# How do you use the chain rule to differentiate log_13(8x^3+8)?

Apr 19, 2018

$\frac{\mathrm{dl} o {g}_{13} \left(8 {x}^{3} + 8\right)}{\mathrm{dx}} = \frac{3 {x}^{2}}{\ln \left(13\right) \left({x}^{3} + 1\right)}$

#### Explanation:

Somehow I find it easier to remember the change-of-base formula for logarithms than remembering how to take the derivative of log functions other than natural log, so I will start by changing the log to the natural log.

${\log}_{13} \left(8 {x}^{3} + 8\right) = \frac{\ln \left(8 {x}^{3} + 8\right)}{\ln} \left(13\right)$

The chain rule says that

$\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Here,

$f \left(x\right) = \ln \left(x\right)$

$f ' \left(x\right) = \frac{1}{x}$

$g \left(x\right) = 8 {x}^{3} + 8$, and

$g ' \left(x\right) = 24 {x}^{2}$.

So

$\frac{\mathrm{dl} o {g}_{13} \left(8 {x}^{3} + 8\right)}{\mathrm{dx}} = \frac{\frac{1}{8 {x}^{3} + 8} \cdot 24 {x}^{2}}{\ln} \left(13\right) = \frac{3 {x}^{2}}{\ln \left(13\right) \left({x}^{3} + 1\right)}$.