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Question

Mathematics Question on limits and derivatives

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

A

−sin⁡ 4x4\frac{-\sin \,4x}{4}

B

−sin⁡ 2x2\frac{-\sin \,2x}{2}

C

sin⁡ 4x4\frac{\sin \,4x}{4}

D

−sin⁡ 4x2\frac{-\sin \,4x}{2}

Answer

−sin⁡ 4x2\frac{-\sin \,4x}{2}

Explanation

Solution

We have y=sin⁡2x+cos⁡4xy={{\sin }^{2}}x+{{\cos }^{4}}x
∴\therefore dydx=2sin⁡xcosx⁡+4cos3x(−sin⁡ x)\frac{dy}{dx}=2\sin x\operatorname{cosx}+4co{{s}^{3}}x(-\sin \,x)
=sin⁡2x−4sin⁡xcos⁡x(cos⁡2x)=\sin 2x-4\sin x\cos x({{\cos }^{2}}x)
=sin⁡2x−2sin⁡2x(cos⁡2x+12)=\sin 2x-2\sin 2x\left( \frac{\cos 2x+1}{2} \right)
=sin⁡2x−sin 2x cos⁡ 2x− sin⁡ 2x=\sin 2x-sin\,2x\,\cos \,2x-\,\sin \,2x
=−sin⁡2x cos⁡ 2x=−sin⁡ 4x2=-\sin 2x\,\cos \,2x=\frac{-\sin \,4x}{2}