d/dx y^3 will equal to zero or 3y^2 ?

Results 1 to 9 of 9

- December 12th 2012, 08:19 PM #1

- Joined
- Jul 2012
- From
- Bangladesh
- Posts
- 55

- December 12th 2012, 08:35 PM #2

- December 12th 2012, 09:13 PM #3

- Joined
- Jul 2012
- From
- Bangladesh
- Posts
- 55

- December 12th 2012, 09:17 PM #4

- Joined
- Jul 2012
- From
- Bangladesh
- Posts
- 55

- December 12th 2012, 09:17 PM #5

- December 12th 2012, 09:38 PM #6

- Joined
- Jul 2012
- From
- Bangladesh
- Posts
- 55

- December 12th 2012, 09:45 PM #7
## Re: what will this d/dx y^3 be?

Normally you would only treat y as a constant when doing partial differentiation. And you don't know what is, this is normally what you are trying to express as a function of x and y.

- December 12th 2012, 09:49 PM #8

- December 13th 2012, 08:43 AM #9

- Joined
- Apr 2005
- Posts
- 16,449
- Thanks
- 1864

## Re: what will this d/dx y^3 be?

Normally, "y" is used as a variable or function of x. You have been told repeatedly that if y is

**independent**of x, then so is f(y) for any f and so df(y)/dx= 0. If, however, y is itself some function of x, then df(y)/dx= (df/dy)(dy/dx), by the chain rule.