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

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- Dec 12th 2012, 07:19 PMameerulislamwhat will this d/dx y^3 be?
d/dx y^3 will equal to zero or 3y^2 ?

- Dec 12th 2012, 07:35 PMMarkFLRe: what will this d/dx y^3 be?
If y is a function of x, then you also need to apply the chain rule to get:

- Dec 12th 2012, 08:13 PMameerulislamRe: what will this d/dx y^3 be?
this kind of confuses me. if there was a constant instead of y like:

d/dx 4^3

it would have been Zero right?

yes I'm doing implicit differentiation and I don't know what it is though, - Dec 12th 2012, 08:17 PMameerulislamRe: what will this d/dx y^3 be?
and what would be

d/dx y

just y. - Dec 12th 2012, 08:17 PMMarkFLRe: what will this d/dx y^3 be?
Yes, if

*y*is a constant, then it would be zero. However, in an implicit relation, it is implied that*y*depends on*x*. So, you would use the form in my first post above. - Dec 12th 2012, 08:38 PMameerulislamRe: what will this d/dx y^3 be?
so if they declare y as constant

this would become zero as well right? d/dx y^3 - Dec 12th 2012, 08:45 PMMarkFLRe: 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.

- Dec 12th 2012, 08:49 PMProve ItRe: what will this d/dx y^3 be?
It depends on whether you are trying to do a total derivative or a partial derivative. If it's a total derivative, use implicit differentiation. If it's a partial derivative, treat y as a constant.

- Dec 13th 2012, 07:43 AMHallsofIvyRe: 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.