# How do you use the chain rule to differentiate 2^(8x^3+8)?

Oct 20, 2016

$= {2}^{8 {x}^{3} + 11} \cdot 3 {x}^{2} \ln \left(2\right)$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({2}^{8 {x}^{3} + 8}\right)$

Applying exponential rule,

${a}^{b} = {e}^{b \ln \left(a\right)}$

$= \frac{d}{\mathrm{dx}} \left({e}^{\left(8 {x}^{3} + 8\right) \ln \left(2\right)}\right)$

Applying chain rule,

$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $\left(8 {x}^{3} + 8\right) \ln \left(2\right) = u$,

$= \frac{d}{\mathrm{du}} \left({e}^{u}\right) \frac{d}{\mathrm{dx}} \left(\left(8 {x}^{3} + 8\right) \ln \left(2\right)\right)$

we know,

=$\frac{d}{\mathrm{du}} \left({e}^{u}\right) = {e}^{u}$

and,

=$\frac{d}{\mathrm{dx}} \left(\left(8 {x}^{3} + 8\right) \ln \left(2\right)\right) = 24 {x}^{2} \ln \left(2\right)$

Substituting back, $u = \left(8 {x}^{3} + 8\right) \ln \left(2\right)$

=${e}^{\left(8 {x}^{3} + 8\right) \ln \left(2\right)} \cdot 24 {x}^{2} \ln \left(2\right)$

Simplifying,
$= {2}^{8 {x}^{3} + 11} \cdot 3 {x}^{2} \ln \left(2\right)$