Question

I need to find the partial derivatives of f(x,y) = xy / sqr(x²+y²).
Thanks for your help.

Answer 1

Task. Find the partial derivatives of

$$f(x, y) = \frac{x y}{\sqrt{x^2 + y^2}}.$$

Solution. We will use the following two formulas:

$$\left(\frac{f}{g}\right)' = \frac{f' g - f g'}{g^2}$$ $$\sqrt{x'} = (x^{1/2})' = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.$$ $$\begin{aligned} f_x'(x, y) &= \left(\frac{x y}{\sqrt{x^2 + y^2}}\right)'_x = \frac{(x y)'_x \sqrt{x^2 + y^2} - x y (\sqrt{x^2 + y^2})'_x}{\sqrt{x^2 + y^2}} \\ &= \frac{y \sqrt{x^2 + y^2} - x y \frac{(x^2 + y^2)'_x}{2 \sqrt{x^2 + y^2}}}{x^2 + y^2} = \frac{y \sqrt{x^2 + y^2} - x y \frac{2 x}{2 \sqrt{x^2 + y^2}}}{x^2 + y^2} \\ &= \frac{y (x^2 + y^2) - x^2 y}{(x^2 + y^2) \sqrt{x^2 + y^2}} = \frac{y x^2 + y^3 - x^2 y}{(x^2 + y^2)^{\frac{3}{2}}} = \frac{y^3}{(x^2 + y^2)^{\frac{3}{2}}} \end{aligned}$$

Since $f(x, y) = f(y, x)$, we have that

$$f_y'(x, y) = f_x'(y, x) = \frac{x^3}{(x^2 + y^2)^{\frac{3}{2}}}.$$

Answer.

$$f_x'(x, y) = \frac{y^3}{(x^2 + y^2)^{\frac{3}{2}}}, \qquad f_y'(x, y) = \frac{x^3}{(x^2 + y^2)^{\frac{3}{2}}}.$$