Question

hw14 the chain rule (target c4; §3.6) due: thu oct 9, 2025 11:59pm attempt 1 in progress next up: submit assignment details no additional details were added for this assignment. hw14 the chain rule (target c4; §3.6) score: 8/11 answered: 9/11 question 10 let ( f(x)=-5e^{xsin x}) ( f(x)=)

Explanation:

Step1: Identify the outer - inner functions

The function $y = f(x)=- 5e^{x\sin x}$ is a composite function. Let $u = x\sin x$, then $y=-5e^{u}$.

Step2: Differentiate the outer function

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=-5e^{u}$ (since the derivative of $e^{u}$ with respect to $u$ is $e^{u}$ and we have a constant factor of - 5).

Step3: Differentiate the inner function

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$ and $v=\sin x$. The derivative of $u = x$ is $u^\prime=1$, and the derivative of $v=\sin x$ is $v^\prime=\cos x$. So $\frac{du}{dx}=\sin x + x\cos x$.

Step4: Apply the chain rule

The chain rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}=-5e^{u}$ and $\frac{du}{dx}=\sin x + x\cos x$ and replacing $u = x\sin x$ back in, we get $f^\prime(x)=-5e^{x\sin x}(\sin x + x\cos x)$.

Answer:

$-5e^{x\sin x}(\sin x + x\cos x)$