Partial differentiation find df/dx and df/dy

[fionamb83]

Homework Statement

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
I think this is wrong please help!

 

[Pere Callahan]

You should use the chain rule

<br /> \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}<br />

 

[Mark44]

I know tan^-1(x)=1/1+X^2 ...

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

 

[HallsofIvy]

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".

 

[fionamb83]

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

 

[Dick]

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

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?

 

[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!