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

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

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

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

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

\(We\quad have:\quad y=3{ e }^{ 2x }+{ 2e }^{ 3x }\)
\(\frac { dy }{ dx } =6{ e }^{ 2x }+6{ e }^{ 3x }=6({ e }^{ 2x }+{ e }^{ 3x })\)
\(and\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =12{ e }^{ 2x }+18{ e }^{ 3x }=6(2{ e }^{ 2x }+3{ e }^{ 3x })\)
\(Now\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ { dx }^{ 2 } } +6y\)
\(=6(2{ e }^{ 2x }+3{ e }^{ 3x })-5.6({ e }^{ 2x }+{ e }^{ 3x })+6(3{ e }^{ 2x }+{ 2e }^{ 3x })\)

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

Chain rule may be extended as follows. Suppose f is a real valued function which is a composite of three functions u, v and w; i.e.,
f = (w o u) o v. If t = v (x) and s = u (t), then
\(\frac{d f}{d x}=\frac{d(w \circ u)}{d t} \cdot \frac{d t}{d x}=\frac{d w \mid}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x}\)
provided all the derivatives in the statement exist. Reader is invited to formulate chain rule for composite of more functions.

Given equations are
\(x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Now, differentiate both w.r.t t
We get,
\(\frac{d x}{d t}=\frac{d\left(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\right)}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\sin ^3 t\right)}{d t}-\sin ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \sin ^2 t \cos t \cdot \sqrt{\cos 2 t}-\sin ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}\)
\(=\frac{3 \sin ^2 t \cos t \cdot \cos 2 t+\sin ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\(=\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}} \quad\left(\because \frac{\cos x}{\sin x}=\cot x\right)\)
Similarly,
\(\frac{d y}{d t}=\frac{d\left(\frac{\cos ^3 t}{\sqrt{\cos 2 t})}\right.}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\cos ^3 t\right)}{d t}-\cos ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \cos ^2 t(-\sin t) \cdot \sqrt{\cos 2 t}-\cos ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{(\sqrt{\cos 2 t})^2}\)
\(=\frac{-3 \cos ^2 t \sin t \cos 2 t+\cos ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\( =\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}} \)
\( \text { Now, } \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}}}=\frac{\cot ^3 t(1-3 \tan t \cot 2 t)}{(3 \cot t \cot 2 t+1)} \)
\(=\frac{\cos ^3 t\left(1-3 \cdot \frac{\sin t}{\cos t} \cdot \frac{\cos 2 t}{\sin 2 t}\right)}{\sin ^3 t\left(3 \cdot \frac{\cos t}{\sin t} \cdot \frac{\cos 2 t}{\sin 2 t}+1\right)}=\frac{\cos ^2 t(\cos t \sin 2 t-3 \sin t \cos 2 t)}{\sin ^2 t(3 \cos t \cos 2 t+\sin t \sin 2 t)} \)
\( =\frac{\cos ^2 t\left(\cos t \cdot 2 \sin t \cos t-3 \sin t\left(2 \cos ^2 t-1\right)\right)}{\sin ^2 t\left(3 \cos t\left(1-2 \sin ^2 2 t\right)+\sin t \cdot 2 \sin t \cos t\right)} \)
\( \left(\because \sin 2 x=2 \sin x \cos x \text { and } \cos 2 x=2 \cos ^2 x-1 \text { and } \cos 2 x=1-2 \sin ^2 x\right) \)
\( =\frac{\cos ^2 t\left(2 \sin t \cos ^2 t-6 \sin t \cos ^2 t+3 \sin t\right)}{\sin ^2 t\left(3 \cos t-6 \cos t \sin ^2 t+2 \sin ^2 \cos t\right)} \)
\( =\frac{\sin t \cos t\left(-4 \cos ^3 t+3 \cos t\right)}{\sin t \cos t\left(3 \sin t-4 \sin ^3 t\right)} \)
\( \frac{d y}{d x}=\frac{-4 \cos ^3 t+3 \cos t}{3 \sin t-4 \sin ^3 t}=\frac{-\cos 3 t}{\sin 3 t}=-\cot 3 t \)
\( \left(\because \sin 3 t=3 \sin t-4 \sin ^3 t \text { and } \cos 3 t=4 \cos ^3 t-3 \cos t\right) \)
Therefore, the answer is \(\frac{d y}{d x}=-\cot 3 t\)

The given relationship is \(f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\)
Taking logarithm on both the sides, we obtain
\(\log f(x)=\log (1+x)+\log \left(1+x^{2}\right)+\log \left(1+x^{4}\right)+\log \left(1+x^{8}\right)\)
Differentiating both sides with respect to x, we obtain
\(\frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log \left(1+x^{2}\right)+\frac{d}{d x} \log \left(1+x^{4}\right)+\frac{d}{d x} \log \left(1+x^{8}\right) \)
\(\Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)=\frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^{2}} \cdot \frac{d}{d x}\left(1+x^{2}\right)+\frac{1}{1+x^{4}} \cdot \frac{d}{d x}\left(1+x^{4}\right)+\frac{1}{1+x^{8}} \cdot \frac{d}{d x}\left(1+x^{8}\right) \)
\(\Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{1+x}+\frac{1}{1+x^{2}} \cdot 2 x+\frac{1}{1+x^{4}} \cdot 4 x^{3}+\frac{1}{1+x^{8}} \cdot 8 x^{7}\right] \)
\(\therefore f^{\prime}(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}\right] \)
\(\text { Hence, } f^{\prime}(1) =(1+1)\left(1+1^{2}\right)\left(1+1^{4}\right)\left(1+1^{8}\right)\left[\frac{1}{1+1}+\frac{2 \times 1}{1+1^{2}}+\frac{4 \times 1^{3}}{1+1^{4}}+\frac{8 \times 1^{7}}{1+1^{8}}\right] \)
\(=2 \times 2 \times 2 \times 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right] \)
\(=16 \times\left(\frac{1+2+4+8}{2}\right) \)
\(=16 \times \frac{15}{2}=120 \)
f'(1) = 120

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.

Let y = a^x_.
Taking log on both sides, we get log y = x log a
On differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\log a \Rightarrow \frac{d y}{d x}=y \log a\) 
Thus,\(\frac{d}{d x}\left(a^{x}\right)=a^{x} \log a\)

Let g (x) = IxI and h(x) = sin x
Now,\(\operatorname{hog}(x)=h[g(x)]=h(|x|)=\sin |x|=f(x)\) 
Since, g (x) and h( x) are both continuous function for all x∈ R,
so composition of g (x) and h( x) is also acontinuous function for all x ∈ R.
Hence, f(x) = sin [x] is a continuous function

For x = 0,
f(x) is continuous and
at x = 1, LHL = 1 and RHL = 5,
hence discontinuous and
at x = 2;
f(x) = 5 which is continuous

As \(f(x)=\frac { log(logx) }{ 2logx } \)
\(\Rightarrow { f }^{ ' }(x)=\frac { 1 }{ 2 } \left[ \frac { logx.\frac { 1 }{ logx } .\frac { 1 }{ x } -log(logx).\frac { 1 }{ x } }{ { (logx) }^{ 2 } } \right] \)
\(\Rightarrow { f }^{ ' }(x)=\frac { 1 }{ 2 } \left[ \frac { 1-log(logx) }{ x{ (logx) }^{ 2 } } \right] \)
\(\Rightarrow { f }^{ ' }(e)=\frac { 1 }{ 2 } \left[ \frac { 1-log(loge) }{ x{ e(loge) }^{ 2 } } \right] \)=\(\frac{1}{2e}\)

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

As f(x) = |x – a| is continuous at x = a but not differentiable thereat.

As differentiable functions is continuous also

As \(\lim _{ x\rightarrow 2 }{ \frac { sin({ e }^{ x-2 }-1) }{ log(x-1) } } =\lim _{ h\rightarrow 0 }{ \frac { sin({ e }^{ h }-1) }{ log(1+h) } } \)
On substituting h = x - 2
= \(\lim _{ h\rightarrow 0 }{ \frac { sin({ e }^{ h }-1) }{ log(1+h) } } .\frac { { e }^{ h }-1 }{ h } \).\(\frac { h }{ log(1+h) } \)
= 1.1.1
= 1 nad f(2) = k

As \(\lim _{ x\rightarrow 0 }{ \frac { \sqrt { \frac { 1 }{ 2 } (1-cosx) } }{ x } } \)
 \(=\lim _{ x\rightarrow 0 }{ \frac { |sin \ x| }{ x } } \)
and LHL ≠ RHL at x = 0

As f(2) is not defined so / is not continuous at x = 2 ‘g’ is a polynomial function, so continuous at x = 2.