Obtaining derivative of log of sigmoid function

Hint:

First, notice that $$ \begin{align} \dfrac{1}{1+e^{-x}} = \dfrac{\mathrm{e}^{x} \cdot 1}{\mathrm{e}^{x} \cdot 1 + \mathrm{e}^{x} \cdot e^{-x}} = \dfrac{\mathrm{e}^{x}}{\mathrm{e}^{x} + 1} \;. \end{align} $$

Second, notice that $$ \begin{align} \ln\left( \dfrac{\mathrm{e}^{x}}{\mathrm{e}^{x} + 1} \right) = \ln\left( \mathrm{e}^{x}\right) - \ln\left( \mathrm{e}^{x} + 1 \right) = x - \ln\left( \mathrm{e}^{x} + 1 \right) \;. \end{align} $$

So, we have $$ \begin{align} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln\left( \dfrac{1}{1+\mathrm{e}^{-x}} \right) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left( x - \ln\left( \mathrm{e}^{x} + 1 \right) \right) = \dfrac{\mathrm{d}x}{\mathrm{d}x} - \dfrac{\mathrm{d}\ln\left( \mathrm{e}^{x} + 1 \right)}{\mathrm{d}x} \;. \end{align} $$

Can you go on from here using the chain rule?

You just have to use the Chain Rule.

$\alpha = 1+e^{-x}$

$\beta = \alpha^{-1}$

$\frac{d\,log(\beta)}{d\,x} = \frac{d\,log(\beta)}{d\,\beta}\,\frac{d\,\beta}{d\,x} = \frac{d\,log(\beta)}{d\,\beta}\,\frac{d\,\alpha^{-1}}{d\,\alpha}\,\frac{d\,\alpha}{d\,x} = \left(\frac{1}{\beta}\right)\,\left(-\frac{1}{\alpha^2}\right)\,\left(-e^{-x}\right) = \frac{e^{-x}}{1+e^{-x}} = \boxed{\frac{1}{e^x + 1}}$

You don't have to worry with signs, because everything in there is always strictly positive.