Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^3\right] \\ & =\left(1+\sin ^2 x\right)^2 \cdot \frac{d}{d x}\left(1+\cos ^2 x\right)^3+\left(1+\cos ^2 x\right)^3 \frac{d}{d x}\left(1+\sin ^2 x\right)^2 \\ & =\left(1+\sin ^2 x\right)^2 \times 3\left(1+\cos ^2 x\right)^2 \cdot \frac{d}{d x}\left(1+\cos ^2 x\right)+\left(1+\cos ^2 x\right)^3 \times 2\left(1+\sin ^2 x\right) \cdot \frac{d}{d x}\left(1+\sin ^2 x\right) \\ & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2 \cdot\left[0+2 \cos x \cdot \frac{d}{d x}(\cos x)\right]+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3 \cdot\left[0+2 \sin x \cdot \frac{d}{d x}(\sin x)\right]\end{aligned}$

$\begin{aligned} & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2 \cdot[2 \cos x(-\sin x)]+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3 \cdot[2 \sin x-\cos x] \\ & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2(-\sin 2 x)+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3(\sin 2 x) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left[-3\left(1+\sin ^2 x\right)+2\left(1+\cos ^2 x\right)\right] \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(-3-3 \sin ^2 x+2+2 \cos ^2 x\right) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left[-1-3 \sin ^2 x+2\left(1-\sin ^2 x\right)\right] \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(-1-3 \sin ^2 x+2-2 \sin ^2 x\right) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(1-5 \sin ^2 x\right) .\end{aligned}$