Question:
$\sin ^{3} x+\cos ^{6} x$
Solution:
Let $y=\sin ^{3} x+\cos ^{6} x$
$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{3} x\right)+\frac{d}{d x}\left(\cos ^{6} x\right)$
$=3 \sin ^{2} x \cdot \frac{d}{d x}(\sin x)+6 \cos ^{5} x \cdot \frac{d}{d x}(\cos x)$
$=3 \sin ^{2} x \cdot \cos x+6 \cos ^{5} x \cdot(-\sin x)$
$=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$