# Differentiation

• Aug 29th 2011, 11:36 PM
Punch
Differentiation
Given that $\displaystyle y=ln[\frac{1}{2}(1+e^{-x})]$, show that $\displaystyle \frac{dy}{dx}=\frac{1}{2}e^{-y}-1$
• Aug 30th 2011, 12:34 AM
Siron
Re: Differentiation
You can do it again this way:
$\displaystyle y=\ln\left(\frac{1+e^{-x}}{2}\right)$
$\displaystyle \frac{dy}{dx}=\frac{d\ln\left(\frac{1}{2}\right)}{ dx}$ $\displaystyle +\frac{d\ln\left(1+e^{-x}\right)}{dx}=\frac{-e^{-x}}{1+e^{-x}}$

If we enter the given information into the equation we get:
$\displaystyle \frac{-e^{-x}}{e^{-x}+1}$ $\displaystyle =\frac{1}{2}e^{-\ln\left(\frac{1+e^{-x}}{2}}\right)}-1$

Because $\displaystyle e^{\ln(x)}=x$ we get:
$\displaystyle \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{2}\left(\frac{2}{e^{-x}+1}\right)-1$
$\displaystyle \Leftrightarrow \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{e^{-x}+1}-\frac{e^{-x}+1}{e^{-x}+1}$
$\displaystyle \Leftrightarrow \frac{e^{-x}}{e^{-x}+1}=\frac{-e^{-x}}{e^{-x}+1}$
• Aug 30th 2011, 03:27 AM
Punch
Re: Differentiation
Quote:

Originally Posted by Siron
You can do it again this way:
$\displaystyle y=\ln\left(\frac{1+e^{-x}}{2}\right)$
$\displaystyle \frac{dy}{dx}=\frac{d\ln\left(\frac{1}{2}\right)}{ dx}$ $\displaystyle +\frac{d\ln\left(1+e^{-x}\right)}{dx}=\frac{-e^{-x}}{1+e^{-x}}$

If we enter the given information into the equation we get:
$\displaystyle \frac{-e^{-x}}{e^{-x}+1}$ $\displaystyle =\frac{1}{2}e^{-\ln\left(\frac{1+e^{-x}}{2}}\right)}-1$

Because $\displaystyle e^{\ln(x)}=x$ we get:
$\displaystyle \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{2}\left(\frac{2}{e^{-x}+1}\right)-1$
$\displaystyle \Leftrightarrow \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{e^{-x}+1}-\frac{e^{-x}+1}{e^{-x}+1}$
$\displaystyle \Leftrightarrow \frac{e^{-x}}{e^{-x}+1}=\frac{-e^{-x}}{e^{-x}+1}$

Sorry but arent we supposed to prove from left to right?