We have: \(y={ \left( { sin }^{ -1 }x \right) }^{ 2 }\)
\({ y }_{ 1 }=2\left( { sin }^{ -1 }x \right) .\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\sqrt { 1-{ x }^{ 2 } } { y }_{ 1 }=2\left( { sin }^{ -1 }x \right) \)
Squaring \(\left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4{ \left( { sin }^{ -1 }x \right) }^{ 2 }\)
\( \left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4y\)
Diff. w.r.t.x, \(\left( 1-{ x }^{ 2 } \right) 2{ y }_{ 1 }{ y }_{ 2 }+(-2x){ { y }_{ 1 } }^{ 2 }=4{ y }_{ 1 }\)
Hence, \(\left( 1-{ x }^{ 2 } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } =2\)

We have \(y={ e }^{ ax }cos\quad bx\) ...(1)
\(\frac { dy }{ dx } ={ e }^{ ax }(-sin\quad bx)(b)+a{ e }^{ ax }cos\quad bx\)
\(={ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx)....(2)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ ax }(-ab\quad sin\quad bx-{ b }^{ 2 }cos\quad bx)+a{ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx)....(3)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-2a\frac { dy }{ dx } +\left( { a }^{ 2 }+{ b }^{ 2 } \right) y\)
\(={ e }^{ ax }\left[ -ab\quad sin\quad bx-{ b }^{ 2 }cos\quad bx+{ a }^{ 2 }cos\quad bx-ab\quad sin\quad bx \right] \)
\(-2a({ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx))+\left( { a }^{ 2 }+{ b }^{ 2 } \right) { e }^{ ax }cos\quad bx]\)

We have: \(y=acos(logx)+bsin(logx)\)
\(\frac { dy }{ dx } =-a\quad sin(log\quad x).\frac { 1 }{ x } +bcos(log\quad x).\frac { 1 }{ x } \)
\(x.\frac { dy }{ dx } =-a\quad sin(log\quad x)+b\quad cos(log\quad x)\)
Again differentiating w.r.t.x, we get 
\(x\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\frac { dy }{ dx } .1=-a\quad cos(log\quad x).\frac { 1 }{ x } -b\quad sin(log\quad x)\frac { 1 }{ x } \)
\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +x\frac { dy }{ dx } =-[aa\quad cos(log\quad x)+b\quad sin(log\quad x)]=-y\)
\( { x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +x\frac { dy }{ dx } +y=0\)

\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ - } }{ \left( 3ax+b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [3a(1-h)+b] } \)
\(=3a(1-0)+b\)
\(=3a+b\)
\(\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ \left( 5ax-2b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [5a(1+0)-2b] } \)
\(=5a-2b\)
\(f(1)=11\)
Also
Since'f' is continuous at x = 1
\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =f(1)\)
From first and third 3a + b = 11 ....(1)
From last two 5a - 2b = 11 .....(2)
Multiplying (1) by 2, 6a + 2b = 22 .....(3)
Adding (2) and (3), 11a = 33 = a = 3
Putting in (1), 3(3) + b = 11
b = 11 - 9 = 2
Hence a = 3 and b = 2.

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

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

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

\(We\quad have:f(x)={ x }^{ 3 }+{ x }^{ 2 }-1\)
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 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.

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

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.

Putting x = \(cos2\theta \left\{ 2tan^{ -1 }\sqrt { \frac { 1-x }{ 1+x } } \right\} \) we get 
\(2tan^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
i,e.,\(2tan^{ -1 }\sqrt { \frac { 2sin^{ 2 }\theta }{ 2cos^{ 2 }\theta } } \)
\(=2\quad tan^{ -1 }\left( tan\theta \right) \)
\(=2\theta =cos^{ -1 }x\)
Hence \(y=e^{ sin^{ 2 } }xcos^{ -1 }x\)
\(\Rightarrow logy=sin^{ 2 }x+log\left( cos^{ 1 }x \right) \)
\(\Rightarrow \frac { 1 }{ y } \times \frac { dy }{ dx } =2sinxcosx+\frac { 1 }{ cos^{ -1 }x } \times \frac { -1 }{ \sqrt { 1-x^{ 2 } } } \)
= sin 2x \(-\frac { 1 }{ cos^{ 1 }x\sqrt { 1-x^{ 2 } } } \)
\(\Rightarrow \frac { dy }{ dx } =e^{ sin^{ 2 } }xcos^{ -1 }x\left[ sin2x-\frac { 1 }{ cos^{ -1 }x\sqrt { 1-x^{ 2 } } } \right] \)

We have y=\(f\left( x \right) =\sqrt { x-2 } \) ...(1)
\(f'(x)=\frac { 1 }{ 2\sqrt { x-2 } } \)
Now \(f(a)=(2)=\sqrt { 2-2 } =0\)
\(f(b)=f(3)=\sqrt { 3-2 } =\sqrt { 1 } =1\)
By Lagrange's Theorem, we have:
\(f'(x)=\frac { f(b)-f(a) }{ b-a } \)
\(=\frac { 1 }{ 2\sqrt { x-2 } } =\frac { 1-0 }{ 3-2 } \Rightarrow \frac { 1 }{ 4(x-2) } =1\)
\(1=4x-8\Rightarrow 4x=9\)
\(x=\frac { 9 }{ 4 } \epsilon \left( 2,3 \right) \)
Putting in (1), \(y=\sqrt { \frac { 9 }{ 4 } -2 } =\sqrt { \frac { 1 }{ 4 } } =\frac { 1 }{ 2 } \)
Hence, \(\left( \frac { 9 }{ 4 } ,\frac { 1 }{ 2 } \right) \)is the required point.

\(\Rightarrow \)x²y'+xy'=-y

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

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.

\(y=e^{x^3} \Rightarrow \frac{d y}{d x},=e^{x^3} 3 x^2=3 x^2 e^{x^3}\)

Let g(x) = sin x and h(x) = |x|
\(\therefore goh(x)=g(h(x))=g(|x|)\)
= sin |x| = f(x)
As g(x) and h(x) are everywhere continuous
So f(x) is also everywhere continuous as composition of two continuous functions is also continuous

f(x) = 5x - 3
LHL = \(\underset { x\rightarrow 3^{ - } }{ lim } \left( 5x-3 \right) =5(-3)-3\)
= -15 - 3
= -18
RHL \(\underset { x\rightarrow 3^{ + } }{ lim } \left( 5x-3 \right) =5(-3)-3\)
= -15-3
= - 18
\(\left[ f(x) \right] _{ atx=-3 }=5(-3)-3=-18\)
Hence F(x) is continuous at x = -3

\(y={ tan }^{ -1 }\frac { 3x+2x }{ 1-3x2x } \)
= \({ tan }^{ -1 }3x+{ tan }^{ -1 }2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3 }{ 1+9{ x }^{ 3 } } +\frac { 2 }{ 1+4{ x }^{ 2 } } \)

On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
\(\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

We have, y = tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)
Put x = cos2\(\theta\)
\(\Rightarrow\)2\(\theta\) = cos^-1x
\(\Rightarrow\)\(\theta\) = 1/2cos^-1x
y = tan^-1\(\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
y = \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \)
(\(\because\)cos2\(\theta\) = 2cos²\(\theta\)-1 = 1-2sin²\(\theta\)
y = tan^-1(tan \(\theta\))
y = \(\theta\) = 1/2cos^-1x
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)​​​​​​​

We have y = \(f({ e }^{ { { sin }^{ -1 } } }2x)\)
dy/dx = f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\) x d/dx \(({ e }^{ { { sin }^{ -1 } } }2x)\)
= f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(({ e }^{ { { sin }^{ -1 } } }2x)\)xd/dx(sin^-12x)
= f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(\frac { 1 }{ \sqrt { 1-4{ x }^{ 2 } } } \times 2\)
= \(\frac { 2{ e }^{ sin-1 }2x }{ \sqrt { 1-4{ x }^{ 2 } } } { f }^{ ' }({ e }^{ sin-1 }2x)\)

y = \({ e }^{ { x }^{ 3 } }\)\(\Rightarrow\)dy/dx = \({ e }^{ { x }^{ 3 } }\).3x² = 3x²\({ e }^{ { x }^{ 3 } }\)

Let f : [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers, then there exists some c in (a, b) such that f^'(c) = 0.

h (x) = x f (x) 
\(\Rightarrow \) h^' (x) = x f '(x)+f(x) 
\(\Rightarrow \) h'(0) = 0 f'(0) + f(0) = 0 - 2 = -2

\(=\frac { 1 }{ log\quad a } .\frac { 1 }{ sin\quad x } .cos\ x=\frac { cot\quad x }{ log\quad a } \)

True, converse may not be true, f (x) = |x| is continuous at x = 0, but not differentiable at x = 0.

f(x) = [x]  is not continuous for integers.Hence not continuous at x = ±2, ±1, 0