# Finding the second derivative of multvariable function

1. Nov 1, 2007

### newguy1234

Hello I am trying to figure out the second derivative of

$$\frac{\partial^2 z}{\partial x^2}$$ and $$\frac{\partial^2 z}{\partial t^2}$$

z(u) and u=x-vt

i found the first derivate of $$\frac{\partial z}{\partial x}$$ to be
$$\frac{\partial z}{\partial x}$$= $$\frac{\partial z}{\partial u}$$ * $$\frac{\partial u}{\partial x}$$ = $$\frac{\partial z}{\partial u}$$ *1

and

$$\frac{\partial z}{\partial t}$$ = $$\frac{\partial z}{\partial u}$$ * $$\frac{\partial u}{\partial t}$$= $$\frac{\partial z}{\partial u}$$*v

after this i am clueless on how to compute
($$\frac{\partial }{\partial t}$$)$$\frac{\partial z}{\partial u}$$*v
and
($$\frac{\partial }{\partial x}$$)$$\frac{\partial z}{\partial u}$$*1

i am new here am hoping to receive some help i will greatly appreciate it!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 1, 2007

### samee

Are you trying to say that z(u)=x-vt?

3. Nov 1, 2007

### newguy1234

i guess the answer should be for $$\frac{\partial^2 z}{\partial t^2}$$ = v^2 * $$\frac{\partial^2t }{\partial u^2}$$

im not sure at all how to get this tho any help?

4. Nov 1, 2007

### newguy1234

i guess im trying to say that z is a function of u and u is a function of x and t

5. Nov 2, 2007

### HallsofIvy

No, since u= x- vt, $\partial u/\partial t= -v$, not v.

The "v" and "1" are constants. So you are just differentiating a function of x and t again:
$$\frac{\partial}{\partial x}\frac{\partial z}{\partial u}= \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial x}$$
and, of course, $\partial u/\partial x$ is 1.

Likewise
$$\frac{\partial}{\partial t}\frac{\partial z}{\partial u}= \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial t}$$
and $\partial u/\partial t$ is -v.