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Find if $$f(x,y)=\sqrt[4]{x^4+y^4}$$ $$g(x,y)=(f(x,y))^2$$ are differentiable in $(0,0)$.

well, $g(x)$ is clearly $\sqrt{x^4+y^4}$, so I guess the answer will be similar to $f(x)$.

$f_x=x^3/(x^4+y^4)^{3/4}$, but what now? $f_x$ has no value in $(0,0)$, so can it be differentiable there?

J.R.
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iTayb
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2 Answers2

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HINT:

The sufficient conditions for differentiability of a function $f(x,y)$ at $(a,b)$ are-

  1. The partial derivatives of $f$ i.e. $f_x$ and $f_y$ must exist at $(a,b)$.
  2. The partial derivatives of $f$ must be continuous at $(a,b)$.
SchrodingersCat
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Using the definition: in both cases, as $(0,0)$ is a local minimum, the (possible) differential will be zero if it exists: $$ \lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-(0,0)(x,y)}{\|(x,y)\|} = \lim_{(x,y)\to(0,0)}\frac{\sqrt[4]{x^4+y^4}}{\sqrt{x^2+y^2}} = \lim_{r\to 0}\frac{r(\cos^4\theta+\sin^4\theta)}r. $$ ($\not\exists$ because depends of the angle) $$ \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)-(0,0)(x,y)}{\|(x,y)\|} = \lim_{(x,y)\to(0,0)}\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}} = \lim_{r\to 0}\frac{r^2(\cos^4\theta+\sin^4\theta)}r = 0. $$ And only $g$ is differentiable.

EDIT: clarification of second limit: $$ \left|\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}\right| = \frac{r^2(\cos^4\theta+\sin^4\theta)}r \le 2 r = 2\|(x,y)\|\to 0, $$ and the squeezing theorem proves that the lim is 0.

Martín-Blas Pérez Pinilla
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  • The last three equalities are all true, but the way the answer is written is at the very best misleading and I dare say wrong. Changing to polar coordinates only proves that certain sublimits exist, not that all sublimits exist, has desired. See this. – Git Gud Jan 23 '16 at 20:33
  • @GitGud, the function is $r\times$ a bounded function of angle. Tomorrow will edit for grater clarity. – Martín-Blas Pérez Pinilla Jan 23 '16 at 22:07