Differentiate the function ^ 3 x + ^6 x, w.r.t to x. - Vidyadip

Question

Differentiate the function $\sin^{3 }x + \cos^6 x,$ w.r.t to x.

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Answer

Let $y = \sin^3 x + \cos^6 x$
Differentiating both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{3} x\right)+\frac{d}{d x}\left(\cos ^{6} x\right)$
$= 3 \sin ^{2} x \times \frac{d}{d x}(\sin x)+6 \cos ^{5} x \times \frac{d}{d x}(\cos x)$
$= 3 \sin ^{2} x \times \cos x+6 \cos ^{5} x \times(-\sin x)$
$[ \because~\left.\frac{d}{d x}(\sin x)=\cos x\ \& \frac{d}{d x}(\cos x)=-\sin x\right]$
$ = 3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$
$\therefore$ $\frac{d y}{d x}=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$

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