Let $$y=\sin \sqrt{x}+(\cos \sqrt{x})^2$$

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

Let $y=\sin ^n\left(a x^2+b x+c\right)$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d}{d x}\left[\sin \left(a x^2+b x+c\right)\right]^n \\ & =n \cdot\left[\sin \left(a x^2+b x+c\right)\right]^{n-1} \cdot \frac{d}{d x} \sin \left(a x^2+b x+c\right) \\ & =n \cdot \sin ^{n-1}\left(a x^2+b x+c\right) \cdot \cos \left(a x^2+b x+c\right) \cdot \frac{d}{d x}\left(a x^2+b x+c\right) \\ & =n \cdot \sin ^{n-1}\left(a x^2+b x+c\right) \cdot \cos \left(a x^2+b x+c\right) \cdot(2 a x+b) \\ & =n \cdot(2 a x+b) \cdot \sin ^{n-1}\left(a x^2+b x+c\right) \cdot \cos \left(a x^2+b x+c\right) \end{aligned}$$

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

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

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

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

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