# Partial Derivatives

1. Feb 14, 2005

I am sorta skipping around at my own pace in Courant's Calculus book and came across partial derivatives. Are they geometrically the intersection of a plane and a surface? Why do we keep only one variable changing and the other variables fixed? Is it basically the definition of the derivative? For example

$$\frac{\partial y}{\partial x}$$ when $$f(x,y) = \sqrt{x^{2} + y^{2}}$$ Ok so would I consider y to be a constant when we want to find $$f_{x}$$ and vice versa for $$f_{y}$$? Ok so this is what I did:

$$f_{x} = \frac{1}{2}\sqrt{x^2+y^2}2x$$. But the answer is:

$$f_{x} = \frac{x}{\sqrt{x^2+y^2}}$$ and the same is true for $$f_{y} = \frac{y}{\sqrt{x^2+y^2}}$$ except the variables are reversed.

Any help is appreciated!

Thanks

Last edited: Feb 14, 2005
2. Feb 14, 2005

### Nylex

Did you mean df/dx and not dy/dx? When you evaluate the partial derivative wrt a particular variable, you keep the others constant as you said.

Your answer is wrong as you've not differentiated with the chain rule properly:

d/dx [(x^2 + y^2)^1/2)] = (1/2)[(x^2 + y^2)^(-1/2)].2x = x/(x^2 + y^2)^1/2 as required.

3. Feb 14, 2005

whoops I must have not noticed that I typed LaTex wrong.

4. Feb 14, 2005

### dextercioby

Please use the notation of Lagrange properly.
$$f'_{x}=:\frac{\partial f}{\partial x}$$

,where the last is C.G.Jacobi's notation.

Daniel.

5. Feb 14, 2005

### hypermorphism

I'm sure you can think of many multivariable functions where you are only interested in what happens when one particular variable is varied (Ie., gas laws, economics, etc.). In addition, partials are useful in general form as they make studying the derivative, and thus properties of a function easier, as the derivative can be written in terms of the partial derivatives of f.

6. Feb 15, 2005

### arildno

That is, $$\frac{\partial{f}}{\partial{x}}\mid_{(\vec{x}=(x_{0},y_{0}))$$ is found by by restricting your attention to f's behaviour along the line $$y=y_{0}$$ (where "y" is obviously a constant!)