Question

y = e ^ {\sin(\cos x)}

Explanation:

Step1: Identify the outer - inner functions

Let $u = \sin(\cos x)$ and $y = e^{u}$.

Step2: Differentiate outer function

The derivative of $y = e^{u}$ with respect to $u$ is $\frac{dy}{du}=e^{u}$.

Step3: Differentiate inner function

Let $v=\cos x$. Then $u = \sin v$. First, $\frac{du}{dv}=\cos v$ and $\frac{dv}{dx}=-\sin x$. By the chain - rule $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=\cos v\cdot(-\sin x)=-\sin x\cos(\cos x)$.

Step4: Use the chain - rule for the whole function

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}=e^{u}$ and $\frac{du}{dx}=-\sin x\cos(\cos x)$ and $u = \sin(\cos x)$ back in, we get $\frac{dy}{dx}=e^{\sin(\cos x)}\cdot(-\sin x\cos(\cos x))=-\sin x\cos(\cos x)e^{\sin(\cos x)}$.

Answer:

$-\sin x\cos(\cos x)e^{\sin(\cos x)}$