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

1 Answer
Mr. Mike
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))#.

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