# Homework Help: Chain rule someone help

1. May 11, 2010

### xstetsonx

chain rule someone help plz

1. let z=y^2-x^2cosy; x=t^3 y=cost, find dz/dt

2. let z=(x-y)^3;x=u+2v,y=2u-v,find dz/dv

my attempt:
so i know the chain rule is (dz/dx)dx+(dz/dy)dy
1. should i substitute the x and y into t first or should i do the partial derivative first?

2. same thing what should i do first?

2. May 11, 2010

### lanedance

Re: chain rule someone help plz

that's not quite an equation...

but guessing at what you're attempting to do, why not write dx & dy in terms of dt for the first?

3. May 11, 2010

### xstetsonx

Re: chain rule someone help plz

k so i am just going to do the first one since they are similar
dx/dt=3t^2------------dy/dt=-sint
and then do dz/dx and dz/dy??? after that substituted the t into x and y?
i am confuse

4. May 11, 2010

### lanedance

Re: chain rule someone help plz

so before doing anything try writing out the chain rule in full
$$\frac{dz(x,y)}{dt} = \frac{\partial z(x,y)}{\partial x}\frac{dx}{dt} + \frac{\partial z(x,y)}{\partial y}\frac{dy}{dt}$$

5. May 11, 2010

### Staff: Mentor

Re: chain rule someone help plz

lanedance is right, xstetsonx. You are likely to become hopelessly confused if you don't distinguish between derivatives and partial derivatives.

In the first problem, z is a function of x and y, so the derivatives of z with respect to x or y are both partial derivatives. Since x and y are functions of a single variable t, then it makes sense to talk about dx/dt and dy/dt. Ultimately, z is a function of a single variable t, so you can write dz/dt, and its formula is as shown by lanedance.

In the second problem, z is a function of x and y, but x and y are each function of u and v. This means that z is ultimately a function of both u and v, so the chain rule in this case will involve only partial derivatives, and there will be two of them - one for the partial of z with respect to u and one for the partial of z with respect to v.

Here's the one for the partial of z with respect to u.
$$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$$

I leave it to you to figure out the partial of z with respect to v, $$\frac{\partial z}{\partial v}$$.

6. May 11, 2010

### xstetsonx

Re: chain rule someone help plz

so i don't substitute t into dz/dx and dz/dy??? and for the second i don't substitute v into dz/dx and dz/dy either??? so my answers will contain x,y,t for the first one and x,y,v?

7. May 11, 2010

### Staff: Mentor

Re: chain rule someone help plz

For 1, you can and should do the problem both ways. Getting the same answer for both methods would tend to confirm that you have done the problem correctly.

Please use the proper notation. There is no dz/dx in the first problem, nor is there dz/dy. Click on the partials that I showed in my previous post to see how it is done in LaTeX.

Alternatively, you can use subscripts to indicate partial derivatives, with zx meaning the same thing as $$\frac{\partial z}{\partial x}$$.