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

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.]
• 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 Apr 7 '18 at 18:03
• 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.) Apr 7 '18 at 18:08
• 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. Apr 7 '18 at 18:38
• 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 $f'(x;d)=\nabla f^T d$ ? Apr 8 '18 at 21:41