How do you find the derivative of (2e^(3x) + 2e^(-2x))^4?

Aug 14, 2017

$4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3} \left(6 {e}^{3 x} - 4 {e}^{- 2 x}\right)$

Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$

$f \left(g \left(x\right)\right) = {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{4}$

$\Rightarrow f ' \left(g \left(x\right)\right) = 4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3}$

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

$\Rightarrow g ' \left(x\right) = 2 {e}^{3 x} . \frac{d}{\mathrm{dx}} \left(3 x\right) + 2 {e}^{- 2 x} . \frac{d}{\mathrm{dx}} \left(- 2 x\right)$

$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = 6 {e}^{3 x} - 4 {e}^{- 2 x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3} \left(6 {e}^{3 x} - 4 {e}^{- 2 x}\right)$