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Partial derivatives can exist while the function is not differentiable

Functions of Several VariablesAnalysis & Linear Algebrafree

The standard counterexample, worked — and why 'both partials exist' is the weakest hypothesis in multivariable calculus.

Partial derivatives can exist while the function is not differentiableinteractivefree

Step 1 / 4The candidate

Let

f(x,y)=xyx2+y2 for (x,y)≠(0,0),f(0,0)=0.f(x,y) = \frac{xy}{x^2+y^2} \text{ for } (x,y) \neq (0,0), \qquad f(0,0)=0.

Is ff differentiable at the origin?

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