Exercise · ZW-DIFFERENTIABILITY-SEVERAL-VARIABLES-007 · ★★☆ Intermediate

Let \(f(x,y)=\sqrt{x^2+y^2}\). Why is it not differentiable at the origin?

Choose one option.

Show answer
  • A. Because \(f_x(0,0)\) does not exist

Explanation

The quotient \(\frac{f(h,0)-f(0,0)}{h}=\frac{|h|}{h}\) is \(1\) from the right and \(-1\) from the left, so \(f_x(0,0)\) does not exist (the graph is a cone with a vertex). It is continuous.

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Review the concept: Differentiability in several variables

A function \(f(x,y)\) is differentiable at a point \((a,b)\) if its graph has a tangent plane there: the error of the linear approximation is negligible compared with the distance to the point.

\[\lim_{(h,k)\to(0,0)}\frac{f(a+h,b+k)-f(a,b)-f_x(a,b)\,h-f_y(a,b)\,k}{\sqrt{h^2+k^2}}=0\]

More practice problems: differentiability in several variables

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