find the


$\sin ^{3} x+\cos ^{6} x$


Let $y=\sin ^{3} x+\cos ^{6} x$

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{3} x\right)+\frac{d}{d x}\left(\cos ^{6} x\right)$

$=3 \sin ^{2} x \cdot \frac{d}{d x}(\sin x)+6 \cos ^{5} x \cdot \frac{d}{d x}(\cos x)$

$=3 \sin ^{2} x \cdot \cos x+6 \cos ^{5} x \cdot(-\sin x)$

$=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$

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