Determine if the derivative $f'(0;a)$ exists for every vector $a$ and compute it's value in terms of the components of $a$.

Question

Let $f(x,y) = \dfrac {xy^3}{x^3 + y^6 }$ if $(x,y) \ne (0,0)$ and define $f(0,0)=0$

Determine if the derivative $f'(0;a)$ exists for every vector $a$ and compute it's value in terms of the components of $a$.

Attempt:

$D_1f(0,0) = \lim_{h\rightarrow 0} \dfrac {h.0}{h^3}=0$.

Similarly : $D_2f(0,0) =0 \implies \nabla f(0) = 0 \widehat i + 0 \widehat j$.

Hence, if $a = (a_1,a_2), f~'(0;a) = \nabla f(0). a = 0$

That means, the directional derivative along any direction is $0$?

Could someone please advise.

Thank you very much for your help in this regard.

Answer 1

Following Apostol, we shall define the directional derivative: $$ f'(\mathbf{a};\mathbf{y}) = \lim_{h \to 0}\frac{f(\mathbf{a} + h\mathbf{y}) - f(\mathbf{a})}{h} $$ where $\mathbf{y}$ is the vector along which we are deriving, and $\mathbf{a}$ is the point about which we are taking the derivative (boldface indicates a vector quantity). In terms of the problem you posted, $\mathbf{a} = (0,0)$ and $\mathbf{y} = (a_{1},a_{2})$. It is just a matter of subbing in these numbers. (Apologies for the confusing notation, but I am trying to follow the one used in the textbook. Just be careful that $\mathbf{a}$ is not $(a_{1},a_{2})$ here, even though that would be nice.) We find: $$ \lim_{h \to 0}\frac{f(\mathbf{a} + h\mathbf{y}) - f(\mathbf{a})}{h} = \lim_{h \to 0}\frac{f((0,0) + h(a_{1},a_{2})) - f(\mathbf{0})}{h} $$ We already know $f(0,0)$, so we can simplify a bit: $$ \lim_{h \to 0}\frac{f((0,0) + h(a_{1},a_{2})) - f(\mathbf{0})}{h} = \lim_{h \to 0}\frac{f(h*a_{1},h*a_{2}))}{h} $$ $$ = \lim_{h \to 0}\frac{h^{4}a_{1}a_{2}^{3}}{h*(h^{3}a_{1}^{3} + h^{6}a_{2}^{6})} $$ Now we can factor out and cancel off the $h$'s (haven't yet taken the limit) and we arrive at: $$ =\lim_{h \to 0}\frac{a_{1}a_{2}^{3}}{a_{1}^{3} + h^{3}a_{2}^{6}} $$ Taking the limit zaps the term containing $h$ in the denominator and you can cancel a bit off to find your value of the derivative: $$ f'(\mathbf{0};\mathbf{a}) = \frac{a_{2}^{3}}{a_{1}^{2}} $$ Since we were able to find it in this fashion and didn't come across any discontinuities or undefined points, we shall say that it clearly exists.