The derivative of sin2x with respect to ecosx is:

Question

The derivative of $\sin^{2} x$ with respect to $e^{\cos x}$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $- \frac{2 \cos x}{e^{\cos x}}$

Explanation:

Let $u (x) = \sin^{2} x$ and $v (x) = e^{\cos x} .$ We want to find $\frac{d u}{d v} .$

Thus, $\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} .$ Clearly, $\frac{d u}{d x} = 2 \sin x \cos x$

and $\frac{d v}{d x} = e^{\cos x} (- \sin x) = - (\sin x) e^{\cos x}$

Therefore, $\frac{d u}{d v} = \frac{2 \sin x \cos x}{- \sin x e^{\cos x}} = - \frac{2 \cos x}{e^{\cos x}}$

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