I need to understand partial derivatives.

[jedjj]

Homework Statement

Find the indicated partial derivative.

[tex]u=e^{r\theta}\sin\theta; \frac{\partial^3u}{\partial r^2\partial\theta}[/tex]


2. The attempt at a solution

I started to derive [tex]u_{\theta}[/tex] and I attained

[tex]r*e^{r\theta}\sin\theta + e^{r\theta}\cos\theta[/tex]

But now I don't know how to take the derivative of the first equation, because there are 3 terms. It has been a few years since I have taken any math (I can assure you I won't take anymore breaks)

I just don't know where to go to attain [tex]u_{\theta r}[/tex]

 

[asd1249jf]

Actually there are two terms for you to derive, since your next objective is to derive with respect to r. In this process, you will treat [tex]\sin(\theta)[/tex] as a constant. Do you recall how to do product rule?

[Edit]Oh of course you do, you just did it on your first derivation lol. Like I said, treat sine term as constant, derive with respect to r and apply product rule.

 

[jedjj]

thank you for the response. so it should be [tex]\theta e^{r\theta}\sin\theta + \theta e^{r\theta}\cos\theta[/tex] ?

If so I'm not sure how I attain an r anywhere from this. The answer in the back of the book is showing a sine term as having an r in front of it.

nevermind, I looked at it for just a few more seconds and found the mistake.
Thanks so much

 

[bob1182006]

you do it the same but remember that you need to split up that derivative into 2 derivatives. and then you do the derivative of each term with respect to r just like you did in the first time with respect to theta.

 

[asd1249jf]

thank you for the response. so it should be [tex]\theta e^{r\theta}\sin\theta + \theta e^{r\theta}\cos\theta[/tex] ?

If so I'm not sure how I attain an r anywhere from this. The answer in the back of the book is showing a sine term as having an r in front of it.

No, I think you might've made some simple mistake

[tex]r * \frac{\partial}{\partial r}(e^{r\theta}\sin\theta) + \frac{\partial}{\partial r}r * (e^{r\theta}\sin\theta)[/tex]

Do this for me and see what you get.

 

[jedjj]

I have another question if you could help me please. I have a derivation done by my professor but I'm having problems getting this.

Question

verify that the function [tex]u=\frac{1}{\sqrt{x^2+y^2+z^2}} [/tex] is a solution of the three-dimensional Laplace Equation [tex]u_{xx}+u_{yy}+u_{zz}=0[/tex].

Attempt at a solution

[tex]u={(x^2+y^2+z^2)}^{-1/2}[/tex]
[tex]u_x=-{(x^2+y^2+z^2)}^{-3/2}x[/tex]
[tex]u_{xx}=-3{(x^2+y^2+z^2)}^{5/2}x^2-{(x^2+y^2+z^2)}^{-3/2}[/tex]

He goes on to answer the question, but I do not understand where [tex]-{(x^2+y^2+z^2)}^{-3/2}[/tex] comes from. And if I didn't catch everything he wrote down, then I'm not seeing what I am missing. [edit] OR why it is there.

Please help.

 

[bob1182006]

In order to do the derivative of [tex]u_x[/tex] you need to use the product rule which gives you the [tex]u_{xx}[/tex]

 

[jedjj]

Thank you so much. I really cannot comprehend product rule!

 

[bob1182006]

well it's exactly the same as the Calc I product rule just that now you take other variables like y,z in this case to be constants.

for your problem: say you had g(x,y,z)*h(x,y,z) and you need the derivative with respect to x. to solve it you need to use the product rule and you'll get:
(g(x,y,z)*h(x,y,z))'=g(x,y,z)*h'(x,y,z)+h(x,y,z)*g'(x,y,z)

in your case h(x,y,z)=-x and g(x,y,z)=[tex](x^2+y^2+z^2)^{-3/2}[/tex]