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

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

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

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

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