(2x – 7)2 (3x + 5)3

Given $y=(2 x-7)^{2}(3 x+5)^{3}$

Applying product rule of differentiation that is

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{t} \cdot \mathrm{y})=\mathrm{y} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow y=(2 x-7)^{2}(3 x+5)^{3}$

$\Rightarrow \frac{d y}{d x}=(3 x+5)^{3} \frac{d}{d x}(2 x-7)^{2}+(2 x-7)^{2} \frac{d}{d x}(3 x+5)^{3}$

On differentiating we get

$\Rightarrow \frac{d y}{d x}=(2)(3 x+5)^{3} 2(2 x-7)^{1}+(3)(2 x-7)^{2} 3(3 x+5)^{2}$

$\Rightarrow \frac{d y}{d x}=4(3 x+5)^{3}(2 x-7)+9(2 x-7)^{2}(3 x+5)^{2}$

On simplification we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(2 \mathrm{x}-7)(3 \mathrm{x}+5)^{2}[4(3 \mathrm{x}+5)+9(2 \mathrm{x}-7)]$

$\Rightarrow \frac{d y}{d x}=(2 x-7)(3 x+5)^{2}(30 x-43)$