Differentiation

${\tan ^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {{{\sin \left( {{\pi \over 2} - x} \right)} \over {1 + \cos \left( {{\pi \over 2} - x} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left[ {{{2\sin \left( {{\pi \over 4} - {\pi \over 2}} \right)\cos \left( {{\pi \over 4} - {\pi \over 2}} \right)} \over {2{{\cos }^2}\left( {{\pi \over 4} - {\pi \over 2}} \right)}}} \right]$

$ = {\tan ^{ - 1}}\left[ {\tan \left( {{\pi \over 4} - {\pi \over 2}} \right)} \right]$

${\tan ^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right) = {\pi \over 4} - {\pi \over 2}$

$\therefore$ ${d \over {dx}}\left( {{{\tan }^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right)} \right) = {d \over {dx}}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {{ - 1} \over 2}$

$x^2+y^2=1$

Again differentiating w.r.t. $x$

$2x+2yy'=0$

or $x+yy'=0$

Again differentiating w.r.t. x

$\begin{aligned} & 1+y\left(y^{\prime \prime}\right)+\left(y^{\prime}\right)^2=0 \\ & y\left(y^{\prime \prime}\right)+\left(y^{\prime}\right)^2+1=0 \end{aligned}$

$y=x+\frac{1}{x}$ ..... (i)

Differentiating with respect to x, we get

$y'=1-\frac{1}{x^2}$

or $x^2y'=x^2-1$ .... (ii)

From Eq. (i), we get

$x^2=xy-1$

$\therefore$ From Eq. (ii), we have

$x^2y'=xy-1-1$ or $x^2y'-xy+2=0$

$3 \sin x y+4 \cos x y=5$

Differentiating with respect to $x$, we get

$\begin{aligned} & 3 \cos x y \cdot\left(x \frac{d y}{d x}+y\right)-4 \sin x y\left(x \frac{d y}{d x}+y\right)=0 \\ & \Rightarrow x \frac{d y}{d x}(3 \cos x y-4 \sin x y) \\ & +y(3 \cos x y-4 \sin x y)=0 \\ & \Rightarrow(3 \cos x y-4 \sin x y)\left(x \frac{d y}{d x}+y\right)=0 \\ & \Rightarrow \quad x \frac{d y}{d x}+y=0 \\ & \text {or}\quad\frac{d y}{d x}=\frac{-y}{x} \\ \end{aligned}$

$\begin{aligned} f(x) & =\sqrt{x^2+1} \\ \Rightarrow f^{\prime}(x) & =\frac{2 x}{2 \sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}} \\ h(x) & =2 x-3 \Rightarrow h^{\prime}(x)=2 \end{aligned}$

So, $\quad h^{\prime}\left(g^{\prime}(x)\right)=2$

$\therefore f^{\prime}\left(h^{\prime}\left(g^{\prime}(x)\right)\right)=f^{\prime}(2)=\frac{2}{\sqrt{2^2+1}}=\frac{2}{\sqrt{5}}$

$f(x)=-x^3+4 a x^2+2 x-5$

$f^{\prime}(x)=-3 x^2+8 a x+2$

Discriminant of $f^{\prime}(x)=(8 a)^2-4(-3)(2)$

$=64 a^2+24>0, \forall a \in R$

Hence, $f(x)$ is increasing for every real value of $a$.