# Math Help - Partial Derivative of Higher Orders

1. ## Partial Derivative of Higher Orders

Hello, everyone. It's been awhile since I've visited the boards. I hope all is well for you and yours.

The problem I'm dealing with is as follows:

-----------------------------------------

$z = u{(v-w)}^{\frac{1}{2}}$

Find the indicated partial derivative:

$\frac{\partial^3z}{{\partial}u{\partial}v{\partial }w}$

----------------------------------------

I'm just wondering if anyone can help explain the problem to me as I'm having trouble getting started. I don't really know where to find a thread. Any tips or hints would be appreciated. I guess the partial derivative it's asking for is the third parital derivative with respect to u, v, and w?

-Austin

2. Originally Posted by auslmar
Hello, everyone. It's been awhile since I've visited the boards. I hope all is well for you and yours.

The problem I'm dealing with is as follows:

-----------------------------------------

$z = u{(v-w)}^{\frac{1}{2}}$

Find the indicated partial derivative:

$\frac{\partial^3z}{\partial u \partial v\partial w}$

----------------------------------------

I'm just wondering if anyone can help explain the problem to me as I'm having trouble getting started. I don't really know where to find a thread. Any tips or hints would be appreciated. I guess the partial derivative it's asking for is the third parital derivative with respect to u, v, and w?

-Austin
Note that $\frac{\partial^3z}{\partial u\partial v\partial w}=\frac{\partial}{\partial u}\left(\frac{\partial^2z}{\partial v\partial w}\right)=\frac{\partial}{\partial u}\left[\frac{\partial}{\partial v}\left(\frac{\partial z}{\partial w}\right)\right]$

So first [partially] differentiate z with respect to w, then [partially] differentiate that result with respect to v, and finally [partially] differentiate that result with respect to u.

Does that make sense?

--Chris

3. Originally Posted by Chris L T521
Note that $\frac{\partial^3z}{\partial u\partial v\partial w}=\frac{\partial}{\partial u}\left(\frac{\partial^2z}{\partial v\partial w}\right)=\frac{\partial}{\partial u}\left[\frac{\partial}{\partial v}\left(\frac{\partial z}{\partial w}\right)\right]$

So first [partially] differentiate z with respect to w, then [partially] differentiate that result with respect to v, and finally [partially] differentiate that result with respect to u.

Does that make sense?

--Chris
It makes sense to me. Tell me if I've gone wrong.

This is what I did with your suggestion:

$[(\frac{-u}{2{{\sqrt{{v-w}}}}})(\frac{u}{2{\sqrt{{v-w}}}})]\sqrt{{u-v}}$ = $\frac{-u^2}{4{\sqrt{{v-w}}}}$

Is this the right idea? Thanks for your help.

4. Originally Posted by auslmar
Hello, everyone. It's been awhile since I've visited the boards. I hope all is well for you and yours.

The problem I'm dealing with is as follows:

-----------------------------------------

$z = u{(v-w)}^{\frac{1}{2}}$

Find the indicated partial derivative:

$\frac{\partial^3z}{{\partial}u{\partial}v{\partial }w}$

----------------------------------------

I'm just wondering if anyone can help explain the problem to me as I'm having trouble getting started. I don't really know where to find a thread. Any tips or hints would be appreciated. I guess the partial derivative it's asking for is the third parital derivative with respect to u, v, and w?

-Austin
Originally Posted by auslmar
It makes sense to me. Tell me if I've gone wrong.

This is what I did with your suggestion:

$[(\frac{-u}{2{{\sqrt{{v-w}}}}})(\frac{u}{2{\sqrt{{v-w}}}})]\sqrt{{u-v}}$ = $\frac{-u^2}{4{\sqrt{{v-w}}}}$

Is this the right idea? Thanks for your help.
$z = u{(v-w)}^{\frac{1}{2}}$

Thus, $\frac{\partial z}{\partial w}=-\frac{u}{2(v-w)^{\frac{1}{2}}}$

Now, $\frac{\partial}{\partial v}\left[\frac{\partial z}{\partial w}\right]=\frac{\partial^2z}{\partial v \partial w}=\frac{u}{4(v-w)^{\frac{3}{2}}}$

Finally, $\frac{\partial}{\partial u}\left[\frac{\partial}{\partial v}\left[\frac{\partial z}{\partial w}\right]\right]=\frac{\partial^3z}{\partial u\partial v \partial w}=\color{red}\boxed{\frac{1}{4(v-w)^{\frac{3}{2}}}}$

Does this make sense?

--Chris

5. Oh yeah, I see now. I was trying to multiply them all together. Thanks for your help.