Question

If $y={\cos }^{2}x$, then find $\frac{dy}{dx}$.

Answer 1

We know that $\cos 2x=2{\cos }^{2}x-1$
$\Rightarrow {\cos }^{2}x=\frac{1+\cos 2x}{2}$
Differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{d}{dx}\left[{\cos }^{2}x\right]=\frac{d}{dx}\left[\frac{1+\cos 2x}{2}\right]$
$=\frac{d}{dx}\left(1/2\right)+\frac{1}{2}\frac{d}{dx}\left(\cos 2x\right)$
$=\frac{1}{2}(-\sin 2x)2=-\sin 2x$
Alternative method: $y={\cos }^{2}x=(\cos x{)}^2$
$\Rightarrow \frac{dy}{dx}=\frac{d\left[\right(\cos x{)}^2]}{dx}=2\left(\cos x{)}^{2-1}\right(-\sin x)$
$=-2\cos x\sin x=-\sin 2x$.