Question

what is the first derivative of $t(x)=\frac{e^{x}}{5x^{2}}$. select the correct answer below: $t(x)=\frac{e^{x}}{10x^{1}}$ $t(x)=\frac{e^{x}(x - 2)}{5x^{3}}$ $t(x)=\frac{e^{x}(-x - 2)}{5x^{3}}$ $t(x)=\frac{e^{x - 1}(2e+x^{2})}{5x^{3}}$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $t(x)=\frac{u(x)}{v(x)}$, then $t^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = e^{x}$, $u^{\prime}(x)=e^{x}$, $v(x)=5x^{2}$, and $v^{\prime}(x)=10x$.

Step2: Apply quotient - rule

$t^{\prime}(x)=\frac{e^{x}\cdot5x^{2}-e^{x}\cdot10x}{(5x^{2})^{2}}=\frac{e^{x}(5x^{2}-10x)}{25x^{4}}=\frac{e^{x}(x - 2)}{5x^{3}}$.

Answer:

$t^{\prime}(x)=\frac{e^{x}(x - 2)}{5x^{3}}$