I’ve been attempting to sketch the function
$y=\frac{\ln(x)}{x + 1}$
However, I’ve been having some trouble calculating the position of the stationary point.
I assume the correct way to calculate this is to differentiate the function and set the result to 0. I have
$\frac{dy}{dx}=\frac{x+1+x \ln(x)}{x(x+1)^2}=0$
This yields
$x+1-x\ln(x) = 0$
Which can also be written as
$x = e^{1+\frac{1}{x}}$
I’m unsure of how to proceed from here though. So what steps do I need to take now in order to find the x-coordinate of the stationary point of the function?
Best Answer
Since
$$y= \dfrac{\ln x}{x+1}$$
we have $$\dfrac{dy}{dx} = \dfrac{1/x\cdot(x+1)-\ln x}{(x+1)^2}$$
If $dy/dx =0$, then
\begin{align} \dfrac{x+1}{x} &= \ln x \\ \iff x &= \exp\bigg(\dfrac{x+1}{x}\bigg) \\ &= \exp(1)\exp(1/x) \\ &= e\exp(1/x) \\ \iff x &= \dfrac{1}{W(1/e)} \end{align}
where $W$ is a Lambert $W$ function. You can find more details about this function here.