<s>[INST] Derivative of $\ln\left(\frac{2x}{1+x}\right)$ I know that the derivative of 
$$f(x)=\ln(x) \ ,\ x>0$$
is just simply
$$f'(x)=\frac{dx}{x}$$
But how do you find the derivative for the function:
$$g(x)=\ln\left(\frac{2x}{1+x}\right)\ , \ x>0$$
 [/INST] The trick hinted at by Peter Tamaroff in another answer is a clever one, but in general you use the chain rule to find the derivative of a composition:
$$(f(g(x)))' = f'(g(x))\cdot g'(x).$$
In your case, $f(x)=\ln x$ and $g(x)=2x/(1+x)$. Let us first find the derivatives of $f$ and $g$:
\begin{align*}
f'(x) &= (\ln x)' = \frac{1}{x},\\
g'(x) &= \left(\frac{2x}{1+x}\right)' = \frac{2(1+x) - 2x\cdot 1}{(1+x)^2} = \frac{2}{(1+x)^2}.
\end{align*}
Now we can apply the chain rule:
$$(f(g(x)))'=\frac{1}{\frac{2x}{1+x}} \cdot \frac{2}{(1+x)^2} = \frac{1+x}{2x}\cdot \frac{2}{(1+x)^2}= \frac{1}{x(1+x)}.$$
Note that $\frac{1}{x(1+x)}=\frac{1}{x} - \frac{1}{1+x},$ which is the answer you would get from Peter's answer.
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