find the derivative of the function $f(x)$, below.\n$f(x)=ln(9 - e^{-8x})\n$f(x)=$

Explanation:

Step1: Apply chain - rule

Let $u = 9 - e^{-8x}$, then $f(x)=\ln(u)$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and we need to multiply by the derivative of $u$ with respect to $x$.

Step2: Find derivative of $u$

The derivative of $u = 9 - e^{-8x}$ with respect to $x$: The derivative of a constant 9 is 0, and the derivative of $-e^{-8x}$ using the chain - rule. Let $v=-8x$, then $y = - e^{v}$. The derivative of $y$ with respect to $v$ is $-e^{v}$, and the derivative of $v$ with respect to $x$ is $-8$. So the derivative of $-e^{-8x}$ is $(-e^{-8x})\times(-8)=8e^{-8x}$.

Step3: Combine results

$f'(x)=\frac{1}{u}\times\frac{du}{dx}$. Substituting $u = 9 - e^{-8x}$ and $\frac{du}{dx}=8e^{-8x}$, we get $f'(x)=\frac{8e^{-8x}}{9 - e^{-8x}}$.

Answer:

$\frac{8e^{-8x}}{9 - e^{-8x}}$