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

\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*}
