How do you use the product rule to differentiate f(x)=e^x(sqrtx+5x^3)?

Feb 21, 2017

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

Explanation:

The function $f \left(x\right) = {e}^{x} \left(\sqrt{x} + 5 {x}^{3}\right)$ can be written as $f \left(x\right) = g \left(x\right) . h \left(x\right)$ so we can use the product rule to find its derivative.

In this case $g \left(x\right) = {e}^{x}$ and $h \left(x\right) = \sqrt{x} + 5 {x}^{3}$.

We differentiate $g \left(x\right) = {e}^{x}$ to get $g ' \left(x\right) = {e}^{x}$

And $h \left(x\right) = \sqrt{x} + 5 {x}^{3} = {x}^{\frac{1}{2}} + 5 {x}^{3}$ to get $h ' \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}} + 15 {x}^{2} = \frac{1}{2 \sqrt{x}} + 15 {x}^{2}$

The product rule states that $f ' \left(x\right) = g ' \left(x\right) . h \left(x\right) + h ' \left(x\right) . g \left(x\right)$

Putting our functions into that equation gives $f ' \left(x\right) = {e}^{x} \left(\sqrt{x} + 5 {x}^{3}\right) + {e}^{x} \left(\frac{1}{2 \sqrt{x}} + 15 {x}^{2}\right)$