Question

compute the derivative of the following function.
h(x) = \frac{x + 4}{5x^{2}e^{x}}
h(x) =

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=x + 4$ and $v(x)=5x^{2}e^{x}$.

Step2: Find $u^{\prime}(x)$

The derivative of $u(x)=x + 4$ using the power - rule ($\frac{d}{dx}(x^n)=nx^{n - 1}$) and the constant rule ($\frac{d}{dx}(c)=0$) is $u^{\prime}(x)=1$.

Step3: Find $v^{\prime}(x)$

Use the product - rule. If $v(x)=5x^{2}e^{x}$, and the product - rule states that if $y = f(x)g(x)$, then $y^{\prime}=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. Here, $f(x)=5x^{2}$ and $g(x)=e^{x}$. The derivative of $f(x)=5x^{2}$ is $f^{\prime}(x)=10x$ (by power - rule), and the derivative of $g(x)=e^{x}$ is $g^{\prime}(x)=e^{x}$. So $v^{\prime}(x)=10xe^{x}+5x^{2}e^{x}=5xe^{x}(2 + x)$.

Step4: Apply quotient - rule

$h^{\prime}(x)=\frac{1\times(5x^{2}e^{x})-(x + 4)\times(5xe^{x}(2 + x))}{(5x^{2}e^{x})^{2}}$.
First, expand the numerator:
\[

$$\begin{align*} &5x^{2}e^{x}-5xe^{x}(x + 4)(x + 2)\\ =&5x^{2}e^{x}-5xe^{x}(x^{2}+6x + 8)\\ =&5x^{2}e^{x}-(5x^{3}e^{x}+30x^{2}e^{x}+40xe^{x})\\ =&5x^{2}e^{x}-5x^{3}e^{x}-30x^{2}e^{x}-40xe^{x}\\ =&-5x^{3}e^{x}-25x^{2}e^{x}-40xe^{x} \end{align*}$$

\]
The denominator is $(5x^{2}e^{x})^{2}=25x^{4}e^{2x}$.
So $h^{\prime}(x)=\frac{-5x^{3}e^{x}-25x^{2}e^{x}-40xe^{x}}{25x^{4}e^{2x}}=\frac{-x^{2}-5x - 8}{5x^{3}e^{x}}$.

Answer:

$\frac{-x^{2}-5x - 8}{5x^{3}e^{x}}$