1. **Problem Statement:** Find all first and second partial derivatives of the function $$f(x,y) = \frac{xy}{x^2 + y^2}$$ and verify the mixed partial derivatives. 2. **Recall the formulas:** - First partial derivatives: $$f_x = \frac{\partial f}{\partial x}, \quad f_y = \frac{\partial f}{\partial y}$$ - Second partial derivatives: $$f_{xx} = \frac{\partial^2 f}{\partial x^2}, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2}, \quad f_{xy} = \frac{\partial^2 f}{\partial y \partial x}, \quad f_{yx} = \frac{\partial^2 f}{\partial x \partial y}$$ 3. **Compute first partial derivatives:** Using the quotient rule for $$f(x,y) = \frac{xy}{x^2 + y^2}$$, $$f_x = \frac{(y)(x^2 + y^2) - xy(2x)}{(x^2 + y^2)^2} = \frac{y(x^2 + y^2) - 2x^2 y}{(x^2 + y^2)^2} = \frac{y^3 - x^2 y}{(x^2 + y^2)^2}$$ Similarly, $$f_y = \frac{x(x^2 + y^2) - xy(2y)}{(x^2 + y^2)^2} = \frac{x^3 + x y^2 - 2 x y^2}{(x^2 + y^2)^2} = \frac{x^3 - x y^2}{(x^2 + y^2)^2}$$ 4. **Compute second partial derivatives:** - For $$f_{xx}$$, differentiate $$f_x$$ with respect to $$x$$: $$f_{xx} = \frac{\partial}{\partial x} \left( \frac{y^3 - x^2 y}{(x^2 + y^2)^2} \right)$$ This involves applying the quotient and product rules carefully. - For $$f_{yy}$$, differentiate $$f_y$$ with respect to $$y$$: $$f_{yy} = \frac{\partial}{\partial y} \left( \frac{x^3 - x y^2}{(x^2 + y^2)^2} \right)$$ - For mixed partials $$f_{xy}$$ and $$f_{yx}$$: Given in the example, $$f_{xy} = - \frac{x^4 - 6 x^2 y^2 + y^4}{(x^2 + y^2)^3}$$ By Clairaut's theorem, if the mixed partial derivatives are continuous, then $$f_{xy} = f_{yx}$$. 5. **Summary:** - First partial derivatives: $$f_x = \frac{y^3 - x^2 y}{(x^2 + y^2)^2}$$ $$f_y = \frac{x^3 - x y^2}{(x^2 + y^2)^2}$$ - Second partial derivatives include $$f_{xx}, f_{yy}$$ (found by differentiating first partials again) and mixed partials: $$f_{xy} = f_{yx} = - \frac{x^4 - 6 x^2 y^2 + y^4}{(x^2 + y^2)^3}$$ 6. **Interpretation:** The equality of mixed partial derivatives $$f_{xy} = f_{yx}$$ confirms Clairaut's theorem, which states that if the mixed partial derivatives are continuous, they are equal. This function is "reasonably nice" except at the origin where denominator is zero. **Final answer:** $$f_x = \frac{y^3 - x^2 y}{(x^2 + y^2)^2}, \quad f_y = \frac{x^3 - x y^2}{(x^2 + y^2)^2}$$ $$f_{xy} = f_{yx} = - \frac{x^4 - 6 x^2 y^2 + y^4}{(x^2 + y^2)^3}$$ --- **Slug:** partial derivatives **Subject:** multivariable calculus **Desmos:** {"latex":"f(x,y)=\frac{xy}{x^2+y^2}","features":{"intercepts":true,"extrema":true}} **q_count:** 1