# Thread: A second order partial derivative

1. ## A second order partial derivative

I'm given this equation:

w= root(u^2 + v^2)

I'm to find all the second order partial derivatives.

Formulas are swirling in my mind and I need someone to show me at least one first and second order partial derivative step by step so I can make sense of the rest of them. I know how to differentiate but its hard to remember all the details. Do I have to use the Chain Rule?

And remember that this is "partial" differentiation so it is close to but not quite the same as ordinary differentiation.

Thanks for the help.

2. Originally Posted by Undefdisfigure
I'm given this equation:

w= root(u^2 + v^2)

I'm to find all the second order partial derivatives.

Formulas are swirling in my mind and I need someone to show me at least one first and second order partial derivative step by step so I can make sense of the rest of them. I know how to differentiate but its hard to remember all the details. Do I have to use the Chain Rule?

And remember that this is "partial" differentiation so it is close to but not quite the same as ordinary differentiation.
First find the first derivatives:
$\displaystyle \frac{\partial w}{\partial u} = \frac{1}{2 \sqrt{u^2 + v^2}} \cdot 2u$
and
$\displaystyle \frac{\partial w}{\partial v} = \frac{1}{2 \sqrt{u^2 + v^2}} \cdot 2v$

You have three second derivatives: $\displaystyle \frac{\partial ^2 w}{\partial u^2}$, $\displaystyle \frac{\partial ^2 w}{\partial u \partial v}$, $\displaystyle \frac{\partial ^2 w}{\partial v^2}$.

The only one that might give you conceptual problems is the mixed derivative so I will mention that
$\displaystyle \frac{\partial ^2 w}{\partial u \partial v} = \frac{\partial}{\partial u} \left ( \frac{\partial w}{\partial v} \right ) = \frac{\partial}{\partial v} \left ( \frac{\partial w}{\partial u} \right )$

-Dan

3. Originally Posted by Undefdisfigure
I'm given this equation:

w= root(u^2 + v^2)

I'm to find all the second order partial derivatives.

Formulas are swirling in my mind and I need someone to show me at least one first and second order partial derivative step by step so I can make sense of the rest of them. I know how to differentiate but its hard to remember all the details. Do I have to use the Chain Rule?

And remember that this is "partial" differentiation so it is close to but not quite the same as ordinary differentiation.
It wants you to find:

$\displaystyle \frac{\partial^2}{\partial u^2}w$,

$\displaystyle \frac{\partial^2}{\partial v^2}w$,

$\displaystyle \frac{\partial^2}{\partial u \partial v}w$

and

$\displaystyle \frac{\partial^2}{\partial v \partial u}w$

RonL

4. I use the quotient rule for the first derivative dw/du = 1/(2root(u^2+v^2)) x 2u or 2u/(2root(u^2+v^2)) and get the strange answer:

(4root(u^2 + v^2) - 8u^2))/8(u^2 + v^2)^3/2

while the book gives the answer: Wuu (or d^2w/du^2) = v^2/(u^2 + v^2)^3/2

I know I jammed up somewhere, but I need someone to show me where.

5. I was able to simplify the answer I got with the quotient rule a little bit. But the answer still appears incorrect. I got: (root(u^2 + v^2) - 2u^2)) / 2(u^2 + v^2)^3/2

Look above to see the book's answer. Something a little fishy here.

6. Originally Posted by Undefdisfigure
I use the quotient rule for the first derivative dw/du = 1/(2root(u^2+v^2)) x 2u or 2u/(2root(u^2+v^2)) and get the strange answer:

(4root(u^2 + v^2) - 8u^2))/8(u^2 + v^2)^3/2

while the book gives the answer: Wuu (or d^2w/du^2) = v^2/(u^2 + v^2)^3/2

I know I jammed up somewhere, but I need someone to show me where.
Originally Posted by topsquark
$\displaystyle \frac{\partial w}{\partial u} = \frac{1}{2 \sqrt{u^2 + v^2}} \cdot 2u$
Let's rewrite this thing a little:
$\displaystyle w = (u^2 + v^2)^{1/2}$
and
$\displaystyle \frac{\partial w}{\partial u} = u(u^2 + v^2)^{-1/2}$

So
$\displaystyle \frac{\partial ^2 w}{\partial u ^2} = 1 \cdot (u^2 + v^2)^{-1/2} + u \cdot \frac{-1}{2} (u^2 + v^2)^{-3/2} \cdot 2u$

$\displaystyle = (u^2 + v^2)^{-1/2} - u^2 (u^2 + v^2)^{-3/2}$

$\displaystyle = (u^2 + v^2)^{-1/2} (1 - u^2(u^2 + v^2)^{-1})$

$\displaystyle = (u^2 + v^2)^{-1/2} \left ( 1 - \frac{u^2}{u^2 + v^2} \right )$

$\displaystyle = (u^2 + v^2)^{-1/2} \left ( \frac{u^2 + v^2 - u^2}{u^2 + v^2} \right )$

$\displaystyle = (u^2 + v^2)^{-1/2} \left ( \frac{v^2}{u^2 + v^2} \right )$

$\displaystyle = \frac{v^2}{(u^2 + v^2)^{3/2}}$

-Dan

7. I guess you just have to be crafty sometimes eh? I guess they were looking for the most simplified answer.

Thanks.

8. Hello, Undefdisfigure!

I'll do this one in baby-steps.
I hope you can use this as a template for the others.

$\displaystyle w \:=\:\sqrt{u^2 + v^2}$

Find all the second-order partial derivatives.

We have: .$\displaystyle w \;=\;\left(u^2+v^2\right)^{\frac{1}{2}}$

. . $\displaystyle \frac{\partial w}{\partial u} \;=\;\frac{1}{2}\left(u^2+v^2\right)^{-\frac{1}{2}}\cdot 2u \;=\;\frac{u}{(u^2+v^2)^{\frac{1}{2}}}$

. . $\displaystyle \frac{\partial w}{\partial v} \;=\;\frac{1}{2}\left(u^2+v^2\right)^{-\frac{1}{2}}\cdot 2v \;=\;\frac{v}{(u^2+v^2)^{\frac{1}{2}}}$

$\displaystyle \frac{\partial^2w}{\partial u^2} \;=\;\frac{(u^2+v^2)^{\frac{1}{2}}\cdot1 - u\cdot\frac{1}{2}(u^2+v^2)^{-\frac{1}{2}}\cdot2u}{u^2+v^2} \;=\;\frac{(u^2+v^2)^{\frac{1}{2}} - u^2(u^2+v^2)^{-\frac{1}{2}}}{u^2+v^2}$

. . Multiply top and bottom by $\displaystyle (u^2+v^2)^{\frac{1}{2}}\!:\;\;\frac{(u^2+v^2) - u^2}{(u^2+v^2)^{\frac{3}{2}}} \quad\Rightarrow\quad\boxed{ \frac{\partial^2w}{\partial u^2} \:=\:\frac{v^2}{(u^2+v^2)^{\frac{3}{2}}}}$

$\displaystyle \frac{\partial^2w}{\partial v^2}\;=\;\frac{(u^2+v^2)^{\frac{1}{2}}\cdot1 - v\cdot\frac{1}{2}(u^2+v^2)^{-\frac{1}{2}}\cdot2v}{u^2+v^2} \;=\;\frac{(u^2+v^2)^{\frac{1}{2}} - v^2(u^2+v^2)^{-\frac{1}{2}}}{u^2+v^2}$

. . multiply top and bottom by $\displaystyle ((u^2+v^2)^{\frac{1}{2}}\!:\;\;\frac{u^2+v^2) - v^2}{(u^2+v^2)^{\frac{3}{2}}}\quad\Rightarrow\quad \boxed{\frac{\partial^2w}{\partial v^2}\;=\;\frac{u^2}{(u^2+v^2)^{\frac{3}{2}}} }$

For the "mixed" partial, we have: .$\displaystyle \frac{\partial w}{\partial u}\;=\;\frac{u}{(u^2+v^2)^{\frac{1}{2}}} \;=\;u\cdot(u^2+v^2)^{-\frac{1}{2}}$

. . and we differentiate with respect to $\displaystyle v$: .$\displaystyle \frac{\partial^2w}{\partial v\partial u} \;=\;u\cdot\left(\text{-}\frac{1}{2}\right)(u^2+v^2)^{-\frac{3}{2}}\cdot 2v$

Therefore: .$\displaystyle \boxed{\frac{\partial^2w}{\partial v\partial u}\;=\;-\frac{uv}{(u^2+v^2)^{\frac{3}{2}}}}$

9. ## Something a little more difficult...

Verify that the function u = 1/root(x^2 + y^2 + z^2) is a solution of the three-dimensional Laplace equation Uxx + Uyy + Uzz = 0.

I used the methods that were stated in this thread however for Uxx I came up with (-2x^2 + y^2 + z^2) / (x^2 + y^2 + z^2)^-5/2

I guess I'll divide that (or multiply it) and see what I get for Uyy and Uzz..tell me if using the above method for the second partial derivative will be good here.