My professor always writes the directional derivative with the condition that the functionis $C^1$ smooth.
i see him write many times the directional derivative written as $f'(x;d)=\nabla f^T d$, but right next to it he writes "because $\beta - C^1 smooth$", where $\beta$ is the Lipschitz constant, and $C^1$ means that it is continuously differentiable up to and including the first derivative.

Is the directional derivative $f'(x;d)$ ALWAYS equal to $\nabla f^T d$? or only under certain conditions?

EDIT And does "differentiable" mean with regards to a certain direction? Or if a function is differentiable, then does that mean it must be differentiable in all directions? If the former, then does $C^1$ mean it is differentiable in ALL directions?


That equality holds when $f$ is differentiable at $x$ (in the sense of having an appropriately good linear approximation at $x$). The standard theorem that one always proves is that $C^1$ implies differentiability. [And, of course, $(\nabla f(x))^\top$ will be the linear map that gives the good approximation.]

  • $\begingroup$ So then which is a stronger form of the directional derivative? the one with the gradient dotted with $d$? or the one with the directional derivative $f'$? I often see the former used instead of the latter $\endgroup$
    – makansij
    Apr 7 '18 at 18:03
  • $\begingroup$ The directional derivative may well exist (and be computed from the definition) even when $f$ fails to be differentiable. (You might find some of the discussion in my lectures 22 and 23 helpful.) $\endgroup$ Apr 7 '18 at 18:08
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    $\begingroup$ An example is $f(x,y) = x^2 + |y|$. It is not differentiable for $y=0$, but the directional derivative along the $x$ axis exists. $\endgroup$
    – LinAlg
    Apr 7 '18 at 18:38
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    $\begingroup$ @LinAlg: More interesting is a function where all directional derivatives exist, but the function isn't even continuous! $\endgroup$ Apr 7 '18 at 18:59
  • $\begingroup$ Right, so then a few clarifications: 1) you're telling me that it is possible that the derivative can exist in ALL directions i.e. ($\forall d$), and yet there might not exist a linear operator to express it (i.e. $\nabla f^T*d$ is invalid/does not exist)? 2) C^1 smoothness implies differentiability in all d irections? If so, then can we say if a function is C^1 smooth, then that means that the linear operator $\nabla f$ DOES EXIST, and we can say $f'(x;d)=\nabla f^T d$ ? $\endgroup$
    – makansij
    Apr 8 '18 at 21:41

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