How do you find f'(x) and f''(x) given f(x)=x^(5/2)e^x?

Sep 28, 2017

$f ' \left(x\right) = {x}^{\frac{5}{2}} {e}^{x} + \frac{5}{2} \cdot {x}^{\frac{3}{2}} {e}^{x}$
$f ' ' \left(x\right) = {x}^{\frac{1}{2}} {e}^{x} \left({x}^{2} + 5 x + \frac{15}{4}\right)$

Explanation:

To find $f ' \left(x\right)$ (and later $f ' ' \left(x\right)$), you can use the Product Rule for derivatives, which says the following:

Given $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$, then $f ' \left(x\right) = g \left(x\right) \cdot h ' \left(x\right) + h \left(x\right) \cdot g ' \left(x\right)$.

In this case, we can break up $f \left(x\right) = {x}^{\frac{5}{2}} {e}^{x}$ into two separate functions:

$\left\{\begin{matrix}g \left(x\right) = {x}^{\frac{5}{2}} \\ h \left(x\right) = {e}^{x}\end{matrix}\right.$

Following the Product Rule pattern, we then have:

$f ' \left(x\right) = {x}^{\frac{5}{2}} {e}^{x} = {x}^{\frac{5}{2}} \cdot {e}^{x} + {e}^{x} \cdot \left(\frac{5}{2}\right) {x}^{\frac{3}{2}}$
$= {x}^{\frac{5}{2}} {e}^{x} + \frac{5}{2} \cdot {x}^{\frac{3}{2}} {e}^{x}$

To find $f ' ' \left(x\right)$, you will technically need to use the Product Rule two more times, once for each term in the preceding $f ' \left(x\right)$. However, the savvy reader should note that the first term of $f ' \left(x\right)$ is an exact match for the original $f \left(x\right)$, and thus there's no need to manually find the derivative of that term again since we already know it (it's $f ' \left(x\right)$ itself!

$f ' ' \left(x\right) = {\underbrace{\frac{d}{\mathrm{dx}} \left({x}^{\frac{5}{2}} {e}^{x}\right)}}_{f ' \left(x\right)} + \frac{d}{\mathrm{dx}} \left(\frac{5}{2} \cdot {x}^{\frac{3}{2}} {e}^{x}\right)$

$= {x}^{\frac{5}{2}} {e}^{x} + \frac{5}{2} \cdot {x}^{\frac{3}{2}} {e}^{x} + \left(\frac{5}{2} {x}^{\frac{3}{2}} \cdot {e}^{x} + {e}^{x} \cdot \left(\frac{5}{2}\right) \left(\frac{3}{2}\right) {x}^{\frac{1}{2}}\right)$

$= {x}^{\frac{5}{2}} {e}^{x} + \frac{5}{2} \cdot {x}^{\frac{3}{2}} {e}^{x} + \left(\frac{5}{2} {x}^{\frac{3}{2}} {e}^{x} + \frac{15}{4} {x}^{\frac{1}{2}} {e}^{x}\right)$

$= {x}^{\frac{5}{2}} {e}^{x} + 5 {x}^{\frac{3}{2}} {e}^{x} + \frac{15}{4} {x}^{\frac{1}{2}} {e}^{x}$

$= {x}^{\frac{1}{2}} {e}^{x} \left({x}^{2} + 5 x + \frac{15}{4}\right)$

Sep 28, 2017

$f ' ' \left(x\right) = \left(\frac{\sqrt{x} . {e}^{x}}{4}\right) \left(15 + 20 x + 4 {x}^{2}\right)$

Explanation:

$u = {x}^{\frac{5}{2}}$
$\mathrm{du} = \left(\frac{5}{2}\right) {x}^{\frac{3}{2}}$
$v = {e}^{x}$
$\mathrm{dv} = {e}^{x}$
$f \left(x\right) = {x}^{\frac{5}{2}} . {e}^{x}$
$f ' \left(x\right) = v . \mathrm{du} + u . \mathrm{dv}$
$f ' \left(x\right) = \left({e}^{x} . \left(\frac{5}{2}\right) . {x}^{\frac{3}{2}}\right) + \left({x}^{\frac{5}{2}} . {e}^{x}\right)$
$f ' ' \left(x\right) = \left(\frac{5}{2}\right) . \left(\left(\left(\frac{3}{2}\right) . \sqrt{x} . {e}^{x}\right) + \left(\frac{5}{2}\right) . {x}^{\frac{3}{2}} . {e}^{x}\right) + \left({e}^{x} . \left(\frac{5}{2}\right) {x}^{\frac{3}{2}}\right) + \left({x}^{\frac{5}{2}} . {e}^{x}\right)$
$f ' ' \left(x\right) = \left(\left(\frac{\sqrt{x} . . {e}^{x}}{4}\right) \left(3 + 5 x + 2 x\right)\right) + \left({x}^{2.} {e}^{x}\right)$
$f ' ' \left(x\right) = \left(\frac{\sqrt{x} . {e}^{x}}{4}\right) \left(15 + 20 x + 4 {x}^{2}\right)$