Partial Differentiation -- y deleted or ignored?

[Prof. 27]

Homework Statement

Given: z = f(x,y) = x^2-y^2
To take the partial derivative of f with respect to x hold y constant then take the derivative of x.
∂f/∂x = 2x

What I don't understand is why such would equal 2x, when the y is still there it just isn't variable and is ignored. Wouldn't it be:

2x - y^2

where y is a constant squared?
In another example this kind of thing occurred:
z = f(x,y) = x^3*y
∂f/∂x = 3x^2*y

where y is a constant unknown.

So why the different pertaining to y in these two answers?

Thanks so much, Bertrand Russell

Homework Equations

None really.

The Attempt at a Solution

http://en.wikipedia.org/wiki/Partial_derivative
http://www.centerofmath.org/mc_pdf/sec2_1.pdf [/B]

 

[cnh1995]

Homework Statement

Given: z = f(x,y) = x^2-y^2
To take the partial derivative of f with respect to x hold y constant then take the derivative of x.
∂f/∂x = 2x

What I don't understand is why such would equal 2x, when the y is still there it just isn't variable and is ignored. Wouldn't it be:

2x - y^2

where y is a constant squared?
In another example this kind of thing occurred:
z = f(x,y) = x^3*y
∂f/∂x = 3x^2*y

where y is a constant unknown.

So why the different pertaining to y in these two answers?

Thanks so much, Bertrand Russell

Homework Equations

None really.

The Attempt at a Solution

http://en.wikipedia.org/wiki/Partial_derivative
http://www.centerofmath.org/mc_pdf/sec2_1.pdf [/B]

We take the partial derivative of the entire function..So for f(x,y)=x²-y², the derivative of x^2 is 2x but the derivative of y^2 w.r.t.x must be 0 since it is constant..Same goes for 3x^2*y..

 

[Prof. 27]

Got it. Thanks so much.

 

[PeroK]

Homework Statement

Given: z = f(x,y) = x^2-y^2
To take the partial derivative of f with respect to x hold y constant then take the derivative of x.
∂f/∂x = 2x

What I don't understand is why such would equal 2x, when the y is still there it just isn't variable and is ignored. Wouldn't it be:

2x - y^2

where y is a constant squared?
In another example this kind of thing occurred:
z = f(x,y) = x^3*y
∂f/∂x = 3x^2*y

where y is a constant unknown.

So why the different pertaining to y in these two answers?

Thanks so much, Bertrand Russell

Homework Equations

None really.

The Attempt at a Solution

http://en.wikipedia.org/wiki/Partial_derivative
http://www.centerofmath.org/mc_pdf/sec2_1.pdf [/B]

First, in both cases y is treated as a constant. For example:

$\frac{d}{dx}(x^2 + a^2) = 2x$

And

$\frac{d}{dx}(x^3a) = 3xa$

On your main point: a partial derivative (wrt x) is the gradient of the curve you get when you hold y constant. You could think of taking y = 0, 1, 2... and see what curves you get. Whatever value of y you choose, you get a parabola $z = x^2 + n^2$ where $n$ is whatever value of y you are considering. This curve has the usual derivative and does not get steeper as y increases: it's the same shape for any value of y.

Whether y appears in the partial derivative depends on whether the value of y affects the derivative. For $z = x^2 + y^2$ it doesn't: the partial derivative is independent of y. But for $z = x^3y$ the partial derivative wrt x does depend on y.

It's exactly the same as whether a constant $a$ affects the ordinary derivative of a function of x.