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 $v(x)=x$.

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 $v(x)$

$v(x)=x$, so $v^{\prime}(x)=\frac{d}{dx}(x)=1$.

Step4: Apply quotient - rule

$F^{\prime}(x)=\frac{2e^{2x}\cdot x - e^{2x}\cdot1}{x^{2}}=\frac{e^{2x}(2x - 1)}{x^{2}}$.

Answer:

$\frac{e^{2x}(2x - 1)}{x^{2}}$