# Product vs. quotient rule

## Main Question or Discussion Point

When differentiating the following equation, $$y=\frac{d}{dx}\frac{e^x}{x}$$, why is it wrong to rearrange the equation to $$y=\frac{d}{dx}\frac{1}{x}e^x$$ and apply the product rule? Doing so gives me a different result than using the chain rule in conjunction with the quotient rule.

Sorry about the latex formatting, I'm still trying to get the hang of it.

arildno
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No, you do not get different results.
If you differentiate correctly, that is.

S_Happens
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They both give the same answer. I suspect it might have something to do with your statement "...the chain rule in conjunction with the quotient rule."

When you split up the fraction to do product, there is no need to treat 1/x as a quotient. Just bring it to the numerator with a negative exponent. It gives you two fractions instead of one, but getting a common denominator results in the same fraction as the normal quotient rule.

hmmm, well my solution was as follows:
1. $$y=e^x*\frac{1}{x}$$ //rearrange equation
2. $$y=e^x*\frac{1}{x}+e^x*-x^{-2}$$ //apply product rule
3. $$y=\frac{e^x}{x}-\frac{e^x}{x^2}$$ //simplify
I'm pretty sure this is incorrect, as Wolfram Alpha comes up with a different solution. Where did I mess up?

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arildno
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It is correct.
What did Wolfram tell you?

To clarify, I used the product rule to find the solution. When I used Alpha to take the derivative, it chose to use the quotient rule and reached a different solution. I did not personally use the quotient rule. Sorry for the confusion.

S_Happens
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There should be no need for Wolfram Alpha. If you multiply the first fraction by x/x to get a common denominator, you get the same single fraction as you would get from the quotient rule.

arildno
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To clarify, I used the product rule to find the solution. When I used Alpha to take the derivative, it chose to use the quotient rule and reached a different solution. I did not personally use the quotient rule. Sorry for the confusion.
No, it did not produce a different solution.
Please post its solution, and I'll show you why it is the same as yours.

I'm going to beat a hasty retreat here... Here's the Wolfram solution: http://www.wolframalpha.com/input/?i=d%2Fdx+e^x%2Fx

It's just arranged differently than my solution.

arildno
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I'm going to beat a hasty retreat here... Here's the Wolfram solution: http://www.wolframalpha.com/input/?i=d%2Fdx+e^x%2Fx

It's just arranged differently than my solution.
We are not here to bite you, but to help you. I'm glad you found out by your own where the problem was.

Welcome to PF!

S_Happens
Gold Member
I wouldn't add anything since you already noticed it's simply a different form of the same thing, but...

just under the two graphs it gives alternate forms, including the exact one you came up with.

Thanks! I really like the forum and wish I had found it years ago. Would have made college a lot easier.

S_Happens, thanks, I didn't notice the alternate forms box before.