partial derivatives of exponential functions

**Fred1956** · November 1st 2009, 02:44 PM

I need assistance is taking the partial derivatives of an exponential function. The partial derivatives need to be taken with respect to t1, t2, and a double partial with respect to t1t2. I posted previously but entered the function incorrectly. Sorry about that but I am new at this. I have attaced a pdf document with the function and would appreciate the help.

Thanks

Fred1956

**adkinsjr** · November 1st 2009, 03:00 PM

Edit: I erased this post since I differentiated the wrong function.

**adkinsjr** · November 1st 2009, 03:10 PM

Hold on, I think I read your document wrong. I think the function was suppose to be:

$\frac{\partial}{\partial t_1}[e^{-2}e^{(1+e^{t_1})e^{t_2}}]$

In this case the derivative is:

$e^{-2}e^{t_2}e^{t_1}e^{(1+e^{t_1})e^{t_2}}$

Which can be simplified by just adding the exponents, and writing it as a single base. I found this derivative by treating $e^{t_2}$ as a constant and using the chain rule. If I define $u=(1+e^{t_1})e^{t_2}$ then $\frac{\partial u}{\partial t_1}=e^{t_2}e^{t_1}$ . The derivative of $e^{u}$ with respect to $u$ is just $e^{u}$ . This is how I obtained: $e^{t_2}e^{t_1}e^{(1+e^{t_1})e^{t_2}}$ . Since we have to keep the constant $e^{-2}$ , the entire derivative looks like this:

$e^{-2}e^{t_2}e^{t_1}e^{(1+e^{t_1})e^{t_2}}=e^{t_1+t_2-2+(1+e^{t_1})e^{t_2}}$

**adkinsjr** · November 1st 2009, 03:37 PM

I'll define the function as $f(t_1,t_2)=e^{-2}e^{(1+e^{t_1})}e^{t_2}$

To find the derivative $\frac{\partial^2 f(t_1,t_2)}{\partial t_1 \partial t_2}$ , just differentiate the derivative we already found with respect to $t_2$ . In other words, we want to differentiate:

$e^{t_1+t_2-2+(1+e^{t_1})e^{t_2}}$

with respect to $t_2$

This shouldn't be very difficult, so I think I will leave it to you. If you can't figure it out, then I'll help. This problem is really difficult to type lol.

**Fred1956** · November 1st 2009, 05:13 PM

$ $ $\frac{\partial}{\partial t_2}[e^{-2}e^{(1+e^{t_1})e^{t_2}}]

So if I use the chain rule to take the partial derivative with respect to t2 then I hold t1 to be a constant. I get the result:

$ $ $[e^{-2}e^{t_2}e^{(1+e^{t_1})e^{t_2}}]

Is this correct for the partial of the function with respect to t2

Now for the second partial: take partial of this with respect to t2 of
$ $ $e^{t_1+t_2-2+(1+e^{t_1})e^{t_2}}

and I get
$ $ $

I tried to use your Latex generated formulas to write out my results. I hope this is the correct way to do it. I saw the tutorial on Latex on the website but am not sure how to input into my messages. Do you go offsite to do that or do you put the equations directly into the message using the Latex code?

Thanks again

Fred1956

**adkinsjr** · November 1st 2009, 05:21 PM

Originally Posted by Fred1956

$ \frac{\partial}{\partial t_2}[e^{-2}e^{(1+e^{t_1})e^{t_2}}] $

So if I use the chain rule to take the partial derivative with respect to t2 then I hold t1 to be a constant. I get the result:

$ [e^{-2}e^{t_2}e^{(1+e^{t_1})e^{t_2}}] $

Is this correct for the partial of the function with respect to t2

Now for the second partial: take partial of this with respect to t2 of
$ 

and I get
$ 

I tried to use your Latex generated formulas to write out my results. I hope this is the correct way to do it. I saw the tutorial on Latex on the website but am not sure how to input into my messages. Do you go offsite to do that or do you put the equations directly into the message using the Latex code?

Thanks again

Fred1956

I type them directly into the box. Don't worry, I can see what you typed by clicking the 'quote' button. I'll respond again in a moment after I've reviewed what you did.

Here is a section where you can ask for latex help if you'de like to learn how to do it. The problem you're working on right now is very difficult to type, so I'm not surprised you made so many latex errors. I made several errors myself when I was typing the derivatives. The exponents inside exponents make it confusing.

http://www.mathhelpforum.com/math-help/latex-help/

**adkinsjr** · November 1st 2009, 05:49 PM

You differentiated the function $f(t_1,t_2)=e^{-2}e^{(1+e^{t_1})e^{t_2}}$ with respect to $t_2$ and obtained:

$\frac{\partial f}{\partial t_2}=e^{t_2}e^{-2}e^{(1+e^{t_1})e^{t_2}}$ .

This is correct.

The other part doesn't seem to be correct. Remember that $\frac{\partial^2f}{\partial t_1 \partial t_2}=\frac{\partial}{\partial t_2}\frac{\partial f}{\partial t_1}$

Let $u=t_1+t_2-2+(1+e^{t_1})e^{t_2}$ .

So we can write: $\frac{\partial f}{\partial t_1}=e^u$

So now we just want to find $\frac{\partial}{\partial t_2}e^u$

So the chain gives this as the product of the derivative with respect to $u$ of the function $e^u$ (which is just $e^u$ ) times the derivative of $\frac{du}{dt_2}$

**Fred1956** · November 1st 2009, 06:11 PM

I think I have it:

e^t1+2t2-4+2(1+e^t1)e^t2

Fred1956

**adkinsjr** · November 1st 2009, 08:23 PM

$u=t_1+t_2-2+(1+e^{t_1})e^{t_2}$

$\frac{\partial}{\partial t_2}e^u=e^u\frac{\partial u}{\partial t_2}$

Where $u=t_1+t_2-2+(1+e^{t_1})e^{t_2}$

$\frac{\partial u}{\partial t_2}=1+(1+e^{t_1})e^{t_2}$

So the derivative you're looking for is $e^u[1+(1+e^{t_1})e^{t_2}]$ . I don't think that this is the same derivative you got, but I didn't simplify my expression since it would be laborious. It should have two terms. I think you should be able to tell if you made a mistake or not. It looks like you didn't calculate $\frac{\partial u}{\partial t_2}$ correctly. It seems you skipped over the $t_2$ term in the expression for $u$ .

**Fred1956** · November 2nd 2009, 04:48 AM

thanks, I see my mistake. I did the partial with respect to t2 only instead of the du/dt2.

Again

thanks for your help.

**Fred1956** · November 2nd 2009, 02:34 PM

I am stuck again trying to take the second partial derivatives of the function with respect to t1 and t2.

Thanks

Fred1956

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