
# 2.4: Directional Derivatives and the Gradient

[ "article:topic", "Gradient Vector", "Directional Derivatives", "authorname:mcorral" ]

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For a function $$z = f (x, y)$$, we learned that the partial derivatives $$\dfrac{∂f}{∂x}\text{ and} \dfrac{∂f}{∂y}$$ represent the (instantaneous) rate of change of $$f$$ in the positive $$x$$ and $$y$$ directions, respectively. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative.

Definition 2.5: directional derivative

Let $$f (x, y)$$ be a real-valued function with domain $$D$$ in $$\mathbb{R}^2$$, and let $$(a,b)$$ be a point in $$D$$. Let $$\textbf{v}$$ be a unit vector in $$\mathbb{R}^2$$. Then the directional derivative of $$\textbf{f}$$ at $$\mathbf{(a,b)}$$ in the direction of $$\mathbf{v}$$, denoted by $$D_v f(a,b)$$, is defined as

$D_v f(a,b)=\lim \limits_{h \to 0}\dfrac{f((a,b)+h\textbf{v})-f(a,b)}{h}\label{Eq2.8}$

Notice in the definition that we seem to be treating the point $$(a,b)$$ as a vector, since we are adding the vector $$h\textbf{v}$$ to it. But this is just the usual idea of identifying vectors with their terminal points, which the reader should be used to by now. If we were to write the vector $$\textbf{v}$$ as $$\textbf{v} = (v_1 ,v_2)$$, then

$D_v f (a,b)=\lim \limits_{h \to 0}\dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h}\label{Eq2.9}$

From this we can immediately recognize that the partial derivatives $$\dfrac{∂f}{∂x}\text{ and} \dfrac{∂f}{∂y}$$ are special cases of the directional derivative with $$\textbf{v} = \textbf{i} = (1,0)\text{ and } \textbf{v} = \textbf{j} = (0,1)$$, respectively. That is, $$\dfrac{∂f}{∂x} = D_i f\text{ and } \dfrac{∂f}{∂y} = D_j f$$ . Since there are many vectors with the same direction, we use a unit vector in the definition, as that represents a “standard” vector for a given direction.

If $$f (x, y)$$ has continuous partial derivatives $$\dfrac{∂f}{∂x}\text{ and }\dfrac{∂f}{∂y}$$ (which will always be the case in this text), then there is a simple formula for the directional derivative:

Theorem 2.2

Let $$f (x, y)$$ be a real-valued function with domain $$D$$ in $$\mathbb{R}^2$$ such that the partial derivatives $$\dfrac{∂f}{∂x}\text{ and }\dfrac{∂f}{∂y}$$ exist and are continuous in $$D$$. Let $$(a,b)$$ be a point in $$D$$, and let $$\textbf{v} = (v_1 ,v_2)$$ be a unit vector in $$\mathbb{R}^2$$ . Then

$D_v f (a,b) = v_1 \dfrac{∂f}{∂x} (a,b)+ v_2 \dfrac{∂f}{∂y} (a,b)\label{Eq2.10}$

Proof: Note that if $$\textbf{v} = \textbf{i}$$ = (1,0) then the above formula reduces to $$D_v f (a,b) = \dfrac{∂f}{∂x} (a,b)$$, which we know is true since $$D_i f = \dfrac{∂f}{∂x}$$ , as we noted earlier. Similarly, for $$\textbf{v} = \textbf{j} = (0,1)$$ the formula reduces to $$D_v f (a,b) = \dfrac{∂f}{∂y} (a,b)$$, which is true since $$D_j f = \dfrac{∂f}{∂y}$$ . So since $$\textbf{i} = (1,0)\text{ and }\textbf{j} = (0,1)$$ are the only unit vectors in $$\mathbb{R}^2$$ with a zero component, then we need only show the formula holds for unit vectors $$\textbf{v} = (v_1 ,v_2)\text{ with }v_1 \neq 0 \text{ and }v_2 \neq 0$$. So fix such a vector $$\textbf{v}$$ and fix a number $$h \neq 0$$.

Then

$f (a+ hv_1 ,b + hv_2)− f (a,b) = f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b)+ f (a+ hv_1 ,b)− f (a,b)\label{Eq2.11}$

Since $$h \neq 0 \text{ and }v_2 \neq 0$$, then $$hv_2 \neq 0$$ and thus any number $$c$$ between $$b \text{ and }b + hv_2$$ can be written as $$c = b+\alpha hv_2$$ for some number $$0 < \alpha < 1$$. So since the function $$f (a+hv_1 , y)$$ is a realvalued function of $$y$$ (since $$a + hv_1$$ is a fixed number), then the Mean Value Theorem from single-variable calculus can be applied to the function $$g(y) = f (a + hv_1 , y)$$ on the interval $$[b,b + hv_2]$$ (or $$[b + hv_2 ,b]$$ if one of $$h \text{ or }v_2$$ is negative) to find a number $$0 < \alpha < 1$$ such that

$\nonumber \dfrac{∂f}{∂y} (a+ hv_1 ,b +\alpha hv_2) = g ′ (b +\alpha hv_2)=\dfrac{g(b + hv_2)− g(b)}{b + hv_2 − b}=\dfrac{f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b)}{hv_2}$

and so

$\nonumber f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b) = hv_2 \dfrac{∂f}{∂y} (a+ hv_1 ,b +\alpha hv_2) .$

By a similar argument, there exists a number $$0 < \beta < 1$$ such that

$\nonumber f (a+ hv_1 ,b)− f (a,b) = hv_1 \dfrac{∂f}{∂x} (a+\beta hv_1 ,b) .$

Thus, by Equation \ref{Eq2.11}, we have

\nonumber \begin{align} \dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h}&=\dfrac{hv_2 \dfrac{∂f}{∂y} (a+ hv_1 ,b +\alpha hv_2)+ hv_1 \dfrac{∂f}{∂x} (a+\beta hv_1 ,b)}{h} \\ \nonumber &=v_2 \dfrac{∂f}{∂y} (a+ hv_1 ,b +\alpha hv_2)+ v_1 \dfrac{∂f}{∂x} (a+\beta hv_1 ,b)\end{align}

so by Equation \ref{Eq2.9} we have

\nonumber \begin{align} D_v f (a,b)&=\lim \limits_{h \to 0}\dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h} \\ \nonumber &=\lim \limits_{h \to 0}\left [v_2 \dfrac{∂f}{∂y} (a+ hv_1 ,b +\alpha hv_2)+ v_1 \dfrac{∂f}{∂x} (a+\beta hv_1 ,b)\right ] \\ \nonumber &= v_2 \dfrac{∂f}{∂y} (a,b)+ v_1 \dfrac{∂f}{∂x} (a,b) \text{ by the continuity of } \dfrac{∂f}{∂x} \text{ and } \dfrac{∂f}{∂y}\text{, so} \\ \nonumber D_v f (a,b) &= v_1 \dfrac{∂f}{∂x} (a,b)+ v_2 \dfrac{∂f}{∂y} (a,b) \end{align}

$\nonumber \text{after reversing the order of summation.}\tag{$$\textbf{QED}$$}$

Note that $$D_v f (a,b) = v \cdot \left (\dfrac{∂f}{∂x} (a,b), \dfrac{∂f}{∂y} (a,b) \right )$$. The second vector has a special name:

Definition 2.6

For a real-valued function $$f (x, y)$$, the gradient of $$f$$ , denoted by $$\nabla f$$ , is the vector

$\nabla f =\left ( \dfrac{∂f}{∂x} , \dfrac{∂f}{∂y} \right ) \label{Eq2.12}$

in $$\mathbb{R}^2$$ . For a real-valued function $$f (x, y, z)$$, the gradient is the vector

$\nabla f = \left ( \dfrac{∂f}{∂x} , \dfrac{∂f}{∂y} , \dfrac{∂f}{∂z}\right ) \label{Eq2.13}$

in $$\mathbb{R}^ 3$$ . The symbol $$\nabla$$ is pronounced “del”.

Corollary 2.3

$\nonumber D_v f = \textbf{v} \cdot \nabla f$

Example 2.15

Find the directional derivative of $$f (x, y) = x y^2 + x^3 y$$ at the point (1,2) in the direction of $$\textbf{v} = \left ( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right )$$.

Solution

We see that $$\nabla f = (y^2 +3x^2 y,2x y+ x^3 )$$, so

$\nonumber D_v f (1,2) = \textbf{v}\cdot \nabla f (1,2) = \left ( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right ) \cdot (2^2 +3(1)^2 (2),2(1)(2)+1^3 ) = \dfrac{15}{\sqrt{2}}$

A real-valued function $$z = f (x, y)$$ whose partial derivatives $$\dfrac{∂f}{∂x}\text{ and }\dfrac{∂f}{∂y}$$ exist and are continuous is called continuously differentiable. Assume that $$f (x, y)$$ is such a function and that $$\nabla f \neq \textbf{0}$$. Let $$c$$ be a real number in the range of $$f$$ and let $$\textbf{v}$$ be a unit vector in $$\mathbb{R}^2$$ which is tangent to the level curve $$f (x, y) = c$$ (see Figure 2.4.1).

Figure 2.4.1

The value of $$f (x, y)$$ is constant along a level curve, so since $$\textbf{v}$$ is a tangent vector to this curve, then the rate of change of $$f$$ in the direction of $$\textbf{v}$$ is 0, i.e. $$D_v f = 0$$. But we know that $$D_v f = \textbf{v} \cdot \nabla f = \left\lVert \textbf{v} \right\rVert \left\lVert \nabla f \right\rVert \cos \theta$$, where $$\theta$$ is the angle between $$\textbf{v} \text{ and }\nabla f$$ . So since $$\left\lVert \textbf{v} \right\rVert = 1 \text{ then }D_v f = \left\lVert \nabla f \right\rVert \cos \theta$$. So since $$\nabla f \neq \textbf{0}\text{ then }D_v f = 0 ⇒ \cos \theta = 0 ⇒ \theta = 90^\circ$$ . In other words, $$\nabla f \perp \textbf{v}$$, which means that $$\nabla f$$ is normal to the level curve.

In general, for any unit vector $$\textbf{v}$$ in $$\mathbb{R}^2$$ , we still have $$D_v f = \left\lVert \nabla f \right\rVert \cos \theta \text{, where }\theta$$ is the angle between $$\textbf{v} \text{ and }\nabla f$$. At a fixed point $$(x, y)$$ the length $$\left\lVert \nabla f \right\rVert$$ is fixed, and the value of $$D_v f$$ then varies as $$\theta$$ varies. The largest value that $$D_v f$$ can take is when $$\cos \theta = 1 (\theta = 0^\circ )$$, while the smallest value occurs when $$\cos \theta = −1 (\theta = 180^\circ )$$. In other words, the value of the function $$f$$ increases the fastest in the direction of $$\nabla f$$ (since $$\theta = 0^\circ$$ in that case), and the value of $$f$$ decreases the fastest in the direction of $$−\nabla f$$ (since $$\theta = 180^\circ$$ in that case). We have thus proved the following theorem:

Theorem 2.4

Let $$f (x, y)$$ be a continuously differentiable real-valued function, with $$\nabla f \neq 0$$. Then:

1. The gradient $$\nabla f$$ is normal to any level curve $$f (x, y) = c$$.
2. The value of $$f (x, y)$$ increases the fastest in the direction of $$\nabla f$$ .
3. The value of $$f (x, y)$$ decreases the fastest in the direction of $$−\nabla f$$ .

Example 2.16

In which direction does the function $$f (x, y) = x y^2 + x^3 y$$ increase the fastest from the point (1,2)? In which direction does it decrease the fastest?

Solution

Since $$\nabla f = (y^2 + 3x^2 y,2xy + x^3 )$$, then $$\nabla f (1,2) = (10,5) \neq \textbf{0}$$. A unit vector in that direction is $$\textbf{v} = \dfrac{\nabla f}{\left\lVert \nabla f \right\rVert} = \left (\dfrac{2}{\sqrt{5}} ,\dfrac{1}{\sqrt{5}}\right )$$. Thus, $$f$$ increases the fastest in the direction of $$\left (\dfrac{2}{\sqrt{5}} ,\dfrac{1}{\sqrt{5}}\right )$$ and decreases the fastest in the direction of $$\left (\dfrac{−2}{\sqrt{5}},\dfrac{−1}{\sqrt{5}}\right )$$ .

Though we proved Theorem 2.4 for functions of two variables, a similar argument can be used to show that it also applies to functions of three or more variables. Likewise, the directional derivative in the three-dimensional case can also be defined by the formula $$D_v f = \textbf{v}\cdot \nabla f$$ .

Example 2.17

The temperature $$T$$ of a solid is given by the function $$T(x, y, z) = e^−x + e^{−2y} + e^{4z}\text{, where }x, y, z$$ are space coordinates relative to the center of the solid. In which direction from the point (1,1,1) will the temperature decrease the fastest?

Solution

Since $$\nabla f = (−e^{−x} ,−2e^{−2y} ,4e^{4z} )$$, then the temperature will decrease the fastest in the direction of $$−\nabla f (1,1,1) = (e^{−1} ,2e^{−2} ,−4e^4 )$$.