calculate the derivative of the following function. y = sin(3 cos x) dy/dx = □

Explanation:

Step1: Identify outer - inner functions

Let $u = 3\cos x$, then $y=\sin u$.

Step2: Differentiate outer function

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

Step3: Differentiate inner function

The derivative of $u = 3\cos x$ with respect to $x$ is $\frac{du}{dx}=- 3\sin x$.

Step4: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=-3\sin x$ into it. Since $u = 3\cos x$, we have $\frac{dy}{dx}=\cos(3\cos x)\cdot(-3\sin x)=- 3\sin x\cos(3\cos x)$.

Answer:

$-3\sin x\cos(3\cos x)$