0

so I was trying to get the derivative of the sigmoid function $\sigma(x)=\frac{1}{1+e^{-x}}$. I know how to get it using the properties of derivatives but I need to do it from limit definition.

Here is what I tried: $\sigma'(x)= \lim_{h \to 0} \frac{\frac{1}{1+e^{-(x+h)}}-\frac{1}{1+e^{-x}}}{h}=\frac{e^{-x}-e^{-(x+h)}}{(1+e^{-(x+h)})(1+e^{-x})h}$.

patamon
  • 67

1 Answers1

1

I guess

$$\cdots=\lim_{h\to 0}\frac{e^{-x-h}\cdot\frac{e^{h}-1}h}{(1+e^{-x-h})(1+e^{-x})}=\left[\frac{e^{-x-0}\cdot 1}{(1+e^{-x-0})(1+e^{-x})}\right]=\frac{e^{-x}}{(1+e^{-x})^2}$$

By continuity of $e^x$ and known limit $\frac{e^x-1}x\stackrel{x\to 0}\longrightarrow 1$