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

Answer: (D)

We have $y={\sin }^{2}x+{\cos }^{4}x$
$\therefore$ $\frac{dy}{dx}=2\sin x\mathrm{cosx}+4co{s}^{3}x(-\sin x)$
$=\sin 2x-4\sin x\cos x({\cos }^{2}x)$
$=\sin 2x-2\sin 2x\left(\frac{\cos 2x+1}{2}\right)$
$=\sin 2x-sin2x\cos 2x-\sin 2x$
$=-\sin 2x\cos 2x=\frac{-\sin 4x}{2}$