Question:
Find $ \frac{dy}{dx}, $ if $ y={{\sin }^{2}}x+{{\cos }^{4}}x $
Find $ \frac{dy}{dx}, $ if $ y={{\sin }^{2}}x+{{\cos }^{4}}x $
Updated On: Jun 23, 2024
- $ \frac{-\sin \,4x}{4} $
- $ \frac{-\sin \,2x}{2} $
- $ \frac{\sin \,4x}{4} $
- $ \frac{-\sin \,4x}{2} $
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The Correct Option is D
Solution and Explanation
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} $
$ \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} $
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