# Partial differentiation find df/dx and df/dy

1. Nov 14, 2008

### fionamb83

1. The problem statement, all variables and given/known data
For the function of two variables f(x,y)=tan^-1(y/x)
find df/dx and df/dy

I know i just differentiate with respect to x and then to y but I'm stuck on the tan^-1(y/x)
I know tan^-1(x)=1/1+X^2 when I applied this with respect to x I get 1/-1+y

2. Nov 14, 2008

### Pere Callahan

You should use the chain rule

$$\frac{\partial}{\partial x}\tan^{-1}\left(\frac{y}{x}\right)=\left.\frac{d}{du}\tan^{-1}(u)\right|_{u=\frac{y}{x}}\frac{\partial}{\partial x}\frac{y}{x}$$

3. Nov 14, 2008

### Staff: Mentor

That's not true.
What is true is that
$$\frac{d}{dx} tan^{-1}(x) = \frac{1}{1 + x^2}$$

4. Nov 14, 2008

### HallsofIvy

Staff Emeritus
The derivative of tan-1(u) is
$\frac{1}{1+ u^2}$
with u= y/x, that is
$\frac{1}{1+ \frac{y^2}{x^2}}$
for the derivative with respect to x or y those are multiplied by the derivative of y/x with respect to x and the derivative of y/x with respect to x "respectively".

5. Nov 14, 2008

### fionamb83

so df/dx = -2x/1+y^2 ?

6. Nov 14, 2008

### Dick

How did you get that? Halls told you it's 1/(1+y^2/x^2) times the x derivative of y/x. What's the x derivative of y/x?

7. Nov 14, 2008

### fionamb83

Sorry just saw what I did there. Oops. the x derivative of y/x is -y/x^2. Sorry confused myself there. Thanks for the help everyone!