# 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?

## 2 Answers

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.

Consider the simplest case, which is a function of a single variable. The plane in this case is the tangent line, and the directional derivatives are in the left and right directions. And they are the same because the tangent line is the same line in both directions. Changing the shape of the function would not change this.

• Ah I see... thank you for the intuition this makes a lot of sense to me! – MasonTep Mar 11 at 19:57

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 $$x$$.

• Ah that explains a lot thank you! – MasonTep Mar 11 at 19:57