Question

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

Answer 1

`(dlog_13(8x^3+8))/(dx)=(3x^2)/(ln(13)(x^3+1))`

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(8x^3+8)=(ln(8x^3+8))/ln(13)`

The chain rule says that

`(df(g(x)))/dx=f'(g(x))*g'(x)`

Here,

`f(x) = ln(x)`

`f'(x) = 1/x`

`g(x)=8x^3+8`, and

`g'(x)=24x^2`.

So

`(dlog_13(8x^3+8))/(dx)=(1/(8x^3+8)*24x^2)/ln(13)=(3x^2)/(ln(13)(x^3+1))`.