a function h(x) is defined in terms of a differentiable function f(x). find an expression for h′(x). h(x)=…

Explanation:

Step1: Apply constant - multiple rule

The constant - multiple rule of differentiation states that if $h(x)=cf(x)$ where $c$ is a constant, then $h'(x)=cf'(x)$. Here $c = - 9$, so $h'(x)=-9\frac{d}{dx}[f(-4x)]$.

Step2: Apply chain rule

Let $u=-4x$. Then $y = f(u)$. By the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{du}{dx}=-4$ and $\frac{dy}{du}=f'(u)=f'(-4x)$. So $\frac{d}{dx}[f(-4x)]=f'(-4x)\cdot(-4)$.

Step3: Substitute and simplify

Substitute $\frac{d}{dx}[f(-4x)]$ into the expression for $h'(x)$ from Step1. We get $h'(x)=-9\cdot f'(-4x)\cdot(-4)=36f'(-4x)$.

Answer:

$36f'(-4x)$