$$\begin{aligned} \text{Let}\quad y & =(2 x-7)^2(3 x+5)^3 \\ \frac{d y}{d x} & =(2 x-7)^2 \frac{d}{d x}(3 x+5)^3+(3 x+5)^3 \frac{d}{d x}(2 x-7)^2 \quad \text{[by product rule]}\\ & =(2 x-7)^2(3)(3 x+5)^2(3)+(3 x+5)^3 2(2 x-7)(2) \quad \text{[by chain rule]}\\ & =9(2 x-7)^2(3 x+5)^2+4(3 x+5)^3(2 x-7) \\ & =(2 x-7)(3 x+5)^2[9(2 x-7)+4(3 x+5)] \\ & =(2 x-7)(3 x+5)^2(18 x-63+12 x+20) \\ & =(2 x-7)(3 x+5)^2(30 x-43) \end{aligned}$$

Let $$ \begin{aligned} y & =x^2 \sin x+\cos 2 x \\ \frac{d y}{d x} & =\frac{d}{d x}\left(x^2 \sin x\right)+\frac{d}{d x} \cos 2 x \\ & =x^2 \cdot \cos x+\sin x 2 x+(-\sin 2 x) \cdot 2 \quad \text{[by product rule] }\\ & =x^2 \cos x+2 x \sin x-2 \sin 2 x\quad \text{[by chain urle]} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad y & =\sin ^3 x \cos ^3 x \\ \therefore\quad \frac{d y}{d x} & =\sin ^3 x \cdot \frac{d}{d x} \cos ^3 x+\cos ^3 x \frac{d}{d x} \sin ^3 x \quad \text{[by product rule] }\\ & =\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cos x \quad \text{[by chain rule] }\\ & =-3 \cos ^2 x \sin ^4 x+3 \sin ^2 x \cos ^4 x \\ & =3 \sin ^2 x \cos ^2 x\left(\cos ^2 x-\sin ^2 x\right) \\ & =3 \sin ^2 x \cos ^2 x \cos 2 x \\ & =\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x \\ & =\frac{3}{4} \sin ^2 2 x \cos 2 x \end{aligned}$$

$$\begin{aligned} \text{Let}\quad y & =\frac{1}{a x^2+b x+c}=\left(a x^2+b x+c\right)^{-1} \\ \therefore\quad \frac{d y}{d x} & =-\left(a x^2+b x+c\right)^{-2}(2 a x+b) \quad \text{[by chain rule]}\\ & =\frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2} \end{aligned} $$

Let $\quad f(x)=\cos \left(x^2+1\right)$ and $f(x+h)=\cos \left\{(x+h)^2+1\right\}$

$\therefore \quad \frac{d}{d x} f(x)=\lim _\limits{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos \left\{(x+h)^2+1\right\}-\cos \left(x^2+1\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \left\{\frac{(x+h)^2+1+x^2+1}{2}\right\} \sin \left\{\frac{(x+h)^2+1-x^2-1}{2}\right\}}{h} \\ & {\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right]} \end{aligned}$$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{(x+h)^2-x^2}{2}\right\}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{x^2+h^2+2 x h-x^2}{2}\right\}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{h^2+2 h x}{2}\right\}\right] \\ & =-2 \lim _{h \rightarrow 0} \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \lim _{h \rightarrow 0}\left\{\frac{\sin h\left(\frac{h+2 x}{2}\right)}{h\left(\frac{h+2 x}{2}\right)}\right\} \times\left(\frac{h+2 x}{2}\right) \\ & =-2 \lim _{h \rightarrow 0} \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \lim _{h \rightarrow 0}\left(\frac{h+2 x}{2}\right)\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ & =-2 x \sin \left(x^2+1\right) \quad {[\because x \rightarrow 0 \Rightarrow k x \rightarrow 0]}\\ \end{aligned}$$