differentiate. f(x) = \\frac{e^{2x}}{x^{8}} f(x) =

Explanation:

Step1: Recall quotient - rule

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

Step2: Differentiate $u(x)$

Using the chain - rule, if $y = e^{2x}$, let $t = 2x$, then $\frac{dy}{dt}=e^{t}$ and $\frac{dt}{dx}=2$. So, $u^{\prime}(x)=\frac{d}{dx}(e^{2x})=2e^{2x}$.

Step3: Differentiate $v(x)$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $v(x)=x^{8}$, $v^{\prime}(x)=8x^{7}$.

Step4: Apply quotient - rule

$F^{\prime}(x)=\frac{2e^{2x}\cdot x^{8}-e^{2x}\cdot8x^{7}}{(x^{8})^{2}}$.

Step5: Simplify the expression

Factor out $2x^{7}e^{2x}$ from the numerator: $F^{\prime}(x)=\frac{2x^{7}e^{2x}(x - 4)}{x^{16}}=\frac{2e^{2x}(x - 4)}{x^{9}}$.

Answer:

$\frac{2e^{2x}(x - 4)}{x^{9}}$