# Homework Help: Finding the derrivative

1. Feb 26, 2006

### Mozart

Hey my teacher worked this problem out the other day and I wrote it down but now im having trouble understanding it. This is what he wrote. Can someone verify if it is correct.

$$y=\frac{e^{2u}}{e^u+e^{-u}}$$

$$y'=\frac{(e^u+e^(-u))(e^(2u))(2)-e^(2u)(e^u+e^-u(-1))}{(e^u+e^-u)^2}$$

$$y'=\frac{e^(2u)[2e^u+2e^-u-e^u+e^-u]}{(e^u+e^-u)^2}$$

$$y'=\frac{e^(2u)[e^u+3e^-u]}{(e^u+e^-u)^2}$$

What I don't understand is how in the step after the quotient rule was used there is in brackets $$-e^u+e^-u$$ Doesn't multiplying it by-1 make it $$e^u+e^-u$$

Last edited: Feb 26, 2006
2. Feb 26, 2006

### Mozart

Oh forget it I can't use latex if my life depended on it. What a waste of time.

3. Feb 26, 2006

### benorin

Here, I tried to clean it up a bit. Use {} for exponents instead of (), for example: e^{2u} instead of e^(2u). You can also double click on the Tex expression to see the code:

Last edited: Feb 27, 2006
4. Feb 26, 2006

### arildno

Would that be:
$$y(u)=\frac{e^{2u}}{e^{u}+e^{-u}}$$

5. Feb 26, 2006

### Mozart

Yes that is exactly what I wanted Benorin except i think for the (2) it should be on the outside of those brackets, and arildno yeah thats the original function. Thanks for fixing it.

So is the final answer correct? I'm confused because of the (-1) in brackets. I don't see how it could give a $$-e^u$$ when there is a negative times a negative.

edit:
Ohhhh nevermind I see it now. Anyways this thread was not in vain. I got some good latex practice in. Thanks again.

6. Feb 26, 2006

### benorin

Just to confirm (since I didn't really folow your question,) I compute the derivative as follows:

$$y=\frac{e^{2u}}{e^u+e^{-u}}\Rightarrow y^{\prime}=\frac{2e^{2u}(e^u+e^{-u})-e^{2u}(e^u-e^{-u})}{(e^u+e^{-u})^2} =\frac{e^{2u}(e^u+3e^{-u})}{(e^u+e^{-u})^2}$$

Yep, same as you.

Last edited: Feb 27, 2006
7. Feb 28, 2006

### Mozart

Thank you for confirming it.