Chain rule of partial derivatives for composite functions.

Question

If I have a function of the form $$f(x^2+y^2)$$ How do I find the partial derivatives $$\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}$$ I am not sure how $f(x^2+y^2)$ behaves. I am assuming it should of the form $$g(x,y)\cdot2y\quad\text{or}\quad h(x,y)\cdot2x$$

Answer 1

Since $f(.)$ is univariate function we have$$\dfrac{\partial f}{\partial x}=2xf'(x^2+y^2)$$similarly$$\dfrac{\partial f}{\partial y}=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$

Answer 2

First, I want to use some different notation: $$f(x^2+y^2)=:f(g(x,y))=(f\circ g)(x,y) $$

Now the partial derivatives can be computed by the chain rule (in multiple dimensions): $$\frac{∂f}{∂x} = \frac{∂f}{∂g}\frac{∂g}{∂x} = \frac{∂f}{∂g}2x. $$ $$\frac{∂f}{∂y} = \frac{∂f}{∂g}\frac{∂g}{∂y} = \frac{∂f}{∂g}2y. $$

Since you don't know anything else about $f$, you can't simplify these terms further.