# Partial Differentiation: second partial derivative

1. Mar 31, 2014

### jellicorse

I am not quite sure how $$\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)$$
$$=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)$$

comes to $$\frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$...

I would have evaluated this to be the same but without the $\frac{\partial z}{\partial x}$ term. i.e.:

$$u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$

I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?

2. Mar 31, 2014

### Staff: Mentor

How are z, u, and v related? I can guess, but that really is information you should supply.

3. Mar 31, 2014

### jellicorse

Sorry. $z =f (x,y)$, with

$$x=\frac{1}{2}(u^2-v^2)$$ and $y =uv$

4. Mar 31, 2014

### Staff: Mentor

Do you understand this part?
$$\frac{\partial z}{\partial u}=\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)$$

Or are you having difficulties with this part?
$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$

$$= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)$$

5. Mar 31, 2014

### jellicorse

This is the part that I don't understand. The first part is OK...

6. Mar 31, 2014

### Staff: Mentor

$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$
$$= \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) + \frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

Now use the product rule on each part.

7. Apr 1, 2014

### jellicorse

Thanks a lot, Mark... I am getting closer but have one problem. Having done it several times over, the same thing keeps happening:

Using the product rule $(uv)' = u'v + uv'$ I get one extra term.

I'll type it piece by piece for clarity:

$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) = (uv)'$$

Gives:

$$\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})$$

And

$$\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} ) = (uv)'$$

Gives:

$$\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )$$

So, in total I get:

$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x})+\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

$$=\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})+\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )$$

8. Apr 1, 2014

### Staff: Mentor

You have z = f(x, y), where x = g(u, v) and y = h(u, v). Here x and y are dependent variables (depending on u and v), and u and v are independent variables. Since u and v are independent (are unrelated to each other), $\frac{\partial v}{\partial u} = 0$, so your third term vanishes. By the same reasoning, $\frac{\partial u}{\partial v}$ would also be zero.

Nice job on the LaTeX formatting!

9. Apr 1, 2014

### jellicorse

Oh, I see now. Thanks a lot, it all makes perfect sense now!