# What is the derivative of  f(x) = (x^5)*(e^x)?

Jun 26, 2018

$\left(5 + x\right) {x}^{4} \cdot {e}^{x}$
$f \left(x\right) = {x}^{5} \cdot {e}^{x}$
Product rule states: $\frac{d}{\mathrm{dx}} u \left(x\right) \cdot v \left(x\right) = u ' \left(x\right) \cdot v \left(x\right) + u \left(x\right) \cdot v ' \left(x\right)$
in this case $u \left(x\right) = {x}^{5}$ ; $v \left(x\right) = {e}^{x}$
$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = 5 {x}^{4} \cdot {e}^{x} + {x}^{5} \cdot {e}^{x}$ note that $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$
thsi can be simplified by factorisation to $f ' \left(x\right) = \left(5 + x\right) {x}^{4} \cdot {e}^{x}$