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If \(f(0,0)=0\) and
f(x, y)=\frac{x y}{x^{2}+y^{2}} \quad \text { if }(x, y) \neq(0,0)
prove that \(\left(D_{1} f\right)(x, y)\) and \(\left(D_{2} f\right)(x, y)\) exist at every point of \(R^{2}\), although \(f\) is not continuous at \((0,0)\).
in Mathematics by Diamond (55.6k points) | 205 views

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At any point \((x, y)\) except \((0,0)\) the differentiability of \(f(x, y)\) follows from the rules for differentiation and the principles of Chapter 5 . At \((0,0)\) it is a routine computation to verify that both partial derivatives equal zero:
\left(D_{1} f\right)(0,0)=\lim _{h \rightarrow 0} \frac{f(h, 0)-f(0,0)}{h}=0 .
However, \(f(x, y)\) is not continuous at \(\left(0,0\right.\), since \(f(x, x)=\frac{1}{2}\) for all \(x \neq 0\), and hence \(\lim _{x \rightarrow 0} f(x, x)=\frac{1}{2} \neq f(0,0)\).
by Diamond (55.6k points)

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