Question

differentiate.
y = \frac{e^{x}}{8 - e^{x}}

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = e^{x}$, $u^\prime=e^{x}$, $v = 8 - e^{x}$, and $v^\prime=-e^{x}$.

Step2: Apply the quotient - rule

Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule formula:
\[

$$\begin{align*} y^\prime&=\frac{e^{x}(8 - e^{x})-e^{x}(-e^{x})}{(8 - e^{x})^{2}}\\ &=\frac{8e^{x}-e^{2x}+e^{2x}}{(8 - e^{x})^{2}} \end{align*}$$

\]

Step3: Simplify the expression

Combine like terms in the numerator. The $-e^{2x}$ and $e^{2x}$ cancel out, so $y^\prime=\frac{8e^{x}}{(8 - e^{x})^{2}}$.

Answer:

$\frac{8e^{x}}{(8 - e^{x})^{2}}$