What is the simplest way to determine if function $f(x,y)=\sqrt[3]{x^3+y^3}$ is differentiable at point $(0,0)$?
Simplest way to determine differentiability at given point
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multivariable-calculus
1 Answers
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Since $f$ has partial derivatives at $(0,0)$, the simplest way to show $f$ is not differentiable is to find a direction $(x,y)$ such that $f(tx,ty)$ is not $f(0,0)+t(x\partial_xf(0,0)+y\partial_yf(0,0))+o(t)$ when $t\to0$.
Here $f(0,0)=0$ and $(x,y)=(1,1)$ yields $f(t,t)=\sqrt[3]2t$ and $x\partial_xf(0,0)+y\partial_yf(0,0)=2$ but $\sqrt[3]2t$ is not $2t+o(t)$. This is enough to prove that $f$ is not differentiable at $(0,0)$.