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Question

Mathematics Question on Differentiability

The derivative of sin⁡ x3\sin \, x^3 w.r.t. cos⁡ x3\cos \, x^3 is equal to

A

−cot⁡x3 - \cot x^3

B

cot⁡x3 \cot x^3

C

tan⁡x3\tan x^3

D

−tan⁡x3 - \tan x^3

Answer

−cot⁡x3 - \cot x^3

Explanation

Solution

Let y=sin⁡x3,∴dydx=cos⁡x3.3x2y =\sin x^{3} , \therefore \frac{dy}{dx} = \cos x^{3} .3x^{2} z=cos⁡x3dzdx=−sin⁡x3.3x2z =\cos x^{3} \frac{dz}{dx} = -\sin x^{3}.3x^{2} ∴dydz=dy/dxdz/dx=−cos⁡x3sin⁡x3=−cot⁡x3\therefore \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = - \frac{\cos x^{3}}{\sin x^{3}} = - \cot x^{3}