Question

How do you differentiate `f(x) = log_3(log_7 (log_5(x + 2)))`?

Answer 1

`f'(x)=1/(ln(3)(x+2)ln(log_5(x+2))ln(x+2))`

Explanation:

The typical method for differentiating a logarithm that doesn't have base `e` requires using the change of base formula to put the expression into terms of the natural logarithm.

For example, the derivative of `log_a(g(x)))` can be found through saying that

`d/dxlog_a(g(x))=d/dx(ln(g(x))/ln(a))=1/ln(a)(d/dxln(g(x)))`

`=1/ln(a)(1/g(x))*g'(x)=1/(g(x)ln(a))*g'(x)`

So, this rule will give us:

`f'(x)=1/(log_7(log_5(x+2))ln(3))*d/dxlog_7(log_5(x+2))`

Using this rule again on the smaller logarithm function (like peeling off its shells):

`f'(x)=1/(log_7(log_5(x+2))ln(3))(1/(log_5(x+2)ln(7)))*d/dxlog_5(x+2)`

Noting that `d/dxlog_5(x+2)=1/((x+2)ln(5))*d/dx(x+2)`, which equals `1/((x+2)ln(5))`, the entire function's derivative equals

`f'(x)=1/(ln(3)ln(5)ln(7)(x+2)log_7(log_5(x+2))log_5(x+2))`

This can be simplified by using the change of base formula on `log_7(log_5(x+2))` and `log_5(x+2))`:

`f'(x)=1/(ln(3)color(red)(cancel(color(black)(ln(5))))color(red)(cancel(color(black)(ln(7))))(x+2)(ln(log_5(x+2))/color(red)(cancel(color(black)(ln(7)))))(ln(x+2)/color(red)(cancel(color(black)(ln(5))))))`

`f'(x)=1/(ln(3)(x+2)ln(log_5(x+2))ln(x+2))`