find f(x).\nf(x)=\frac{8 - e^{x}}{7 + e^{x}}\nf(x)=square

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 8 - e^{x}$, $u^\prime=-e^{x}$, $v = 7 + e^{x}$, and $v^\prime=e^{x}$.

Step2: Substitute into quotient - rule formula

$f^\prime(x)=\frac{(-e^{x})(7 + e^{x})-(8 - e^{x})(e^{x})}{(7 + e^{x})^{2}}$.

Step3: Expand the numerator

[ \begin{align*} (-e^{x})(7 + e^{x})-(8 - e^{x})(e^{x})&=-7e^{x}-e^{2x}-8e^{x}+e^{2x}\ &=(-7e^{x}-8e^{x})+(-e^{2x}+e^{2x})\ &=-15e^{x} \end{align*} ]

Answer:

$\frac{-15e^{x}}{(7 + e^{x})^{2}}$