1

Let's says $f(x,y,z)$ is differentiable real function at point $(0,0,0)$.

We know that $f_y(0,0,0) = f_x(0,0,0) = 0$, and $f(t^2,2t^2,3t^2) = 4t^2$ for every $t>0$.

what can we say about $f_z(0,0,0)$ ? It seems to me that we can say $f_z(0,0,0) = \dfrac{4}{3} $, but I struggle in proving this officially..

d_e
  • 1,565

1 Answers1

2

Since $f$ is differentiable at the origin, it is continuous, so letting $t \to 0$ shows that $f(0,0,0) = 0$.

If $s>0$, we have $f(s,2s,3s) = 4s$, and so $\lim_{s \downarrow 0} {f(s,2s,3s) - f(0,0,0) \over s} = 4$. That is, the directional derivative in the direction $(1,2,3)^T$ is $4$.

Since $f$ is differentiable at the origin, we have $\lim_{s \to 0} {f((0,0,0)+s(1,2,3)) - f(0,0,0) \over s} = Df((0,0,0))((1,2,3)^T)$, where $Df((0,0,0))$ is the derivative at $(0,0,0)$.

Since $Df((0,0,0)) = (f_x(0,0,0), f_y(0,0,0), f_z(0,0,0))$, this gives $3f_z(0,0,0) = 4$.

copper.hat
  • 172,524
  • when you write: $Df((0,0,0))((1,2,3))$ you mean directional derivative dot product of $(1,2,3)$ ? – d_e Jun 21 '15 at 17:03
  • I am a little sloppy above with row/columns. I mean the derivative multiplied (as in matrix multiply) by the direction. The derivative is a row. – copper.hat Jun 21 '15 at 17:09
  • ok. I think I understand. thanks. – d_e Jun 21 '15 at 17:10