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

Limits and Derivatives

Solution:

Let $y =\sin x^{3} , \therefore \frac{dy}{dx} = \cos x^{3} .3x^{2} $
$z =\cos x^{3} \frac{dz}{dx} = -\sin x^{3}.3x^{2}$
$ \therefore \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = - \frac{\cos x^{3}}{\sin x^{3}} = - \cot x^{3} $