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$\\f(x,y) = \begin{cases} \frac{xy}{x^2 + y^2}, & \text{if $(x,y) \ne (0,0)$} \\ 0, & \text{if $(x,y) = (0,0)$} \end{cases}$

Using the definition of partial derivatives: \begin{align*} f_x(x,y) &= \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h} \\&= \lim_{h\to 0} \frac{\frac{(x+h)y}{(x+h)^2 + y^2}-\frac{xy}{x^2 + y^2}}{h}\\&= \lim_{h\to 0} \frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)} \end{align*}

I'm not sure how to get rid of the variable $h$. I wanted to know exactly how I could further simplify in order to get $\frac{y(y^2-x^2)}{(x^2 + y^2)^2}$?

Skm
  • 2,296

2 Answers2

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Notice that \begin{align}&\phantom{ = }\frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\ &= \frac{\color{red}{xy(x^2 +y^2)} + hy(x^2 +y^2) \color{red}{-xy(x^2 + y^2)} - xy(2hx+ h^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\ &=\frac{hx^2y+hy^3 -2hx^2y-h^2xy}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\ &=\frac{hy(y^2 -x^2-hx)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\ &=\frac{y(y^2 -x^2-hx)}{(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\ &\to \frac{y(y^2-x^2)}{(x^2+y^2)^2}, \end{align} as $h\to 0$.

MSDG
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Continuing your calculation, we get $$f_x(x,y)=\lim_{h\rightarrow 0}\frac{hy(y^2-x^2-hx)}{h(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$ Or $$f_x(x,y)=\lim_{h\rightarrow 0}\frac{y(y^2-x^2-hx)}{(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$

Can you finish from here?

pi66
  • 7,164