Question

Find $\frac{dy}{dx},$ if $y={{\sin }^{2}}x+{{\cos }^{4}}x$

• A. $\frac{-\sin \,4x}{4}$
• B. $\frac{-\sin \,2x}{2}$
• C. $\frac{\sin \,4x}{4}$
• D. $\frac{-\sin \,4x}{2}$

Answer 1

The Correct Option is D

We have $y={{\sin }^{2}}x+{{\cos }^{4}}x$
$\therefore$ $\frac{dy}{dx}=2\sin x\operatorname{cosx}+4co{{s}^{3}}x(-\sin \,x)$
$=\sin 2x-4\sin x\cos x({{\cos }^{2}}x)$
$=\sin 2x-2\sin 2x\left( \frac{\cos 2x+1}{2} \right)$
$=\sin 2x-sin\,2x\,\cos \,2x-\,\sin \,2x$
$=-\sin 2x\,\cos \,2x=\frac{-\sin \,4x}{2}$