f'(x) = 4x + 3
f'(0) = 3
f'(-1) = -1
\(\therefore\) f'(0) + 3f(-1) = 3 + 3(-1) = 3 - 3 = 0

\(Let\quad u=\log { (1+{ x }^{ 2 } } )\quad and\quad v=\tan ^{ -1 }{ x } \)
\( \Rightarrow \frac { du }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } .\frac { d }{ dx } (1+{ x }^{ 2 })\quad \frac { dv }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\frac { du }{ dx } =\frac { 2x }{ 1+{ x }^{ 2 } } \)

\(\therefore \frac { du }{ dv } =2x\)

\(x=a\sec ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dx }{ d\theta } =3a\sec ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\sec { \theta } )=3a\sec ^{ 2 }{ \theta } .\sec { \theta } \tan { \theta } =3a\sec ^{ 3 }{ \theta } \tan { \theta } \)
\(y=a\tan ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dy }{ d\theta } =a.3\tan ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\tan { \theta } )=3a\tan ^{ 2 }{ \theta } .\sec ^{ 2 }{ \theta } \)
\(\frac { dy }{ dx } =\frac { dy }{ d\theta } /\frac { dx }{ d\theta } =\frac { 3a\tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } }{ 3a\sec ^{ 3 }{ \theta } \tan { \theta } } =\frac { \tan { \theta } }{ \sec { \theta } } =\frac { \sin { \theta } }{ \cos { \theta \times \frac { 1 }{ \cos { \theta } } } } =\sin { \theta } \)
\(\therefore \frac { dy }{ dx } \quad at\quad \theta =\frac { \pi }{ 3 } =\sin { \frac { \pi }{ 3 } } =\frac { \sqrt { 3 } }{ 2 } \)

\({ x }^{ 2 }+2xy+{ y }^{ 3 }=42\)
Differentiating both sides with respect to 'x' we get,
\(2x+2\left[ x.\frac { dy }{ dx } +y(1) \right] +3{ y }^{ 2 }\frac { dy }{ dx } =0 \Rightarrow 2x+2x\frac { dy }{ dx } +2y+3{ y }^{ 2 }\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } (2x+3{ y }^{ 2 })=-2x-2y \Rightarrow \frac { dy }{ dx } =\frac { -2\left( x+y \right) }{ 2x+3{ y }^{ 2 } } \)

Let y = \(\sec ^{ -1 }{ \left( \frac { 1 }{ 2{ x }^{ 2 }-1 } \right) } \)
Putting x = \(\cos { \theta } ,\quad we\quad get\quad \theta =\cos ^{ -1 }{ x } \)
\(\therefore y=\sec ^{ -1 }{ \left( \frac { 1 }{ 2\cos ^{ 2 }{ \theta } -1 } \right) } =\sec ^{ -1 }{ \left( \frac { 1 }{ \cos { 2\theta } } \right) } =\sec ^{ -1 }{ \left( \sec { 2\theta } \right) } \)
\(\Rightarrow y=2\theta \)
\(\Rightarrow y=2\cos ^{ -1 }{ x } \)
Differentiating with respect to 'x' both sides we get,
\(\frac { dy }{ dx } =2\left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -2 }{ \sqrt { 1-{ x }^{ 2 } } } \)

Given xy = 4
Differentiating both sides with respect to 'x' we get,
\(x.\frac { dy }{ dx } =y(1)=0\quad \Rightarrow x\frac { dy }{ dx } =-y ...(1)\)
\(LHS= x\left( \frac { dy }{ dx } +{ y }^{ 2 } \right) =x\frac { dy }{ dx } +x{ y }^{ 2 }= -y+(xy)y=-y+4y\quad \left[ \because \quad xy=4 \right] \)
\(=3y=RHS\)
Hence proved

Given \(y=\sqrt { x+1 } +\sqrt { x-1 } ={ \left( x+1 \right) }^{ 1/2 }+{ \left( x-1 \right) }^{ 1/2 }\)
\( \therefore \frac { dy }{ dx } =\frac { 1 }{ 2 } { \left( x+1 \right) }^{ 1/2-1 }+\frac { 1 }{ 2 } { \left( x-1 \right) }^{ 1/2-1 }=\frac { 1 }{ 2 } { \left( x+1 \right) }^{ -1/2 }+\frac { 1 }{ 2 } { \left( x-1 \right) }^{ -1/2 }\)
\(=\frac { 1 }{ 2\sqrt { x+1 } } +\frac { 1 }{ 2\sqrt { x-1 } } =\frac { \sqrt { x-1 } +\sqrt { x+1 } }{ 2\sqrt { (x+1)(x-1) } } =\frac { y }{ 2\sqrt { { x }^{ 2 }+1 } } \Rightarrow \sqrt { { x }^{ 2 }+1 } .\frac { dy }{ dx } =\frac { y }{ 2 } \)
Hence proved

Let \(f\left( x \right) ={ e }^{ 2x }\)
Then \(f\left( x+h \right) ={ e }^{ 2x+h }\)
\(\therefore \frac { d }{ dx } (f\left( x \right) )=\lim _{ h\rightarrow 0 }{ \frac { f\left( x+h \right) -f\left( x \right) }{ h } } =\lim _{ h\rightarrow 0 }{ \frac { { e }^{ 2\left( x+h \right) }-{ e }^{ 2x } }{ h } = } \lim _{ h\rightarrow 0 }{ \frac { { e }^{ 2x }.{ e }^{ h }-{ e }^{ 2x } }{ h } }\)
\( =2{ e }^{ 2x }\lim _{ 2h\rightarrow 0 }{ \frac { { e }^{ 2h }-1 }{ 2h } } =2{ e }^{ x }.\lim _{ y\rightarrow 0 }{ \frac { { e }^{ y }-1 }{ y } } where\quad y=2h\)
\( \frac { d }{ dx } (f\left( x \right) )=2{ e }^{ x }\times 1\quad \left[ \because \lim _{ y\rightarrow 0 }{ \frac { { e }^{ y }-1 }{ y } =1 } \right] \)
\(\frac { d }{ dx } ({ e }^{ 2x })={ 2e }^{ 2x }\)

\(f^{ ' }\left( 1^{ - } \right) =\lim _{ x\rightarrow 1^{ - } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ (x+1) } =1+1=2\quad ...(1)\)
\(f^{ ' }\left( 1^{ + } \right) =\lim _{ x\rightarrow 1^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ x+1 } =1+1=2 ...(2)\)
From (1)and (2), \(f^{ ' }\left( 1^{ - } \right) =f^{ ' }\left( 1^{ + } \right) \)
\(\therefore f\left( x \right) \)is differentiable at x = 1 and \(f^{ ' }\left( 1 \right) =2\)

\(f^{ ' }\left( { 2 }^{ - } \right) =\lim _{ x\rightarrow { 2 }^{ - } }{ \frac { f\left( x \right) -f\left( 2 \right) }{ x-2 } } =\lim _{ x\rightarrow { 2 }^{ - } }{ \frac { (x-1)-(1) }{ x-2 } } \quad \quad \left[ \because \quad f\left( x \right) =2x-3;f\left( 2 \right) =4-3=1 \right] \)
\(=\lim _{ x\rightarrow { 2 }^{ - } }{ \frac { x-2 }{ x-2 } } =1 ...(1)\)
\(f^{ ' }\left( { 2 }^{ + } \right) =\lim _{ x\rightarrow { 2 }^{ + } }{ \frac { f\left( x \right) -f\left( 2 \right) }{ x-2 } } =\lim _{ x\rightarrow { 2 }^{ + } }{ \frac { 2x-3-1 }{ x-2 } } =\lim _{ x\rightarrow { 2 }^{ + } }{ \frac { 2x-4 }{ x-2 } } =\lim _{ x\rightarrow { 2 }^{ + } }{ \frac { 2(x-2) }{ x-2 } } =2\quad ...(2)\)
\(From(1)and(2),f^{ ' }\left( { 2 }^{ - } \right) \neq f^{ ' }\left( { 2 }^{ + } \right) \)
\( \therefore f(x)is\ not\ differentiable\quad at\quad x=2.\)