Question 14

Maths Question: 

$\text{Differentiate }y=8{{x}^{2}}(1+\sin x)(1+\cos x)\text{ wrt }x$

Maths Solution: 

$\begin{align}  & y=8{{x}^{2}}(1+\sin x)(1+\cos x) \\ & \frac{dy}{dx}=8{{x}^{2}}(1+\sin x)\frac{d}{dx}(1+\cos x)+8{{x}^{2}}(1+\cos x)\frac{d}{dx}(1+\sin x)+(1+\sin x)(1+\cos x)\frac{d}{dx}8{{x}^{2}} \\ & \frac{dy}{dx}=8{{x}^{2}}(1+\sin x)(-\sin x)+8{{x}^{2}}(1+\cos x)(\cos x)+(1+\sin x)(1+\cos x)(16x) \\ & \frac{dy}{dx}=-8{{x}^{2}}\sin x(1+\sin x)+8{{x}^{2}}(\cos x+{{\cos }^{2}}x)+16x(1+\sin x)(1+\cos x) \\\end{align}$

University mathstopic: