what will this d/dx y^3 be?

• Dec 12th 2012, 07:19 PM
ameerulislam
what will this d/dx y^3 be?
d/dx y^3 will equal to zero or 3y^2 ?
• Dec 12th 2012, 07:35 PM
MarkFL
Re: 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:

$\displaystyle \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}$
• Dec 12th 2012, 08:13 PM
ameerulislam
Re: 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 PM
ameerulislam
Re: what will this d/dx y^3 be?
and what would be
d/dx y
just y.
• Dec 12th 2012, 08:17 PM
MarkFL
Re: 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 PM
ameerulislam
Re: 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 PM
MarkFL
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
Prove It
Re: 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 AM
HallsofIvy
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.