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$f(x,y)= xy^3/(x^3+y^6)$ if $(x,y)\neq 0$, $f(x,y)=0$ if $(x,y)=0$.

Prove that $f'(0; a)$ exists for every vector $a$. I know how to find the directional derivative from limit equation, but don't know how to prove it. Please help.

Tomás
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Ashika
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1 Answers1

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So, you should fix a generic vector $(a_1,a_2) \in\mathbb{R}^2$ and prove that $$ \lim_{\epsilon \to 0} \frac{f(0+\epsilon a_1,0+\epsilon a_2)-f(0,0)}{\epsilon} $$ exists as a real number. Now, $$ f(0+\epsilon a_1,0+\epsilon a_2)-f(0,0) = \frac{\epsilon a_1 \epsilon^3 a_2^3}{\epsilon^3 a_1^3 + \epsilon^3 a_2^3} = \frac{\epsilon^4 a_1 a_2^3}{\epsilon^3 ( a_1^3 + a_2^3)} = \epsilon \frac{a_1 a_2^3}{ a_1^3 + a_2^3}. $$ Now divide by $\epsilon$ and you find that $$ \frac{\partial f}{\partial (a_1,a_2)}(0,0)=\frac{a_1 a_2^3}{ a_1^3 + a_2^3}. $$ Since $(a_1,a_2) \neq (0,0)$, the limit exists as a real number.

Siminore
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