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

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

Given, y = tan^-1\(\sqrt { \frac { sinx }{ 1+cosx } } \)
y = tan^-1\(\sqrt { \frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } } \)
y = tan^-1\(\left( \sqrt { tan\frac { x }{ 2 } } \right) \)
y = tan^-1

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.

\(\frac { d }{ dx } \left( { e }^{ m\quad tan^{ -1 } }x \right) ={ e }^{ m\quad tan^{ -1 }x }.\frac { m }{ 1+{ x }^{ 2 } } =\frac { me^{ m\quad tan^{ -1 }x } }{ 1+{ x }^{ 2 } } \)

MTV verified as e^x is continuous and differentiable in [0, 1].
Ans. c = log (e-1)

\(\frac { d }{ dx } \left( { e }^{ \sqrt { x } +3 } \right) ={ e }^{ \sqrt { x } +3 },\frac { 1 }{ 2\sqrt { x } } =\frac { 1 }{ 2\sqrt { x } } { e }^{ \sqrt { x } +3\quad \quad }\)

Function is not defined for x = 5. Hence discontinuous at x = 5.

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

Absolute value function f(x) = Ix - 1I is continuous at x = 1 but not differentiable at x = 1.

For x = -3 function is not defined. Hence, not continuous for x ∈ R.

f( -1 ) = 1 + 5 = 6,
\(\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}\left(x^{2}+5\right)=1+5=6 \)
\(\text { As } \lim _{x \rightarrow-1} f(x)=f(-1)\)
Hence, continuous at x = -1.

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, \(\left( 1-{ x }^{ 2 } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } =2\)

\(x=a\left( cos\quad t+log\quad tan\frac { t }{ 2 } \right) ,\)
\(y=a\quad sin\ t\)
\(\frac { dx }{ dt } =a\left( -sin\quad t+\frac { 1 }{ tan\frac { t }{ 2 } } { sec }^{ 2 }\frac { t }{ 2 } \frac { 1 }{ 2 } \right) \)
\(=a\left( -sin\quad t+\frac { cos\frac { t }{ 2 } }{ sin\frac { t }{ 2 } } \frac { 1 }{ { cos }^{ 2 }\frac { t }{ 2 } } \frac { 1 }{ 2 } \right)\)
\(=a\left( -sin\quad t+\frac { 1 }{ { 2sin\frac { t }{ 2 } cos }\frac { t }{ 2 } } \right) \)
\(=a\left( -sin\quad t+\frac { 1 }{ sin\quad t } \right) \)
\(=a\left( \frac { 1-{ sin }^{ 2 }t }{ sin\quad t } \right) =a\frac { { cos }^{ 2 }t }{ sin\quad t } \)
\(\frac { dy }{ dt } =a\quad cos\ t\)
\(\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =-a\quad sin\ t\)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { a\quad cos\quad t }{ \frac { { acos }^{ 2 }t }{ sin\quad t } } =\frac { sin\quad t }{ cos\quad t } =tan\ t\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ sec }^{ 2 }t\frac { dt }{ dx } \)
\(={ sec }^{ 2 }t\frac { 1 }{ \frac { { acos }^{ 2 }t }{ sin\quad t } } =\frac { 1 }{ a } { sec }^{ 4 }t\ sin\ t\)

\(\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { 1-cos\quad 4x }{ { x }^{ 2 } } } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { 2{ sin }^{ 2 }2x }{ { x }^{ 2 } } } \)
\(=\lim _{ x\rightarrow { 0 }^{ - } }{ { \left( \frac { sin\quad 2x }{ 2x } \right) }^{ 2 }=8{ \left( 1 \right) }^{ 2 }=8 } \)
\(=\lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \sqrt { x } }{ \sqrt { 16+\sqrt { x } } -4 } } \)
\(=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \sqrt { x } \left( \sqrt { 16+\sqrt { x } } +4 \right) }{ 16+\sqrt { x } -16 } } \)
\(=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \sqrt { x } \left( \sqrt { 16+\sqrt { x } } +4 \right) }{ \sqrt { x } } } \)
\(=\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \sqrt { 16+\sqrt { x } } +4 \right) } \)
\(=\sqrt { 16+0 } +4=4+4=8\)
Also f(0)=a
For 'f' to be continuous at x=0,
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =f(0)\)
8 = 8 = a
Hence, a = 8.

\(\lim _{ x\rightarrow { 4 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 4 }^{ - } }{ \frac { X-4 }{ \left| X-4 \right| } +a } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \left( 4-h \right) -4 }{ \left| 4-h-4 \right| } } +a=\lim _{ h\rightarrow 0 }{ \frac { -h }{ \left| -h \right| } } +a\)
\(=\lim _{ h\rightarrow 0 }{ \frac { -h }{ \left| -h \right| } } +a=\lim _{ h\rightarrow 0 }{ -1+a } =-1+a\)
\(\lim _{ x\rightarrow { 4 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 4 }^{ - } }{ \frac { x-4 }{ \left| x-4 \right| } +b } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -h }{ h } } +b=\lim _{ h\rightarrow 0 }{ \frac { -h }{ \left| -h \right| } } +b\)
\(Also\ f(4)=a+b\)
For continuity, \(\lim _{ x\rightarrow { 4 }^{ - } }{ f\left( x \right) = } \lim _{ x\rightarrow { 4 }^{ + } }{ f\left( x \right) = } f(4)\)
-1 + a = 1 + b = a + b
Taking first and last, -1 + a = a + b
Taking last two, 1 + b = a + b
1 = a
a = 1
Hence, a = 1, b = -1.

Let \(y=(\sin x)^x+\sin ^{-1} \sqrt{x}\)
Let \(u=(\sin x)^x \ v=\sin ^{-1} \sqrt{x}\)
y=u+v
Differentiating both sides w.r.t. x.
\( \frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)

\(\lim _{ x\rightarrow -\frac { \pi }{ 6 } }{ f\left( x \right) =\lim _{ x\rightarrow -\frac { \pi }{ 6 } }{ \frac { 2sin\left( x+\frac { \pi }{ 6 } \right) }{ x+\frac { \pi }{ 6 } } } } \)= 2
\(f\left( -\frac { \pi  }{ 6 }  \right) =k\)
For the continuity of f(x) at x = -\(\pi/6\)
\(f\left( -\frac { \pi }{ 6 } \right) =\lim _{ x\rightarrow -\frac { \pi }{ 6 } }{ f\left( x \right) } \)
k = 2