Moredimensional Derivative of Trigonometric Functions with Substitution

Question

I've searched for a similiar question/problem, but I haven't found one. The closest one was this question. Though it is just about substitution in more dimension but not in connection with trigonometric functions.

I have a two-dimensional function: $$f(x,y) = \arctan\left(\frac{x}{y}\right) $$

I have to find the following partial derivatives: $$\frac{\partial f(x,y)}{\partial x}$$ $$\frac{\partial f(x,y)}{\partial y}$$

Now I have some trouble doing this in the "classic" way.

My idea was to substitute the term or rather substitute the following: $$\frac{x}{y} = u$$ This seems quite nice, because that way I can easily build the derivative of $arctan(x)$.

I don't know how to go from there, though.

The question(s):

• Is it even possible to substitute that way?
• Is this even an efficent way?
• How to solve that eqation at all?

Thank you in advance,

Max K.

Answer 1

This is just a matter of using the chain rule. Since $\arctan'(x)=\frac1{1+x^2}$ and since $\frac\partial{\partial x}\left(\frac xy\right)=\frac1y$, you get that$$\frac\partial{\partial x}\arctan\left(\frac xy\right)=\frac1{1+\left(\frac xy\right)^2}\times\frac1y=\frac y{x^2+y^2}.$$Can you take it from here?

Answer 2

by the Chaine roule we obtain $$\frac{\partial f(x,y)}{\partial x}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\cdot \frac{1}{y}$$ and $$\frac{\partial f(x,y)}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\cdot \left(-\frac{x}{y^2}\right)$$