I need help finding the derivative of this natural logarithm function. Okay so $$f(x)=\ln[x\ln(x+2)]$$
so $$\ln(x)+\ln(\ln(x+2))$$so $$1.a\frac{dy}{dx}\ln(x)=\frac{1}{x}$$and I thought by chain rule that $$1.b\frac{dy}{dx}\ln(\ln(x+2))=\frac{1}{\ln(x+2)}\cdot\frac{1}{x}$$ so $$2.af'(x)=\frac{1}{x}+\frac{1}{\ln(x+2)}*\frac{1}{x}$$But the site says the following is step 2 $$2.bf'(x)=\frac{1}{x}+\frac{\frac{1}{x+2}}{\ln(x+2)}$$and I'm having problems going from $$\ln(x)+\ln(\ln(x+2))$$ to $$f'(x)=\frac{1}{x}+\frac{\frac{1}{x+2}}{\ln(x+2)}$$.
A step by step explanation would be greatly appreciated. Thanks in advance
 A: \begin{align*}
f'(x) &= \frac{d}{dx} f(x) \\
&= \frac{d}{dx} ln[xln(x+2)] \\
&= \frac{d}{dx}(ln(x) + ln(ln(x+2))) \\
&= \frac{d}{dx}ln(x)+\frac{d}{dx}ln(ln(x+2)) \\
&= \frac{\frac{d}{dx}x}{x}+\frac{\frac{d}{dx}ln(x+2)}{ln(x+2)} \;\; \text{(Chain Rule)} \\
&= \frac{1}{x}+\frac{\frac{1}{x+2}}{ln(x+2)}
\end{align*}
A: Recall that, in general,
$$\frac{d}{dx} \log f(x) = \frac{f'(x)}{f(x)},$$
which follows from the chain rule. Using this formula we have
\begin{align*}
\frac{d}{dx} \log(x\log(x + 2)) &= \frac{d}{dx} \log x + \frac{d}{dx}\log(\log(x + 2)) \\
&= \frac{1}{x} + \frac{(\log(x + 2))'}{\log(x + 2)} \\
&= \frac{1}{x} + \frac{1/(x + 2)}{\log(x + 2)}.
\end{align*}
A: From what I understand, you know that $\dfrac{d}{dx} \ln x = \dfrac{1}{x}$, and you also know the chain rule $\dfrac{d}{dx} f(u(x)) = \dfrac{d}{du}f(u) \times \dfrac{du}{dx}$.
So $\dfrac{d}{dx}\ln(x + 2) = \dfrac{1}{x + 2}\quad$ (it is not $\dfrac{1}{x}$)
And therefore, $\dfrac{d}{dx} \ln \ln (x + 2) = \dfrac 1 {\ln(x + 2)} \times\dfrac d {dx} \ln(x + 2) = \dfrac 1 {\ln(x + 2)} \times \dfrac 1 {x + 2}$
$\therefore \boxed{f'(x) = \dfrac{1}{x} + \dfrac 1 {(x + 2)\ln(x + 2)}}$
