# Differentiation

• August 30th 2011, 12:36 AM
Punch
Differentiation
Given that $y=ln[\frac{1}{2}(1+e^{-x})]$, show that $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1$
• August 30th 2011, 01:34 AM
Siron
Re: Differentiation
You can do it again this way:
$y=\ln\left(\frac{1+e^{-x}}{2}\right)$
$\frac{dy}{dx}=\frac{d\ln\left(\frac{1}{2}\right)}{ dx}$ $+\frac{d\ln\left(1+e^{-x}\right)}{dx}=\frac{-e^{-x}}{1+e^{-x}}$

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

Because $e^{\ln(x)}=x$ we get:
$\frac{-e^{-x}}{e^{-x}+1}=\frac{1}{2}\left(\frac{2}{e^{-x}+1}\right)-1$
$\Leftrightarrow \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{e^{-x}+1}-\frac{e^{-x}+1}{e^{-x}+1}$
$\Leftrightarrow \frac{e^{-x}}{e^{-x}+1}=\frac{-e^{-x}}{e^{-x}+1}$
• August 30th 2011, 04:27 AM
Punch
Re: Differentiation
Quote:

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

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

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

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