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

1 Answer
Apr 19, 2018

(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)).