Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\frac{d y}{d x}=\frac{d}{d x}\left[(1+4 x)^5\left(3+x-x^2\right)^8\right]$

$\begin{aligned} & =(1+4 x)^5 \cdot \frac{d}{d x}\left(3+x-x^2\right)^8+\left(3+x-x^2\right)^8 \cdot \frac{d}{d x}(1+4 x)^5 \\ & =(1+4 x)^5 \times 8\left(3+x-x^2\right)^7 \cdot \frac{d}{d x}\left(3+x-x^2\right)+\left(3+x-x^2\right)^8 \times 5(1+4 x)^4 \cdot \frac{d}{d x}(1+4 x) \\ = & 8(1+4 x)^5\left(3+x-x^2\right)^7 \cdot(0+1-2 x)+5(1+4 x)^4\left(3+x-x^2\right)^8 \cdot(0+4 \times 1) \\ = & 8(1-2 x)(1+4 x)^5\left(3+x-x^2\right)^7+20(1+4 x)^4\left(3+x-x^2\right)^8 .\end{aligned}$