Directional Derivatives

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In order to find how a function $f:\mathbb {R}^n\rightarrow \mathbb {R}$ changes with each of the input variables, we defined the partial derivatives of $f$ . For example, when $n=2$ , we defined the partial derivative of $f$ with respect to $x$ to be $f_x(x,y)=\lim _{h\rightarrow 0}\frac {f(x+h,y)-f(x,y)}{h}.$ Here, we thought of $y$ as a constant, which made $f$ only a function of $x$ , and reduced us to a single variable derivative. This told us how a small change in $x$ would after the value of $f$ , if we kept $y$ constant. In other words, the partial derivatives described how the function $f$ was changing in the positive $x$ -direction and in the positive $y$ -direction.

But what if we want to find how $f$ changes if we change both $x$ and $y$ ? One possible way to do this would be to increase $x$ and $y$ by the same amount, which would be equivalent to finding how $f$ changes as we increase $x$ and $y$ along the line $y=x$ .

Alternatively, we could decrease $y$ by twice as much as $x$ . This would be equivalent to finding how $f$ changes as decreasing $x$ and $y$ along the line $y=2x$ .

As you can see, there are many different ways that we can change $x$ and $y$ , corresponding to different directions in the $xy$ -plane. In order to determine how $f$ changes as we move in all of these different directions, we will now define directional derivatives.

Directional derivatives

We would like to compute the instantaneous rate of change of a function $f:\mathbb {R}^n\rightarrow \mathbb {R}$ at a point as we move in some given direction in $\mathbb {R}^n$ . We will model our definition after partial derivatives and single variable derivatives, and use a unit vector $\vec {v}$ to describe the direction.

Consider a function $f:\mathbb {R}^n\rightarrow \mathbb {R}$ , a point $\vec {a}\in \mathbb {R}^n$ , and a direction given by a unit vector $\vec {v}\in \mathbb {R}^n$ . Then we define the directional derivative of $f$ at $\vec {a}$ in the direction of $\vec {v}$ to be $D_{\vec {v}}f(\vec {a}) = \lim _{h\rightarrow 0}\frac {f(\vec {a}+h\vec {v})-f(\vec {a})}{h},$ provided this limit exists.

Noticing that by looking at $f(\vec {a}+h\vec {v})$ , we are finding the value of $f$ when we move a small distance, $h$ , in the direction of $\vec {v}$ from the point $\vec {a}$ .

When computing directional derivatives, it’s important to remember that the direction must be given by a unit vector. Otherwise, the length of the vector will change the value of the limit above. If you’d like to find a directional derivative in a direction given by a non-unit vector $\vec {w}$ , you should normalize $\vec {w}$ to unit length.

We’ll compute the directional derivative of $f(x,y) = x^2y+y^2$ at $\vec {a}=(2,0)$ , in the direction of $(3,4)$ .

Since $(3,4)$ isn’t a unit vector, we need to normalize it. Since $\|(3,4)\| = \sqrt {3^2+4^2}=5$ , we’ll use the vector $\vec {v}=\left (\frac {3}{5},\frac {4}{5}\right )$ to compute our desired directional derivative.

$\begin{align*} D_{\vec{v}}f(\vec{a}) &= \lim_{h\rightarrow 0}\frac{f(\vec{a}+h\vec{v})-f(\vec{a})}{h}\\ &= \lim_{h\rightarrow 0}\frac{f\left((2,0) + h\left(\frac{3}{5},\frac{4}{5}\right)\right)-f(2,3)}{h}\\ &= \lim_{h\rightarrow 0}\frac{f\left(2+\frac{3}{5}h,\frac{4}{5}h \right)-f(2,0)}{h}\\ &= \lim_{h\rightarrow 0}\frac{\left(2+\frac{3}{5}h\right)^2\cdot \frac{4}{5}h + \left(\frac{4}{5}h\right)^2-0}{h}\\ &= \lim_{h\rightarrow 0}\left(2+\frac{3}{5}h\right)^2\cdot \frac{4}{5} + \left(\frac{4}{5}\right)^2h\\ &= 4\cdot \frac{4}{5}\\ &= \frac{16}{5}. \end{align*}$

Fortunately, we won’t always need to resort to evaluating directional derivatives using the limit definition. We’ll soon see how we can use the gradient to compute directional derivatives.