# How do you differentiate g(x) = e^(3x) ( 5x-2) using the product rule?

Mar 5, 2016

${e}^{3 x} \left(15 x - 1\right)$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ Product rule }}$

If g(x) = f(x).h(x) then g'(x) = f(x).h'(x) + h(x).f'(x)

here$f \left(x\right) = {e}^{3 x} \Rightarrow f ' \left(x\right) = {e}^{3 x} \frac{d}{\mathrm{dx}} \left(3 x\right) = 3 {e}^{3 x}$

and h(x) = (5x - 2 ) $\Rightarrow h ' \left(x\right) = 5$

substitute these results into g'(x) above

$g ' \left(x\right) = 5 {e}^{3 x} + 3 {e}^{3 x} \left(5 x - 2\right)$

take out common factor of ${e}^{3 x}$

$\Rightarrow g ' \left(x\right) = {e}^{3 x} \left(5 + 15 x - 6\right) = {e}^{3 x} \left(15 x - 1\right)$