Derivative of $\ln\left(\frac{2x}{1+x}\right)$

Question

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:

$$g(x)=\ln\left(\frac{2x}{1+x}\right)\ , \ x>0$$

Answer 1

Hint For $x>0$, we have $2x>0,x+1>0$. It is then the case $$\begin{align}\log\frac{2x}{x+1}&=\log 2x-\log (1+x)\\&=\log x+\log 2-\log(1+x)\end{align} $$

Answer 2

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.