$$\begin{aligned} \text { Let } \quad y & =\cos (\tan \sqrt{x+1}) \\ \therefore \quad \frac{d y}{d x} & =\frac{d}{d x} \cos (\tan \sqrt{x+1})=-\sin (\tan \sqrt{x+1}) \cdot \frac{d}{d x}(\tan \sqrt{x+1}) \\ & =-\sin (\tan \sqrt{x+1}) \cdot \sec ^2 \sqrt{x+1} \cdot \frac{d}{d x}(x+1)^{1 / 2} \quad\left[\because \frac{d}{d x}(\tan x)=\sec ^2 x\right] \\ & =-\sin (\tan \sqrt{x+1}) \cdot(\sec \sqrt{x+1})^2 \cdot \frac{1}{2}(x+1)^{-1 / 2} \cdot \frac{d}{d x}(x+1) \\ & =\frac{-1}{2 \sqrt{x+1}} \cdot \sin (\tan \sqrt{x+1}) \cdot \sec ^2(\sqrt{x+1}) \end{aligned}$$

Let $\quad y=\sin x^2+\sin ^2 x+\sin ^2\left(x^2\right)$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d}{d x} \sin \left(x^2\right)+\frac{d}{d x}(\sin x)^2+\frac{d}{d x}\left(\sin x^2\right)^2 \\ & =\cos \left(x^2\right) \frac{d}{d x}\left(x^2\right)+2 \sin x \cdot \frac{d}{d x} \sin x+2 \sin x^2 \cdot \frac{d}{d x} \sin x^2 \\ & =\cos x^2 2 x+2 \cdot \sin x \cdot \cos x+2 \sin x^2 \cos x^2 \cdot \frac{d}{d x} x^2 \\ & =2 x \cos (x)^2+2 \cdot \sin x \cdot \cos x+2 \sin x^2 \cdot \cos x^2 \cdot 2 x \\ & =2 x \cos (x)^2+\sin 2 x+\sin 2(x)^2 \cdot 2 x \\ & =2 x \cos \left(x^2\right)+2 x \cdot \sin 2\left(x^2\right)+\sin 2 x \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad y & =\sin ^{-1} \frac{1}{\sqrt{x+1}} \\ \therefore\quad\frac{d y}{d x} & =\frac{d}{d x} \sin ^{-1} \frac{1}{\sqrt{x+1}} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^2}} \cdot \frac{d}{d x} \frac{1}{(x+1)^{1 / 2}} \quad\left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}\right] \\ & =\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \cdot \frac{d}{d x} \cdot(x+1)^{-1 / 2} \\ & =\sqrt{\frac{x+1}{x}} \cdot \frac{-1}{2}(x+1)^{-\frac{1}{2}-1} \cdot \frac{d}{d x}(x+1) \\ & =\frac{(x+1)^{1 / 2}}{x^{1 / 2}} \cdot\left(-\frac{1}{2}\right)(x+1)^{-3 / 2}=\frac{-1}{2 \sqrt{x}} \cdot\left(\frac{1}{x+1}\right) \end{aligned}$$

Let $\quad y= (\sin x)^{\cos x}$

$$\begin{array}{ll} \Rightarrow & \log y=\log (\sin x)^{\cos x}=\cos x \log \sin x \\ \therefore & \frac{d}{d y} \log y \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x \cdot \log \sin x) \end{array}$$

$$\begin{aligned} \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\cos x \cdot \frac{d}{d x} \log \sin x+\log \sin x \cdot \frac{d}{d x} \cos x \\ & =\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x} \sin x+\log \sin x \cdot(-\sin x) \end{aligned}$$

$$ \begin{aligned} & =\cot x \cdot \cos x-\log (\sin x) \cdot \sin x \quad \left[\because \cot x=\frac{\cos x}{\sin x}\right]\\ \therefore\quad \frac{d y}{d x} & =y\left[\frac{\cos ^2 x}{\sin x}-\sin x \cdot \log (\sin x)\right] \\ & =\sin x^{\cos x}\left[\frac{\cos ^2 x}{\sin x}-\sin x \cdot \log (\sin x)\right] \end{aligned}$$

Let, $y=\sin ^m x \cdot \cos ^n x$

$\therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left[(\sin x)^m \cdot(\cos x)^n\right]$

$$\begin{aligned} & =(\sin x)^m \cdot \frac{d}{d x}(\cos x)^n+(\cos x)^n \cdot \frac{d}{d x}(\sin x)^m \\ & =(\sin x)^m \cdot n(\cos x)^{n-1} \cdot \frac{d}{d x} \cos x+(\cos x)^n m(\sin x)^{m-1} \cdot \frac{d}{d x} \sin x \\ & =(\sin x)^m \cdot n(\cos x)^{n-1}(-\sin x)+(\cos x)^n \cdot m(\sin x)^{m-1} \cos x \\ & =-n \sin ^m x \cdot \cos ^{n-1} x \cdot(\sin x)+m \cos ^n x \cdot \sin ^{m-1} x \cdot \cos x \\ & =-n \cdot \sin ^m x \cdot \sin x \cdot \cos ^n x \cdot \frac{1}{\cos x}+m \cdot \sin ^m x \cdot \frac{1}{\sin x} \cdot \cos ^n x \cdot \cos x \\ & =-n \cdot \sin ^m x \cdot \cos ^n x \cdot \tan x+m \sin ^m x \cdot \cos ^n x \cdot \cot x \\ & =\sin ^m x \cdot \cos ^n x[-n \tan x+m \cot x] \end{aligned}$$