathrm{D}_{1} \cup \mathrm{D}_{2} \text { where } \mathrm{D}_{1}=\{x \in \mathbf{R}: x<0\} \text { and }\)
\(\mathrm{D}_{2}=\{x \in \mathbf{R}: x>0\}\)
Case 1 \(\text { If } c \in \mathrm{D}_{1}, \text { then } \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+2)\)
= c + 2 = f (c) and hence f is continuous in D₁
Case 2 \(\text { If } c \in \mathrm{D}_{2}, \text { then } \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-x+2)\)
= – c + 2 = f (c) and hence f is continuous in D2.
Since f is continuous at all points in the domain of f, we deduce that f is continuous. Note that to graph this function we need to lift the pen from the plane of the paper, but we need to do that only for those points where the function is not defined.

We may rewrite f as:
\(f(x)=\begin{cases} -x,\quad \quad if\quad x<0 \\ x,\quad \quad \quad if\quad x\ge 0 \end{cases}\)
we know that f is continuous at x = 0
Let c be a real number such that c < 0. Then f(c) = – c. Also
\( \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ (-x) } =-c\)
\(Thus\ \lim _{ x\rightarrow c }{ f(x) } =f(c).\) f is continuous at all negative real numbers.
Now, let c be a real number such that c > 0. Then f(c) = c. Also
\(\lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ (x) } =c\)
Since \(\lim _{ x\rightarrow c }{ f(x) } =f(c).\) f is continuous at all positive real numbers.
Hence, f is continuous at all points

\(\lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ \frac { 1 }{ x } =\frac { 1 }{ c } } \)
\(Also\quad f(c)=\frac { 1 }{ c } \)
\(Thus\quad \lim _{ x\rightarrow c }{ f(x) } =f(c)\)
Thus f is continuous at each part in D_f
Hence, f is a continuous function.

\(We\quad have:f(x)={ x }^{ 3 }+{ x }^{ 2 }-1\)
Which is polynomial function
\(and\ { D }_{ f }=R\)
\(Let\quad c\in { D }_{ f }\)
\(Then\ \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ ({ x }^{ 3 }+{ x }^{ 2 }-1) } \)
\(={ c }^{ 3 }+{ c }^{ 2 }-1=f(c)\)
\(\Rightarrow \) f is continuous at x = c.
But c is arbitrary.
Hence, f is continuous at each of its domains.

The function is clearly defined at every point and f(c) = c for every real number c. Also
\(\lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ x } =c.\)
\( \lim _{ x\rightarrow c }{ f(x) } =f(c) \) and hence the function is continuous at every real number.
Having defined continuity of a function at a given point, now we make a natural extension of this definition to discuss continuity of a function

Let cϵR be arbitrary point.
\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ (k) } =k.\)
\(Also\quad f(c)=k.\)
\(Thus\quad \lim _{ x\rightarrow c }{ f(x) } =f(c).\)
Thus ′f′ is continuous at x = c.
But c is arbitrary.
Hence, f(x) is continuous at each real number.

The function is defined at x = 0 and its value at x = 0 is 1. When x \(\ne\) 0, the function is given by a polynomial. Hence
\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ { x }^{ 3 }+3 } =0+3=3\)
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function.

\( f(x)=\begin{cases} -x,if\quad x<0 \\ x,\quad if\quad x\ge 0. \end{cases}\)
Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (-x) } =0\)
Similarly, the right hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (x) } =0\)
Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0.
Hence, f is continuous at x = 0.

First note that the function is defined at the given point x = 0 and its value is 0.
Then find the limit of the function at x = 0. Clearly
\(\lim _{ x\rightarrow 1 }{ f(x) } =\quad { x }^{ 2 }={ 0 }^{ 2 }=0\)
\( f(0)={ 0 }^{ 2 }=0\)
\(\lim _{ x\rightarrow 0 }{ f(x) } =f(0).\)
Hence,f is continuous at x = 0

First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly
\(\lim _{ x\rightarrow 1 }{ f(x) } =\lim _{ x\rightarrow 1 }{ \left( { 2x }+3 \right) } =2(1)+3=5\)
\(f(1)=2(1)+3=5.\)
\(Thus\quad \lim _{ x\rightarrow 1 }{ f(x) } =f(1)\)
Hence,f is continuous at x = 1.

\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ \left( { x }^{ 3 }+3 \right) } =0+3=3\)
\( f(0)=1\)
\( Thus\ \lim _{ x\rightarrow 0 }{ f(x) } \neq f(0)\)
Hence, f is not continuous at x = 0.

The function y = (sin x)^{sin x} is defined for all positive real numbers. Taking logarithms, we have
log y = log (sin x)^{sin x} = sin x log (sin x)
Then \(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(\sin x \log (\sin x))\)
\( =\cos x \log (\sin x)+\sin x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x) \)
\(=\cos x \log (\sin x)+\cos x \)
\(=(1+\log (\sin x)) \cos x\)
Thus, \(\frac{d y}{d x}=y((1+\log (\sin x)) \cos x)=(1+\log (\sin x))(\sin x)^{\sin x} \cos x\)

Since f is differentiable at c, we have
\(\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}=f^{\prime}(c)\)
But for x ≠ c, we have
\(f(x)-f(c)=\frac{f(x)-f(c)}{x-c} \cdot(x-c)\)
Therefore \(\lim _{x \rightarrow c}[f(x)-f(c)]=\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c} \cdot(x-c)\right]\)
\( \lim _{x \rightarrow c}[f(x)]-\lim _{x \rightarrow c}[f(c)] =\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c}\right] \cdot \lim _{x \rightarrow c}[(x-c)] \\ =f^{\prime}(c) \cdot 0=0 \)
\(\lim _{x \rightarrow c} f(x)=f(c)\)
Hence f is continuous at x = c.

We are investigating continuity of (f + g) at x = c. Clearly it is defined at
x = c. We have
\( \lim _{x \rightarrow c}(f+g)(x) =\lim _{x \rightarrow c}[f(x)+g(x)] \\ =\lim _{x \rightarrow c} f(x)+\lim _{x \rightarrow c} g(x) \\ =f(c)+g(c) \\ =(f+g)(c) \)
Hence, f + g is continuous at x = c.
Proofs for the remaining parts are similar and left as an exercise to the reader.

Let u = \(tan^{ -1 }\left( \frac { x }{ \sqrt { 1-x^{ 2 } } } \right) \) 
and v = \(sin^{ -1 }(2x\sqrt { 1-x^{ 2 } } )\quad x\neq 0\)
Let x = sin \(\theta \)
\(\therefore \theta =sin^{ -1 }x\)
\(\therefore u=tan^{ -1 }\left( \frac { sin\theta }{ \sqrt { 1-sin^{ 2 }\theta } } \right) \)
\(=tan^{ -1 }(tan\theta )=\theta =sin^{ -1 }x\)
\(\Rightarrow \frac { du }{ dx } =\frac { 1 }{ \sqrt { 1-x^{ 2 } } } \)
and \(v=sin(2x\sqrt { 1-x^{ 2 } } )\)
\(sin^{ -1 }(sin2\theta )=2\theta =2sin^{ -1 }x\)
\(\Rightarrow \frac { dv }{ dx } =\frac { 2 }{ \sqrt { 1-x^{ 2 } } } \)
\(\therefore \frac { du }{ dv } =\frac { 1 }{ \sqrt { 1-x^{ 2 } } } \times \frac { \sqrt { 1-x^{ 2 } } }{ 2 } \)
\(=\frac { 1 }{ 2 } \)

\(y=\sin^{-1}(\frac{2^{x+1}}{1+4^x})\)
\(\Rightarrow y=\sin^{-1}(\frac{2.2^x}{1+2^{2x}})\)
Let  \(2^x=\tan \theta ,then\)
\(y=\sin^{-1}(\sin 2\theta)\)
\(\because\ \frac{2\tan \theta}{1+\tan^2\theta}=\sin2\theta\)
\(y=2\theta\)
\(y=2\tan^{-1}(2^x)\)
Diff. w.r.t x, 
\(\frac{dy}{dx}=2.\frac{1}{1+(2^x)^2}.\frac{d}{dx}(2^x)\)
\(=\frac{2}{1+4^x}.2^x.\log_e2.\)
\(=\frac{2^{x+1}}{1+4^x}\log_e2\)

\(x=sin\alpha ,\sqrt { x } =sin\beta \)
\(cos\alpha =\sqrt { 1-x^{ 2 } } \)
\(cos\beta =\sqrt { 1-x } \)
\(y=sin^{ -1 }(sin\alpha cos\beta -cos\alpha sin\beta )\)
\(y=sin^{ -1 }[sin\alpha -\beta ]=\alpha -\beta \)
\(y=sin^{ -1 }x-sin^{ -1 }\sqrt { x } \)
\(\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-x^{ 2 } } } -\frac { 1 }{ 2\sqrt { x } \sqrt { 1-x } } \)

y = e^ax
\(\frac { dy }{ dx } \)= e^ax(-sinbx)b+cosbx \(\times \) e^ax\(\times \) a
\(\frac { dy }{ dx } -be^{ ax }sinbx+ay\Rightarrow \left[ \frac { -1 }{ b } \left( \frac { dy }{ dx } -ay \right) \right] =e^{ ax }sinbx\)
\(\frac { dy }{ dx } =-b\left[ e^{ a^{ x }cosb }\times a \right] +\frac { ady }{ dx } \)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } -b\left[ by\frac { -a }{ -b } \left( \frac { dy }{ dx } -ay \right) \right] +\frac { ady }{ dx } \) from (1)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } -b^{ 2 }y+a\frac { dy }{ dx } -a^{ 2 }y+a\frac { dy }{ dx } \)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +\left( a^{ 2 }+b^{ 2 } \right) y-2a\frac { dy }{ dx } =0\)
Hence proved

\(f(x)=\{2x-1, \ if\ x<0 \ 2x+1,\ if\ x\ge0\)
L.H.L = \(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ 2x-1 } c\) = -1
R.H.L = \(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ 2x-1 } c\) = 1
L.H.L. ≠ R.H.L
ஃ f(x) = is not continues at x = 0
ஃ The dangerous point on the path is at x = 0
The rider should not pass through the point at x = 0.

Then sec y = sec (sec^-1x)
Differentiating both sides with respect to x, we have 
\(\frac{d}{dx}\sec y=\frac{d}{dx}(x)\)
\(\Rightarrow \frac{d}{dy}(\sec y)=\frac{dy}{dx}=1\)
\(\Rightarrow \sec y \tan y \frac{dy}{dx}=1\)
If x >1, then \(y\in(0,\frac{\pi}{2})\)
\(\therefore\ \sec y>, \tan y>0\)
\(\Rightarrow |\sec|.|\tan y|=\sec y\tan y\)
If x < -1, then
\(y\in (\frac{\pi}{2},\pi) \ \ \therefore \sec y<0,\tan y<0\)
\(\Rightarrow |\sec y|\tan y|\)
\(\Rightarrow |-\sec|.|-\tan y|=\sec y\tan y\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\sec y \tan y}=\frac{1}{|\sec y|.|\tan y|}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{|\sec y| \sqrt{\tan^2 y}}\)
\(=\frac{1}{|\sec y|\sqrt{\sec^2y-1}}\)
\(=\frac{1}{|x|\sqrt{x^2-1}}\)

f(x) is continuous in [-1, 1] as exponential function is always continuous
\(f^{\prime}(x)=-2 x e^{1-x^{2}} \text { exists in }(-1,1)\)
\(\therefore f \text { is derivable. }\)
\(f(-1)=1, f(1)=1 \text { , conditions satisfy. }\)
\(\text { Hence, } \quad f^{\prime}(c)=0 \text { for } c \in(-1,1)\)
\(\Rightarrow -2 c \cdot e^{1-c^{2}} =0 . \text { But } e^{1-c^{2}} \neq 0 \)
\(\therefore c =0 \in(-1,1)\)

(i) f is continuous in [0, \(\pi\)] sine function is always continuous.
(ii) f(x) = cos x, exists in (0, \(\pi\)). Hence f is derivable.
(iii) f(0) = sin 0 = 0
f(\(\pi\)) = sin \(\pi\) = 0
conditions of Rolle's Theorem are satisfied.
Hence, there exists at least one point C \(\in\) (0, \(\pi\)) such that f(c) = 0
\(\Rightarrow \cos c=0 \Rightarrow c=\frac{\pi}{2} \in(0, \pi)\)

f(x) = x² 5x + 6 in [2, 3]
(i) f is continuous in [2,3] as polynomial function is always continuous
(ii) f(x) = 2x - 5, exists in (2, 3). Hence f is derivable in (2, 3).
(iii) f(2) = 0,f(3) = 0 => f(2) =f(3)
.'. conditions of Rolle's Theorem are satisfied.
Hence, there exists, at least one point C \(\in\) (2, 3) such that
\(f^{\prime}(c)=0 \Rightarrow 2 c-5=0 \Rightarrow c=\frac{5}{2} \in(2,3)\)

\(\frac { dy }{ dx } =\frac { { cos }^{ 2 }\left( a+y \right) }{ sin\quad a } \)

e^y =\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\left( \frac { dy }{ dx } \right) ^{ 2 }\)

\(\frac { dy }{ dx } =\frac { { sin }^{ 2 }(a+y) }{ sin\quad a } \)

\(y'=\frac { -x }{ y } \)

\(\frac { -3 }{ \sqrt { 1-{ x }^{ 2 } } } \)

\(y'=\frac { -1 }{ 1+x } .\frac { 1 }{ 2\sqrt { x } } =\frac { -1 }{ 2\sqrt { x } (1+x) } \)

y=x \(\Rightarrow \)y'=1

\(y'=\frac { 2 }{ 1+\left( { 2 }^{ x } \right) ^{ 2 } } .{ 2 }^{ x }log_{ e }2=\frac { { 2 }^{ x+1 }{ log }_{ e }2 }{ 1+{ 4 }^{ x } } \)

y'=cos (tan^-1e^-x).\(\frac { 1 }{ 1+{ e }^{ -2x } } \).(-e^-x)=\(\frac { { -e }^{ -x }cos\left( { tan }^{ -1 }{ e }^{ -x } \right) }{ 1+{ e }^{ -2x } } \)

\(\frac { d }{ dx } \left( { 2 }^{ cos^{ 2 }x } \right) ={ 2 }^{ cos^{ 2 }x }.log_{ e }2.\left( -2sinx\quad cosx \right) =-{ 2 }^{ cos^{ 2 }x }log_{ e }2. sin2x.\)

(i) f (x) = |x|, continuous in [-1, 1]
(ii) LHD at x = 0 \(\neq \) RHD at x = 0.
\(\therefore \)Not derivable at x = 0. Theorem is not applicable.

f (x) = cos x + sin x, [0, 2\(\pi \)]
(i) f (x) is continuous as sine and cosine functions are continuous and sum is also continuous.
(ii) f'(x) = -sin x + cos x, exists in [0, 2\(\pi \)] Hence, derivable.
(iii) f (0) = f ( 2\(\pi \))=1, conditions satisfied
\(\Rightarrow \) f'(c)=0 for 0< c<2\(\pi \)
\(\Rightarrow \) -sin c + cos c = 0 \(\Rightarrow \)tan c = 1 \(\Rightarrow \)\(c=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \)

\(c=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \)

\(y'-y+\frac { { x }^{ n } }{ n! } =0\)

\(\frac { -y(y+x\ \ \ log\ y) }{ x(x+y\quad log\ x) } \)

\(\Rightarrow \frac { dy }{ dx } +{ sec }^{ 2 }\left( \frac { \pi }{ 4 } -x \right) =0\)

\(\frac { dy }{ dx } =-cosec^{ 2 }\ x\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)=2 cosec² x cot x,
substitute for \(\frac { dy }{ dx } ,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) in LHS.

\(\Rightarrow (1-{ x }^{ 2 })\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\frac { dy }{ dx } -{ a }^{ 2 }y=0\)

=m²y

\(\frac { dy }{ dx } \Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-A{ n }^{ 2 }y=0\)

Let y=2 tan^-1 x and t=2 tan^-1 x           
\(\Rightarrow y=t\quad \Rightarrow \frac { dy }{ dt } =1\)

\(e^{x}+e^{y} \frac{d y}{d x}=e^{x+y} \cdot\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \left(e^{y}-e^{x+y}\right) \frac{d y}{d x}=e^{x+y}-e^{x} \)
\(\Rightarrow \frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}} \)
\(\Rightarrow \frac{d y}{d x}=\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(1-e^{x}\right)} \)
\(\Rightarrow \frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}\)

\(\Rightarrow \left( \frac { 1 }{ y } -1 \right) y'=1\quad \Rightarrow y'=\frac { y }{ 1-y } \)

\(y'=\frac { 1 }{ 1+(a+bx)^{ 2 } } .b\Rightarrow 1=\frac { b }{ 1+{ a }^{ 2 } } \Rightarrow 1+{ a }^{ 2 }=b\)

\(\Rightarrow y'=\frac { \frac { 2 }{ a } }{ 1+\frac { { 4x }^{ 2 } }{ { a }^{ 2 } } } +\frac { \frac { 3 }{ a } }{ 1+\frac { { 9x }^{ 2 } }{ { a }^{ 2 } } } \)

y=tan^-1\(\left[ tan\frac { \theta }{ 2 } \right] =\frac { 1 }{ 2 } { cot }^{ -1 }x\Rightarrow \frac { dy }{ dx } =\frac { -1 }{ 2\left( 1+{ x }^{ 2 } \right) } \)

\(\Rightarrow \frac { dy }{ dx } ={ x }^{ { x }^{ x } }.{ x }^{ x }.\ log\ x\left[ 1+log \ \ x\ +\frac { 1 }{ x\quad log\quad x } \right] \)

\(\Rightarrow y'=\frac { -1.\left( 2x \right) }{ 2\times \sqrt { 1-{ x }^{ 4 } } } =\frac { -x }{ \sqrt { 1-{ x }^{ 4 } } } \)

\(\Rightarrow y'=\frac { 1 }{ 2 } .\frac { 1 }{ 1+{ a }^{ 2 }{ x }^{ 2 } } .a=\frac { a }{ 2\left( 1+{ a }^{ 2 }{ x }^{ 2 } \right) } \)

\(\Rightarrow y'=-1\)

\(\Rightarrow y'=-1\)

\(\Rightarrow y=\frac { \pi }{ 4 } +\frac { x }{ 2 } \Rightarrow y'=\frac { 1 }{ 2 } \)

\(\Rightarrow y=\frac { x }{ 2 } \Rightarrow y'=\frac { 1 }{ 2 } \)

\(\Rightarrow { y }^{ ' }=\frac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)

y'=\({ e }^{ { -x }^{ 2 } }\)\(\left[ -2xsin\quad (logx)+\frac { cos(log\quad x) }{ x } \right] \)

y'=sec x

\(\Rightarrow { y }^{ ' }=\frac { 1 }{ 2 } \)

\(\frac { dy }{ dx } ={ e }^{ x }\left[ 2cot2x+log\quad (sin2x) \right] \)

y^'=-2x tan x²

y=\(\frac { 1 }{ 2\sqrt { x } (1+x) } \)

\(\frac { dy }{ dx } =\frac { 1 }{ 4\sqrt { a+x } \sqrt { a+\sqrt { a+x } } } \)

\(\Rightarrow y=\frac { 2a }{ \sqrt { 1-{ a }^{ 2 }{ x }^{ 2 } } } \)

\(\Rightarrow { y }^{ ' }=0+\frac { 1 }{ 2 } \times \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)

\({ y }^{ ' }=\frac { 2 }{ 1+x^{ 2 } } \)

\(\Rightarrow \) \(\left( { 1-x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\frac { { d }y }{ d{ x } } =0\)\(\left[ On\quad dividing\quad by\quad 2\left( \frac { dy }{ dx } \right) \right] \)

\(\Rightarrow \left( { 1+x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x({ 1+x }^{ 2 })\frac { { d }y }{ d{ x } } =2\)

\(\Rightarrow \) (1+x²) y₂ + (2x-1) y₁=0

\({ y }^{ ' }=Ame^{ mx }+{ Bne }^{ nx }\Rightarrow { y }^{ ' }={ Am }^{ 2 }{ e }^{ mx }+{ Bn }^{ 2 }{ e }^{ nx }\),
substitute for y^'', y^', y in LHS.

\(\Rightarrow { x }^{ 2 }{ y }^{ ' }+xy^{ ' }+y=0\)

\(\Rightarrow y'=\frac { -2 }{ 1+\left( { 2 }^{ X } \right) ^{ 2 } } .{ 2 }^{ X }{ log }_{ e }2=\frac { { -2 }^{ x+1 }{ log }_{ e }2 }{ 1+{ 2 }^{ 2x } } \)

\(\text { Let } \quad x=\cos \theta \)
\(\text { then } \quad y=\cos ^{-1}\left(\frac{3 \cos \theta+4 \sin \theta}{5}\right) \)
\(\Rightarrow y=\cos ^{-1}\left(\frac{3}{5} \cos \theta+\frac{4}{5} \sin \theta\right) \)
\(\text { Let } \frac{3}{5}=\cos \alpha \text { and } \frac{4}{5}=\sin \alpha \)
\(\text { then } \sin ^{2} \alpha+\cos ^{2} \alpha=1 \)
\(\therefore y =\cos ^{-1}(\cos \alpha \cos \theta+\sin \alpha \cdot \sin \theta) \)
\(=\cos ^{-1}\{\cos (\theta-\alpha)\} \)
\(=\theta-\alpha=\cos ^{-1} x-\cos ^{-1} \frac{3}{5} \)
\(\Rightarrow \frac{d y}{d x} =\frac{-1}{\sqrt{1-x^{2}}}\)

\(\text { Given }y =\sqrt{\frac{\sec x-1}{\sec x+1}}=\sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}} \)
\(=\sqrt{\frac{1-\cos x}{1+\cos x}} \)
\(=\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}=\sqrt{\tan ^{2} \frac{x}{2}}\)
\(\Rightarrow y=\tan \frac{x}{2} \Rightarrow \frac{d y}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2}\)
On simplifying we get y= tan \(\frac { x }{ 2 } \)\(\Rightarrow \)y'=\(\frac { 1 }{ 2 } \)sec²\(\frac { x }{ 2 } \)

\(y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\}, \)
\(\text { Let } \text x=\sin \theta, \)
\(\text { then } y=\sin ^{-1}\left\{\frac{5 \sin \theta+12 \cos \theta}{13}\right\} \)
\(=\sin ^{-1}\left\{\frac{5}{13} \sin \theta+\frac{12}{13} \cos \theta\right\} \)
\(\text { Let } \frac{5}{13}=\cos \alpha, \frac{12}{13}=\sin \alpha, \)
\(\text { We notice } \sin ^{2} \alpha+\cos ^{2} \alpha=1 .\)
\(\Rightarrow y =\sin ^{-1}\{\sin \theta \cos \alpha+\cos \theta \sin \alpha\} \)
\(=\sin ^{-1}\{\sin (\theta+\alpha)\}=\theta+\alpha \)
\(=\sin ^{-1} x+\cos ^{-1} \frac{5}{13} \)
\(\Rightarrow \frac{d y}{d x} =\frac{1}{\sqrt{1-x^{2}}}\)
\(=\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)

Consider,h(x) = sin x
Let us check continuity at x = a
h(a) = sin a
\(\underset{x=a}{\operatorname{LHL}} =\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0} \sin (a-h) \)
\(=\lim _{h \rightarrow 0}(\sin a \cos h-\cos a \sin h) \)
\(=\sin a \cos 0-\cos a \sin 0 \)
\(=\sin a-0=\sin a \)
\(\mathrm{RHL} x=a =\lim _{h \rightarrow 0} f(a+h)=\lim _{h \rightarrow 0} \sin (a+h) \)
\(=\lim _{h \rightarrow 0}(\sin a \cos h+\cos a \sin h) \)
\(=\sin a \cos 0+\cos a \sin 0=\sin a+0 \)
\(=\sin a \)
\(\mathrm{LHL} =\mathrm{RHL}=f(a)\)
\(\therefore\) h is continuous at x = a.
Hence, sine function is continuous.

If function is continuous at x = 2
\(\text { then } \lim _{x \rightarrow 2} f(x)=f(2) \)
\(\Rightarrow \lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=k \)
\(\lim _{x \rightarrow 2} \frac{4.2^{x}-16}{4^{x}-16}=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}=\frac{4}{8}=\frac{1}{2}=k \)

For continuity at x = 0,
\(\lim _{x \rightarrow 0} f(x)=f(0)\)
Consider
\( \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+3 x)}{x} \)
\(=\lim _{x \rightarrow 0} 3 \cdot \frac{\log (1+3 x)}{3 x} \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right] \)
\(=3 \times 1=3 \text { and } f(0)=3 \)
\(\text { we have } \lim _{x \rightarrow 0} f(x)=f(0)=3\)
\(\therefore f \text { is continuous at } x=3 \text { . }\)

Consider \(\lim _{x \rightarrow 0} f(x) =\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}} \)
\(=\lim _{x \rightarrow 0} \frac{e^{2 x}+1-2 e^{x}}{e^{x} \cdot x^{2}} \)
\(=\lim _{x \rightarrow 0} \frac{1}{e^{x}}\left(\frac{e^{x}-1}{x}\right)^{2} \)
\(=\frac{1}{1} \times(1)^{2}=1 \)
\(f(0) =4 k\)
If continuous then 1 = 4 \(k\Rightarrow k=\frac { 1 }{ 4 } \)

8 From (i), we notice, continuous.

From (i) , (ii) and (iii), we note that function is continuous at x=\(\frac { 1 }{ 2 } \).

If function is continuous at x = 0, then
\( \lim _{x \rightarrow 0} f(x)=f(0) \)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 5 x}{3 x}=k \)
\(\Rightarrow \frac{5}{3} \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}=k \Rightarrow \frac{5}{3}=k .\)

\(\begin{matrix} Lim \\ x\rightarrow 0 \end{matrix}\left[ \frac { sin\quad x }{ x } +cos\quad x \right] =k\Rightarrow 2=k\)

\(\mathrm{LHL}=\lim _{x=0} f(0-h)=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1} \)
\(=\lim _{h \rightarrow 0} \frac{\frac{1}{e^{1 / h}}-1}{\frac{1}{e^{1 / h}}+1}=\frac{0-1}{0+1}=-1\)
\(\mathrm{RHL} =\lim _{h=0} f(0+h)=\lim _{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{e^{1 / h}}}{1+\frac{1}{e^{1 / h}}} \)
\(=\frac{1-0}{1+0}=1\)
\(\begin{matrix} LHL \\ x=0 \end{matrix}\neq \begin{matrix} RHL \\ x=0 \end{matrix} \)
\(\therefore \) Discontinuous.

Solving (i) and (ii), we get a=3, b=-8.

\(\underset{x=1}{\mathrm{LHL}}= \lim _{x \rightarrow 1^{-}} \frac{[x]-1}{x-1}=\lim _{h \rightarrow 0} \frac{[1-h]-1}{1-h-1}\)
\(=\lim _{h \rightarrow 0} \frac{0-1}{-h}=\lim _{h \rightarrow 0} \frac{1}{h} \)
\(\Rightarrow \text { does not exist. } \)
\(\mathrm{RHL} =\lim _{x=1} \frac{[x]-1}{x-1}=\lim _{h \rightarrow 0} \frac{[1+h]-1}{1+h-1} \)
\(=\lim _{h \rightarrow 0} \frac{1-1}{h}=0\)
\(\mathrm{LHL}_{x=1} \neq \mathrm{RHL}_{x = 1}\) 
\(\therefore \) Discontinuous.

\(\mathrm{LHL}=\lim _{x=0} f(0-h)=\lim _{h \rightarrow 0} \frac{|-h|}{-h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}=\lim _{h \rightarrow 0}(-1)=-1 \)
\(\underset{x=0}{\mathrm{RHL}} =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{|0+h|}{0+h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)=1 \)
\(\text { As } \underset{x=0}{\mathrm{LHL}} \neq \mathrm{RHL}_{x=0}\)
hence functions is not continuous at x = 0

\(\begin{matrix} LHL \\ x=0 \end{matrix}=\begin{matrix} Lim \\ h\rightarrow 0 \end{matrix}\frac { -h }{ h+2h^{ 2 } }\)
\(=\begin{matrix} Lim \\ h\rightarrow 0 \end{matrix}\frac { -1 }{ 1+2h } =-1;\begin{matrix} RHL \\ x=0 \end{matrix}=\begin{matrix} Lim \\ h\rightarrow 0 \end{matrix}\frac { h }{ h+{ 2h }^{ 2 } } \)
\(=\begin{matrix} Lim \\ h\rightarrow 0 \end{matrix}\frac { 1 }{ 1+2h } =1\)

\(\begin{matrix} LHL \\ x=0 \end{matrix}\neq \begin{matrix} RHL \\ x=0 \end{matrix}\)  , limit does not exist.  
Hence, the function is not continuous for any value of k. 

Given function \(f(x)=|x-3|=\left\{\begin{aligned} x-3, & x \geq 3 \\ -x+3, & x<3 \end{aligned}\right.\)
\(\underset{x=3}{\mathrm{LHL}} =\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}\{-(3-h)+3\} \)
\(=\lim _{h \rightarrow 0} h=0 \)
\(\mathrm{RHL} =\lim _{x=3} f(3+h)=\lim _{h \rightarrow 0}\{(3+h)-3\} \)
\(=\lim _{h \rightarrow 0} h=0 \)
\(f(3)=3-3=0\)
\(\text { As } \underset{x=3}{\mathrm{LHL}}=\underset{x=3}{\mathrm{RHL}}=f(3) \text { , }\)
For sontinuity at x = 3,
Hence, function is continuous at x = 3.
\(\underset{x=3}{\operatorname{LHD}} =\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}=\lim _{h \rightarrow 0} \frac{(-3+h+3)-(0)}{-h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}=\lim _{h \rightarrow 0}(-1)=-1 \)
\(\operatorname{RHD}_{x=3} =\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h-0} \frac{(3+h-3)-(0)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)=1 \)
For differentiability at x = 3,
As Hence, function is not derivable (differentiable) at x = 3.

Here, \(f(x)=\left\{\begin{array}{cl} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.\) 
At \(x=0, \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)\) 
\(\therefore \mathrm{LHL}=\lim _{h \rightarrow 0} \lambda\left[(0-h)^{2}-2(0-h)\right]=\lim _{h \rightarrow 0}\left[\lambda\left(h^{2}+2 h\right)\right]=0\) 
\(\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)\) 
\(\therefore \mathrm{RHL}=\lim _{h \rightarrow 0}[4(0+h)+1]=\lim _{h \rightarrow 0}[4 h+1]=0+1=1\)
\(\text { [put } x=0+h \text { ; when } x \rightarrow 0^{+} \text {, then } \left.h \rightarrow 0\right] \) 
\(\therefore \mathrm{LHL} \neq \mathrm{RHL}\) 
Thus, f(x) is not continuous at x = 0 for any value of λ.
At x = 1,
 \( \mathrm{LHL} =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(4 x+1) \)
\(\therefore \mathrm{LHL} =\lim _{h \rightarrow 0}[4(1-h)+1]=\lim _{h \rightarrow 0}[5-4 h]=5-0=5 \)
[put x=1−h; when x→1−,then h→0]   
\( \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x+1) \)
\(\therefore \lim _{h \rightarrow 0}[4(1+h)+1]=\lim _{h \rightarrow 0}(5+4 h)=5+0=5 \)
[put x=1+h; when x→1, then h→0]   
Also, f(1)=4×1+1=5
\([\because f(x)=4 x+1]\)
Thus f(x) is continuous at x=1 for all values of λ.

For function to be continuous at x = 2, we have
\(\mathrm{LHL}_{x=2}=\mathrm{RHL}_{x=2}=f(2)\)
\(\Rightarrow \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \)
\(\Rightarrow \lim _{h \rightarrow 0}\{2(2-h)+1\}=\lim _{h \rightarrow 0}\{3(2+h)-1\}=k \)
\(\Rightarrow 4+1=6-1=k \Rightarrow k=5 . \)
k = 5

For continuity at x - 1
\(\lim _{1} f(x)=\lim _{1} f(x)=f(1)\)
\(\Rightarrow \lim _{x \rightarrow 1}(4)=\lim _{x \rightarrow 1}\left(k x^{2}\right)=k \)
\(\Rightarrow 4=k=k \Rightarrow k=4 \)
k = 4