# Product Rule

1. Oct 23, 2005

### Midas_Touch

How do I carry out the product rule for

xy'z'

Is it possible to do the product rule with y'z' and after that multiply it by x?

2. Oct 23, 2005

### Jameson

What variable is this respect to? Is it $$\frac{d}{dx},\frac{d}{dy}$$,etc.

3. Oct 23, 2005

### Midas_Touch

It's not respect to any variable. It's just three separate variable... for instance it can be xyz.... so i was thinking that i first do the product rule for yz which is y'z + z'y and then i use this result and multiply it by x and x'. I am not sure if I am allowed to do this.

4. Oct 23, 2005

### Jameson

The product rule extends for three variables as follows. Let's say we have three functions that are in terms of x, we'll call them f(x),g(x), and h(x).

$$\frac{d(f(x)g(x)(h(x))}{dx}=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$$.

5. Oct 23, 2005

### Midas_Touch

Thank you, I really appreciate your help.

6. Feb 10, 2008

### digital19

I have a question regarding the product rule.

Our modern physics textbook asks us to derive the following:

$$d(\gamma mu)=m(1- \frac{u^2}{c^2})^{-3/2} du$$

Is it implied that this is with respect to u? I can see the chain rule here, but I'm not sure precisely how this differentiation is done.

7. Feb 10, 2008

### HallsofIvy

Staff Emeritus
First, please do not post a new question in someone else's thread. That is very rude- start your own thread.

Second, strictly speaking, the right hand side is a differential with respect to u while the left side is just the differential of $\gamma mu$ and is not "with respect to" anything. If you were to rewrite it as
$$\frac{d(\gamma mu)}{du}= m\left(1-\frac{u^2}{c^2}\right)^{-3/2}$$
then the derivative on the left is with respect to u.

I can't tell you how to derive it since you haven't said what it is to be derived from- which, hopefully, you will do in a separate thread.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook