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

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- Dec 12th 2012, 07:19 PM #1

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- Dec 12th 2012, 07:35 PM #2

- Dec 12th 2012, 08:13 PM #3

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- Dec 12th 2012, 08:17 PM #4

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- Dec 12th 2012, 08:38 PM #6

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- Dec 12th 2012, 08:45 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 $\displaystyle \frac{dy}{dx}$ is, this is normally what you are trying to express as a function of x and y.

- Dec 12th 2012, 08:49 PM #8

- Dec 13th 2012, 07:43 AM #9

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## 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.

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