Exercise 2186Calculus › Multivariable calculus › Partial derivativesL1Added 3 Nov 2026

Partial derivatives exist but function is not differentiable

#differentiability#counterexample#partial derivatives

Statement

Show that $f(x,y)=\dfrac{xy}{x^2+y^2}$ for $(x,y)\neq(0,0)$ , $f(0,0)=0$ , has partial derivatives at $(0,0)$ but is not continuous there.

Hint

Compute partial derivatives along the axes; check continuity along $y=x$ .

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Solution

Partial derivatives. $f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-0}{h}=0$ ; $f_y(0,0)=0$ .

Continuity. Along $y=x$ : $f(x,x)=\frac{x^2}{2x^2}=\frac12$ , but along axes $f=0$ . Limit doesn't exist.

Conclusion. $f$ is not continuous, hence not differentiable, though partials exist at $(0,0)$ . This shows partial derivatives alone are insufficient for differentiability.

Partial derivatives exist but function is not differentiable

Statement

Solution

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