Deriative of $\ln( \ln(2x+5)) $

Question

I used chain rule on this one:

$\frac{2}{2x+5}$ that is the internal deriative.

I'm a bit confused on how to do the external.

answer: $\frac{2}{(2x+5)\ln (2x+5)}$

Answer 1

The general rule for differentiating logarithms is:

$$\dfrac{d}{dx}\ln(f(x))= \frac{f'(x)}{f(x)}$$

So

\begin{align} \dfrac{d}{dx} \ln(\ln(2x+5)) & = \frac{\frac{2}{2x+5}}{\ln(2x+5)} \\ & = \frac{2}{(2x+5) \ln (2x+5)} \end{align}

I differentiated $\ln(2x+5)$ to get $\frac{2}{2x+5}$. Then that is divided by $\ln(2x+5)$ since that is $f(x)$, so I have $\frac{f'(x)}{f(x)}$.

Take each step slowly.

Answer 2

use the chain rule $$u=\ln( \ln(2x+5))$$ $$u'=\frac 1 {\ln(2x+5))}{(\ln(2x+5))'}$$ $$u'=\frac 1 {\ln(2x+5))}\frac 1 {(2x+5))}(2x+5)'$$ $$u'=\frac 1 {\ln(2x+5))}\frac 1 {(2x+5))}2$$ $$u'=\frac 2 {(2x+5)\ln(2x+5)}$$