# Understanding directional derivatives

I am confusing myself when it comes to directional derivatives and gradients. The gradient of a function shows the direction of the greatest change. So when we chose a unit vector as the direction to find the directional derivative, from my understanding, what we are doing is projecting the the gradient vector onto the directional vector. That projection is the directional derivative at that give point say $(a,b)$. Say also that the gradient is the vector g and that the vector u is the unit direction vector. Why is it that the gradient vector is projected on the unit vector thus giving us a the directional derivative in the direction of the unit vector u and not vice versa, that is the unit vector projects on the gradient giving us the directional derivative in that direction (of the gradient vector).

I have mixed everything up in my head and any help to clear things up is appreciated! Thanks :)

• The two projection methods give the same result: $|g||u|\cos\theta$. – Yves Daoust Feb 1 '15 at 11:34
• But isn't that in the direction of the gradient vector if you do it the other way round? – Nash Feb 1 '15 at 11:48
• It doesn't matter, you are taking a dot product so the direction disappears. The directional derivative is a scalar. – Yves Daoust Feb 1 '15 at 15:39

We start with the definition of directional derivative. Let $f:\mathbb{R}^n\to\mathbb{R}$ be a function, and let $x,v\in\mathbb{R}^n$. One can define another function $f_{x,v}:\mathbb{R}\to\mathbb{R}$ by$$f_{x,v}(t)=f(x+tv).$$Intuitively, $f_{x,v}$ is the restriction of $f$ to the line through $x$ in the direction $v$. The directional derivative of $f$ at $x$ in the direction $v$ is then given by$$df_x(v)=\left.\frac{d}{dt}\right|_{t=0}f_{x,v}(t).$$The differential of $f$ at $x$ is the linear functional $df_x:\mathbb{R}^n\to\mathbb{R}$ that eats a vector and returns the directional derivative in the direction of this vector.
The gradient of $f$ at $x$, $\nabla f_x$, is defined to be the unique vector that satisfies for every $v$ $$df_x(v)=\langle\nabla f_x,v\rangle.$$So, whenever one uses the gradient to calculate a directional derivative, the right operation is the inner product.