# Are directional derivatives symmetric around the gradient?

I have read that the definition of the directional derivative of a function $$f$$ on a vector space at a point $$x$$ along a vector $$\vec{v}$$ is given by:

$$D_\vec{v}f(x)=\nabla f(x) \cdot\vec{v}=|\nabla f(x)||\vec{v}|\cos(\theta)$$

where $$D_\vec{v}f$$ is the directional derivative of $$f$$ in the direction $$\vec{v}$$, $$\nabla f$$ is the gradient of $$f$$ and $$\theta$$ is the angle between $$\vec{v}$$ and $$\nabla f(x)$$. The last equality follows by the definition of the dot product.

I don't get this... does this mean that for any vector $$\vec{a}$$ with equal length as $$\vec{v}$$ such that the angle between $$\nabla f(x)$$ and $$\vec{a}$$ is $$\theta$$, $$D_\vec{v}f(x)=D_\vec{a}f(x)$$?!

This seems crazy! The directional derivative is kind of symmetric about the gradient?! Surely this can't be true - if I'm defining a function can't I define it to be anything I like with no symmetry like this?

This makes no sense to me... have I misunderstood?

I think what you are missing is that for the purpose of the derivative you can replace the function with a (hyper-)plane. In other words, the first-order Taylor approximation $$f(x) \approx f(x_0) + \nabla f(x_0)(x-x_0)$$ has the same value and derivatives as $$f(x)$$ at $$x = x_0$$. The plane is entirely determined by the gradient and you cannot change the shape of the plane to be some arbitrary shape.
You can define a function with arbitrary directional derivatives that don't satisfy the given equation, but such a function is not considered differentiable at $$x$$ (even if $$D_{\vec v}f(x)$$ exists for every nonzero $$\vec v$$). The symmetry you noticed results from the linear relationship between $$D_{\vec v}f(x)$$ and $$\vec v$$, which is essentially the definition of differentiability of $$f$$ at $$x$$.