Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\left( x ^2+4 x +1\right)^3+\left( x ^3-5 x -2\right)^4\right] \\ & =\frac{d}{d x}=\left( x ^2+4 x +1\right)^3+\frac{d}{d x}\left( x ^3-5 x -2\right)^4 \\ & =3\left( x ^2+4 x +1\right)^2 \cdot \frac{d}{d x}\left( x ^2+4 x +1\right)+4\left( x ^3-5 x -2\right)^4 \cdot \frac{d}{d x}\left( x ^3-5 x -2\right) \\ & =3\left( x ^2+4 x +1\right)^3 \cdot(2 x +4 \times 1+0)+4\left( x ^3-5 x -2\right)^3 \cdot\left(3 x ^2-5 \times 1-0\right) \\ & =6( x +2)\left( x ^2+4 x +1\right)^2+4\left(3 x ^2-5\right)\left( x ^3-5 x -2\right)^3 .\end{aligned}$