Directional derivative: Better explained than in Serge Lang’s book

This is an attempt to explain directional derivatives better than how it is explained in Serge Lang’s seminal book “A Second Course in Calculus”.

The directional derivative of $f(X)$ is $($ grad $f(X)).A$ , where $A$ is a unit vector in the direction that we’re interested in.

Let us suppose we need to find the derivative of function $f:\Bbb{R}^n\to \Bbb{R}$ along the direction $X+tR$ , where points $X,R\in \Bbb{R}^n$ and $t\in \Bbb{R}$ is the parameter. The derivative will obviously be $\lim\limits_{\|tR\|\to 0}\frac{f(X+tR)-f(X)}{\|tR\|}$ .

Lang’s book mentions it as $\frac{d f(X(t))}{dt}$ , where $X(t)$ is obviously $X+tR$ . This is theoretically an inaccurate assertion. But we will now work out why the formula of the directional derivative comes out to be the same.

Let us determine why $\frac{d f(X(t))}{dt}=\lim\limits_{t\to 0}f(X+tR)-f(X)$ .

We know that $f(X+tR)-f(X)=(D_1(s)\times\|R\|\times t)+\|tR\|g(X,R,t)$ such that $\lim\limits_{t\to 0}g(X,R,t)=0$ . Here, $s\in (X,X+tR)$ (the Mean Value Theorem has been used here). Note that $\lim\limits_{t\to 0}D_1(s)=\frac{\partial f}{\partial X}$ .

Although the above formula is valid, it is not very helpful, as determining $\frac{\partial f}{\partial X}$ would be difficult. What should we do now?

Let $X=(x_1,x_2,x_3,\dots,x_n)$ and $R=(r_1,r_2,r_3,\dots,r_n)$ . Then $\lim\limits_{t\to 0}f(X+tR)-f(X)=D_1(X).r_1.t+D_2(X).r_2.t+\dots+D_n(x).r_n.t$ . Hence, $\lim\limits_{t\to 0}\frac{f(X+tR)-f(X)}{t}=($ grad $f(X)).R$ . Divide both sides by $\|R\|$ to get a unit vector in the direction of $\|R\|$ . You get the required formula.

Summing up the above argument, if $R$ is the required direction in which you want to determine the slope, $\lim\limits_{\|tR\|\to 0}\frac{f(X+tR)-f(X)}{\|tR\|}$ is indeed what you’re looking for! Your intuition was correct. The world is happy and pink again. For calculational purposes, we use $\frac{1}{\|R\|}($ grad $f(X)).R$ , as $\lim\limits_{\|tR\|\to 0}\frac{f(X+tR)-f(X)}{\|tR\|}=($ grad $f(X)).R$