We start with the graph of a surface defined by the equation [latex]z=f(x, y)[/latex]. Given a point [latex](a, b)[/latex] in the domain of [latex]f[/latex], we choose a direction to travel from that point. We measure the direction using an angle [latex]\theta[/latex] which is measured counterclockwise in the [latex]x, y[/latex]-plane, starting at zero from the positive [latex]x[/latex]-axis (Figure 1). The distance we travel is [latex]h[/latex] and the direction we travel is given by the unit vector [latex]{\bf{u}}=(\cos\theta){\bf{i}}+(\sin\theta)\bf{j}[/latex]. Therefore, the [latex]z[/latex]-coordinate of the second point on the graph is given by [latex]z=f(a+h\cos\theta,\,b+h\sin\theta)[/latex].

Figure 1. Finding the directional derivative at a point on the graph of [latex]z=f(x,y)[/latex]. The slope of the black arrow on the graph indicates the value of the directional derivative at that point.

We can calculate the slope of the secant line by dividing the difference in [latex]z[/latex]-values by the length of the line segment connecting the two points in the domain. The length of the line segment is [latex]h[/latex]. Therefore, the slope of the secant line is

[latex]\LARGE{m_{\text{sec}}=\frac{f(a+h\cos\theta,\,b+h\sin\theta)-f(a,\,b)}{h}}[/latex]

To find the slope of the tangent line in the same direction, we take the limit as [latex]h[/latex] approaches zero.

The directional derivative of [latex]f[/latex] in the direction of [latex]\bf{u}[/latex] provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.

Example: finding the Directional derivative from the definition

Let [latex]\theta=\arccos(3/5)[/latex]. Find the directional derivative [latex]D_{\bf{u}}f(x,y)[/latex] of [latex]f(x,y)=x^2-xy+3y^2[/latex] in the direction of [latex]{\bf{u}}=(\cos\theta){\bf{i}}+(\sin\theta)\bf{j}[/latex]. What is [latex]D_{\bf{u}}f(-1,2)[/latex]?

Show Solution

First of all, since [latex]\cos\theta=3/5[/latex] and [latex]\theta[/latex] is acute, this implies

[latex]\large{\sin\theta=\sqrt{1-\left(\frac35\right)^2}=\sqrt{\frac{16}{25}}=\frac45}[/latex]

Using [latex]f(x,y)=x^2-xy+3y^2[/latex], we first calculate [latex]f(x+h\cos\theta,y+h\sin\theta)[/latex]:

[latex]\begin{align} f(x+h\cos\theta,y+h\sin\theta)&=(x+h\cos\theta)^2-(x+h\cos\theta)(y+h\sin\theta)+3(y+h\sin\theta)^2 \\ &=x^2+2xh\cos\theta+h^2\cos^2\theta-xy-xh\sin\theta-yh\cos\theta-h^2\sin\theta\cos\theta+3y^2 \\ &\;\;\;+6yh\sin\theta+3h^2\sin^2\theta \\ &=x^2+2xh\left(\frac35\right)+\frac{9h^2}{25}-xy-\frac{4xh}5-\frac{3yh}5-\frac{12h^2}{25}+3y^2 +6yh\left(\frac45\right)+3h^2\left(\frac{16}{25}\right) \\ &=x^2-xy+3y^2+\frac{2xh}5+\frac{9h^2}5+\frac{21yh}5. \end{align}[/latex]

We substitute this expression into the directional derivative of [latex]f[/latex] in the direction of [latex]\bf{u}[/latex]:

[latex]\hspace{4cm}\begin{align} D_{\bf{u}}f(a,b)&=\displaystyle\lim_{h\to0}\frac{f(a+h\cos\theta,b+h\sin\theta)-f(a,b)}{h} \\ &=\displaystyle\lim_{h\to0}\frac{(x^2-xy+3y^2+\frac{2xh}5+\frac{9h^2}5+\frac{21yh}5)-(x^2-xy+3y^2)}{h} \\ &=\displaystyle\lim_{h\to0}\frac{\frac{2xh}5+\frac{9h^2}5+\frac{21yh}5}{h} \\ &=\displaystyle\lim_{h\to0}\frac{2x}5+\frac{9h}5+\frac{21y}5 \\ &=\frac{2x+21y}5. \end{align}[/latex]

To calculate [latex]D_{\bf{u}}f(-1,2)[/latex], we substitute [latex]x=-1[/latex] and [latex]y=2[/latex] into this answer:

[latex]\hspace{4cm}\begin{align} D_{\bf{u}}f(-1,2)&=\frac{2(-1)+21(2)}5 \\ &=\frac{-2+42}5 \\ &= 8. \end{align}[/latex]

(See the following figure.)

Figure 2. Finding the directional derivative in a given direction [latex]\small{\bf{u}}[/latex] at a given point on a surface. The plane is tangent to the surface at the given point [latex]\small{(-1,2,15)}[/latex]

Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.

Proof

The directional derivative of [latex]f[/latex] in the direction of [latex]\bf{u}[/latex] states that the directional derivative of [latex]f[/latex] in the direction of [latex]{\bf{u}}=\cos\theta{\bf{i}}+\sin\theta\bf{j}[/latex] is given by

[latex]D_{\bf{u}}f(a,b)=\displaystyle\lim_{h\to0}\frac{f(a+h\cos\theta,b+h\sin\theta)-f(a,b)}{h}[/latex]

Let [latex]x=a+t\cos\theta[/latex] and [latex]y=b+t\sin\theta[/latex], and define [latex]g(t)=f(x, y)[/latex]. Since [latex]f_x[/latex] and [latex]f_y[/latex] both exist, and therefore [latex]f[/latex] is differentiable, we can use the chain rule for functions of two variables to calculate [latex]{g}'(t)[/latex]:

[latex]\hspace{8.5cm}\begin{align} g'(t)&=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} \\ &=f_x(x,y)\cos\theta+f_y(x,y)\sin\theta. \end{align}[/latex]

If [latex]t=0[/latex], then [latex]x=x_0(=a)[/latex] and [latex]y=y_0(=b)[/latex], so

[latex]g'(0)=f_x(x_0,y_0)\cos\theta+f_y(x_0,y_0)\sin\theta[/latex]

By the definition of [latex]{g}'(t)[/latex], it is also true that

[latex]\hspace{8.5cm}\begin{align} g'(0)&=\displaystyle\lim_{t\to0}\frac{g(t)-g(0)}t \\ &=\displaystyle\lim_{t\to0}\frac{f(x_0+t\cos\theta,y_0+t\sin\theta)-f(x_0,y_0)}t. \end{align}[/latex]

Therefore, [latex]D_{\bf{u}}f(x_0,y_0)=f_x(x,y)\cos\theta+f_y(x,y)\sin\theta[/latex].

[latex]_\blacksquare[/latex]

Watch the following video to see the worked solution to the above Try It

If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in Example: Finding a Directional Derivative: Alternative Method in the direction of the vector [latex]\langle-5,12\rangle[/latex] we would first divide by its magnitude to get [latex]\bf{u}[/latex]. This gives us [latex]{\bf{u}}=\langle-(5/13),12/13\rangle[/latex]. Then

[latex]\hspace{8.5cm}\begin{align} D_{\bf{u}}f(x,y)&=\nabla{f}(x,y)\cdot\bf{u} \\ &=-\frac5{13}(2x-y)+\frac{12}{13}(-x+6y) \\ &=-\frac{22}{13}x+\frac{17}{13}y. \end{align}[/latex]