Derivative of 14x/(1+12.6e^(-0.73x)

**AZach** · July 16th 2012, 04:23 PM

$f(x)=\frac{14x}{(1+12.6e^{-0.73x})}$

Hello, I'm having some brain freeze on finding the derivative for this problem.

First, I changed the denominator to ${(1+12.6e^{-0.73x})^{-1}$ , and then used the Product Rule. $f(x)=\frac{d}{dx}{[14x]}*{(1+12.6e^{-0.73x})^{-1}+\frac{d}{dx}{[(1+12.6e^{-0.73x})^{-1}]*{14x}$

That left me with: $f(x)={[14]}*{(1+12.6e^{-0.73x})^{-1}+{[(12.6e^{-0.73x})^{-1}]*{14x}$

This is where I'm getting brain frozen. I checked the back off the book it says the answer is: $f(x)={[14]}*{(1+12.6e^{-0.73x})^{-1}+ {128.772x(1+12.6e^{-0.73x})^{-2}e^{-0.73x}$ .

I'm not sure where the value of ${128.772x(1+12.6e^{-0.73x})^{-2}e^{-0.73x}$ came from. I see parts of the Chain Rule, but the 128.772 has me dumbfounded. If there's a coefficient in front of ${e}$ what happens when taking the derivative of ${e}^{x}$ ?

**daigo** · July 16th 2012, 04:56 PM

Sorry about my handwriting

The last step I just used the distributive property

**daigo** · July 16th 2012, 04:59 PM

During your step here:

$f(x)={[14]}*{(1+12.6e^{-0.73x})^{-1}+{[(12.6e^{-0.73x})^{-1}]*{14x}$

You should show how you used the Chain Rule to see where you made the error, since you never finished the Chain rule on the second term's $[(12.6e^{-0.73x})^{-1}]$

**daigo** · July 16th 2012, 05:41 PM

If you feel insistent about the Product rule:

I thought this might not be clear enough and I was going to make some changes but it's late and I have to sleep so bye for now

**Soroban** · July 16th 2012, 07:21 PM

Hello, AZach!

$f(x)\:=\:\frac{14x}{(1+12.6e^{-0.73x})}$

$\text{Hello, I'm having some brain freeze on finding the derivative for this problem.}$

$\text{First, I changed the denominator to }(1+12.6e^{-0.73x})^{-1},\,\text{ and then used the Product Rule.}$

$f'(x)\:=\:\frac{d}{dx}(14x)\cdot(1+12.6e^{-0.73x})^{-1}+\frac{d}{dx}[(1+12.6e^{-0.73x})^{-1}]\cdot(14x)$

$\text{That left me with: }\:f'(x)\:=\:(14)\cdot{(1+12.6e^{-0.73x})^{-1}+\underbrace{[(12.6e^{-0.73x})^{-1}]}_{\text{This is wrong!}}\cdot(14x)$

The derivative of: . $g(x) \;=\;\left(1+12.6\:\!e^{-0.73x}\right)^{-1}$

. . . . . . . . . . .is: . $g'(x) \;=\;-\left(1 + 12.6\:\!e^{-0.73x}\right)^{-2}\cdot 12.6\:\!e^{-0.73x}\cdot (-0.73)$

. . . . . . . . . . . . . . . . . . $=\;9.198\:\!e^{-0.73x}\left(1 + 12.6\:\!e^{-0.73x}\right)^{-2}$

. . . . . . . . . . . . . . . . . . $=\;\frac{9.198}{e^{0.73x}\left(1 + 12.6\:\!e^{-0.73x}\right)^2}$

Unless you are very good at handling negative exponents,
. . I do not recommend switching to the Product Rule.

**AZach** · July 16th 2012, 07:51 PM

Originally Posted by Soroban

. . . . . . . . . . .is: . $g'(x) \;=\;-\left(1 + 12.6\:\!e^{-0.73x}\right)^{-2}\cdot 12.6\:\!e^{-0.73x}\cdot (-0.73)$

First off, thanks for the thorough responses, Daigo and Soroban. For this quoted part, I think this is where one of my main problems has been. I knew the derivative of ${e}^{x}$ was itself, so I ignorantly assumed that a coefficient in front of the 'x' variable wouldn't alter that. Looking at this from a function point of view, is $12.6e^{-0.73x}$ and ${-0.73x}$ both considered inside functions of $(1+12.6e^{-0.73x})^{-1}$ ?

**tom@ballooncalculus** · July 17th 2012, 01:48 AM

Originally Posted by AZach

Looking at this from a function point of view, is $12.6e^{-0.73x}$ and ${-0.73x}$ both considered inside functions of $(1+12.6e^{-0.73x})^{-1}$ ?

Absolutely - it's a glass-onion / russian doll situation!

Just in case a picture helps...

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Spoiler:

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