Question

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

• A. $\frac{2\cos x}{e^{\cos x}}$
• B. $-\frac{2\cos x}{c^{\cos x}}$
• C. $\frac{2}{e^{\cos x}}$
• D. None of the above

Answer 1

Answer: (B)

Let $u(x)={\sin }^{2}x$ and $v(x)=e^{\cos x}$.
We want to find $\frac{du}{dv}=\frac{du/dx}{dv/dx}.$
Clearly, $\frac{du}{dx}=2\sin x\cos x$
and $\frac{dv}{dx}=e^{\cos x}(-\sin x)=-(\sin x)e^{\cos x}$
$\frac{du}{dv}=\frac{2\sin x\cos x}{-\sin x{e}^{\cos x}}=-\frac{2\cos x}{e^{\cos x}}$