y = tan 3x
Take u = 3x ⇒ \(\frac{d u}{d x}=3\)
\(y=\tan u\)
\( \frac{d y}{d x} =\frac{d y}{d u} \times \frac{d u}{d x}=\sec ^2 u(3)=\sec ^2(3 x) \cdot 3 \)
\(=3 \sec ^2(3 x)\)

Let y = 2^x = e^xlog2.
Take u = (log 2)x so that
y = e^u
\({dy\over dx}={dy\over du}\times {du\over dx}=e^u \times log2=e^{xlog 2}\)
= (log2)2^x.

Take u = sin x so that
y = e^u
\({dy\over dx}={d(e^u)\over du}\times {du\over dx}=e^x \times cos \ x=cos \ x e^{sin \ x}\)

Given g(t) = 4 sec t + tan t
\(g^{\prime}(t)=4 \frac{d}{d t}(\sec t)+\frac{d}{d t}(\tan t)\)
= 4 sec t tan t + sec² t

f(x) = x sin x
f'(x) = \(x\frac { d }{ dx } (sinx)+sinx\frac { d }{ dx } (x)\)
= x cosx + sinx(1) = x cosx + sinx

\({dy\over dx}=3x^2+10x+3.\)

The greatest integer function \(f(x)=\lfloor x\rfloor\) is not continuous at every integer point n, since \(\left.\lim _{x \rightarrow n^{-}} \mid x\right\rfloor=n-1\) and \(\lim _{x \rightarrow n^{+}}\lfloor x\rfloor=n .\) Thus f'(n) does not exist.

Given y = sin^-1x
y' = \(\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } =(1-x^{ 2 })^{ -\frac { 1 }{ 2 } }\)
 y' '= \(-\frac { 1 }{ 2 } (1-{ x }^{ 2 })^{ \frac { 1 }{ 2 } -1 }\)(-2x)
= x(1 - x²)^-3/2 = \(\frac { x }{ (1-x^{ 2 })^{ \frac { 3 }{ 2 } } } \)

y = Cosec x . cot x
\(y^{\prime}=\operatorname{cosec} x \frac{d}{d x}(\cot x)+\cot x \frac{d}{d x}(\operatorname{cosec} x)\)
\(=\operatorname{cosec} x\left(-\operatorname{cosec}^2 x\right)+\cot x(-\operatorname{cosec} x \cot x)\)
\(=-\operatorname{cosec}^3 x-\operatorname{cosec} x \cot ^2 x\)
\(=\frac{1}{\sin ^3 x}-\frac{1}{\sin x} \cdot \frac{\cos ^2 x}{\sin ^2 x}\)
\(=\frac{-1}{\sin ^3 x}-\frac{\cos ^2 x}{\sin ^3 x}=\frac{-1-\cos ^2 x}{\sin ^3 x}\)

\(=\frac{-\left(1+\cos ^2 x\right)}{\sin ^3 x}\)

y = e^x sin^x
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^x\right) \sin x_1+e^x \frac{d}{d x}(\sin x)\)
\(=e^x \sin x+e^x \cos x\)
\(=e^x(\sin x+\cos x)\)