How can the product rule be applied to a function with 3 variables?

[bumclouds]

hey there,

At the moment at school I'm doing Implicit Differentiation.

If i had for Instence 2y'yx + 6x = 0

how can I use the product rule on the first step when there are 3 variables?

Cheers-
Andy

 

[Curious3141]

It's simple, you can derive it yourself using the rule for 2 variables :

[tex]\frac{d}{dx}(uvw) = uv\frac{dw}{dx} + w\frac{d}{dx}(uv) = uv\frac{dw}{dx} + w(u\frac{dv}{dx} + v\frac{du}{dx}) = uv\frac{dw}{dx} + vw\frac{du}{dx} + uw\frac{dv}{dx}[/tex]

Doesn't the form look simple ? Differentiate one variable at a time, multiply by all other variables, then take it through all the combinations, and add everything up. Of course, it applies for any number of variables.

Can you do your problem now ? (But to tell you the truth, the problem you gave is unclear - if you're trying to find out y' in terms of y and x, all you need to do is rearrange the terms).

 

[bumclouds]

Thankyou very much. Yes, I was trying to find y' in terms of the other stuff. Previously, I had only learned the product rule for two variables.. like u'.v + u.v'.

 

[Astronuc]

If I had for instance 2y'yx + 6x = 0

how can I use the product rule on the first step when there are 3 variables?

Well, unless something is mis-typed, there are two variables if one assumes y = y(x) and y' = dy(x)/dy, and y would be dependent on x which is an independent variable.

In the above equation, "2x" factors out leaving y'y + 3 = 0, or y' = dy/dx = -3/y, or y dy = -3 dx.

Curious3141 provided the correct chain rule for 3 dependent variables, the fourth x being independent.

 

[benorin]

That would be three dependent variables.

 

[HallsofIvy]

It would have been better to say "3 factors".

(uvw)'= u'vw+ uv'w+ uvw'.