Thank you for reporting, we will resolve it shortly
Q.
The derivative of $\sin ^2 x$ with respect to $e^{\cos x}$ is
Continuity and Differentiability
Solution:
Let $u(x)=\sin ^2 x$ and $v(x)=e^{\cos x}$.
We want to find $\frac{d u}{d v}=\frac{d u / d x}{d v / d x} . $
Clearly, $ \frac{d u}{d x}=2 \sin x \cos x $
and $\frac{d v}{d x}=e^{\cos x}(-\sin x)=-(\sin x) e^{\cos x}$
$\frac{d u}{d v}=\frac{2 \sin x \cos x}{-\sin x e^{\cos x}}=-\frac{2 \cos x}{e^{\cos x}}$