# Why is the dotproduct of direction and gradient the directional derivative? ($\nabla_\hat{v} f = \nabla f \bullet \hat{v}$)

I have seen the claim that the directional derivative of $$f$$ in the direction $$\hat{v}$$, where $$||\hat{v}|| = 1$$, (denoted $$\nabla_\hat{v} f$$) is equal to the gradient of $$\nabla f$$ dotted with $$\hat{v}$$.


Let $$f : \R^n \to \R$$

I accept that

$$\nabla_{\hat{v}} f = \limit{h}{0} \frac{f(x + h\hat{v}) - f(x)}{h}$$

for making intuitive sense. Furthermore I know that

$$\pderiv{f}{x_i} = \limit{h}{0} \frac{f(x + h\hat{i}) - f(x)}{h}$$

where $$\hat{i}$$ is the unit vector of the $$i$$-th dimension. I also know that

$$\nabla f = \left( \pderiv{f}{x_1}, \dots, \pderiv{f}{x_n} \right)$$

Now I want to show that $$\nabla_\hat{v} f = \nabla f \bullet \hat{v}$$:

\begin{align*} \nabla f \bullet \hat{v} & = \left( \pderiv{f}{x_1}, \dots, \pderiv{f}{x_n} \right) \bullet v\hat{v}\\ &= \sum_{i = 1}^{n} \pderiv{f}{x_i} \cdot \hat{v}_i\\ &= \sum_{i = 1}^{n} \limit{h}{0} \frac{f(x + h\hat{i}) - f(x)}{h} \cdot \hat{v}_i\\ &= \sum_{i = 1}^{n} \limit{h}{0} \frac{\hat{v}_if(x + h\hat{i}) - \hat{v}_if(x)}{h}\\ \end{align*}

And now I am stuck. I don't see a way to transform the last line into $$\limit{h}{0} \frac{f(x + h\hat{v}) - f(x)}{h}$$ to reach $$\nabla_{\hat{v}} f$$

Can you help me out here?

# Edit

I now conceptually understand why the dotproduct of direction and gradient is the directional derivative:

Let's say we have a differentiable function $$f$$ from $$\mathbb{R}^2$$ to $$\mathbb{R}^3$$ mapping the $$xy$$-plane into the $$xyz$$-space and we want to know the directional derivative of $$f$$ at a point $$p = (x', y')$$ for some vector $$u$$.

First what we need to know is, how much $$f$$ changes in $$x$$ direction and how much it changes in $$y$$ direction.

Then we need to realize that for small distances whatever direction we go along the surface of $$f$$, the total change in height is the sum of the change in height in the $$x$$ component of our direction and the change in height in $$y$$ component in our direction.

But now we can weight the partial derivatives for the $$x$$ and $$y$$ direction with the components of $$u$$ to get their individual contributions for the direction in which $$u$$ is pointing!

So if $$u$$ has an $$x$$-component of $$u_x$$ we weight the partial derivative of $$f$$ for the $$x$$ direction accordingly. When we do the same for $$y$$ we get:

$$f'(x') \cdot u_x + f'(y') \cdot u_y$$

which is in fact $$\nabla f \bullet u$$

I understood this after listeing to this lecture:

• hint you will have to use that f is (totally) differentiable! – VanillaThunder Mar 28 '18 at 17:35
• Could you hint a little more? This is not an assignment or anything... I tried this proof out of my own interest. – user3578468 Mar 28 '18 at 17:39

There is more in the gradient than the $n$-tuple of partial derivatives!

The function $f:\>{\mathbb R}^n\to{\mathbb R}$ is differentiable at $x$ if there is a linear map $L:\>{\mathbb R}^n\to{\mathbb R}$ such that $$f(x+X)-f(x)=LX\ +o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{1}$$ This map is then uniquely determined, and is denoted by $df(x)$ (or similar). Since $df(x)$ in this case is a linear functional, by linear algebra there is a vector $a\in{\mathbb R}^n$ such that $$df(x).X= a\cdot X\qquad(X\in{\mathbb R}^n)\ .$$ This vector $a$ is called the gradient of $f$ at $x$, and is denoted by $\nabla f(x)$. Coordinatewise we have $$\nabla f(x)=\left({\partial f\over\partial x_1},\ldots,{\partial f\over\partial x_n}\right)_x\ ,$$ but we shall not need this. Anyway, we now can write $(1)$ in the form $$f(x+X)-f(x)=\nabla f(x)\cdot X\ +o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{2}$$ Now let a unit vector $e$ be given. Letting $X:=t\,e$ in $(2)$ implies $$f(x+ t e)-f(x)=t\>\nabla f(x)\cdot e +\ o\bigl(|t|\bigr)\qquad(t\to0)\ ,$$ hence $$D_e f(x):=\lim_{t\to0+}{f(x+ t e)-f(x)\over t}=\nabla f(x)\cdot e\ ,$$ as claimed.

Let $v:=(v_1,v_2,...,v_n)$ and $x:=(x_1,...,x_n)$. Define $y$ with $y_k:=x_k+hv_k$ for $h>0$ and $k=1,...,n$ so that $y=x+hv$. Then $$\frac{df(y_1,...,y_n)}{dh}=\frac{\partial f}{\partial y_1}\frac{dy_1}{dh}+...+\frac{\partial f}{\partial y_n}\frac{dy_n}{dh}=\frac{\partial f}{\partial y_1}v_1+...+\frac{\partial f}{\partial y_n}v_n=\langle\nabla_y f,v\rangle$$ On the other hand $$\frac{df(y_1,...,y_n)}{dh}=\frac{df(x+hv)}{dh}:=g'(h)$$ where $g(h):=f(x+hv)$. Therefore $$g'(0)=\lim_{h\to 0}\frac{f(x+hv)-f(x)}{h}=\frac{df}{dh}\Big|_{h=0}=\langle\nabla_y f,v\rangle\Big|_{h=0}=\langle\nabla_x f,v\rangle$$

• Why is the introduction of $g$ necessary? And what does $\langle\nabla_y f,v\rangle\Big|_{h=0}$ denote? – user3578468 Mar 28 '18 at 18:08
• $g$ is to make exposition easier since at the end what you really want is $g'(0)$. The other thing you asked follows from the first line since $y=x+hv$. – Arian Mar 28 '18 at 18:23