How is the Product Rule Applied to Derivatives in the Form xy'z'?

[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?

 

[Jameson]

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

 

[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.

 

[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).

 

[Midas_Touch]

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).

Thank you, I really appreciate your help.

 

[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.

 

[HallsofIvy]

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.