Exercise 1795Calculus › Multivariable calculus › Partial derivativesL1Added 3 Nov 2026

Differentiability of a piecewise function

#differentiability#continuity#piecewise

Statement

Let
$f(x,y)=\begin{cases}\dfrac{xy}{\sqrt{x^2+y^2}}&\text{if }(x,y)\neq(0,0),\\0&\text{if }(x,y)=(0,0).\end{cases}$
Show that $f$ is continuous at $(0,0)$ and compute its partial derivatives there.

Hint

Bound $|xy|\le(x^2+y^2)/2$ and use the definition of partials.

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Solution

Continuity. $|f(x,y)|\le\frac{1}{2}\sqrt{x^2+y^2}\to0$ ; so $f$ is continuous at the origin.

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

Conclusion. $f$ is continuous with $f_x(0,0)=f_y(0,0)=0$ . (However, $f$ is not differentiable at $(0,0)$ : check along $y=x$ .)

Differentiability of a piecewise function

Statement

Solution

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