When you studied partial derivatives, you learned that a function [latex]z=f(x,y)[/latex] has two partial derivatives: [latex]\partial{z}/\partial{x}[/latex] and [latex]\partial{z}/\partial{y}[/latex]. These derivatives correspond to each independent variable and represent instantaneous rates of change—slopes of tangent lines. For example, [latex]\partial{z}/\partial{x}[/latex] gives the slope of a tangent line parallel to the [latex]x[/latex]-axis, and [latex]\partial{z}/\partial{y}[/latex] gives the slope of a tangent line parallel to the [latex]y[/latex]-axis.

But what if you want to find the slope of a tangent line that’s not parallel to either axis? That’s where directional derivatives come in.

Directional Derivatives

Let’s start with a surface defined by [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 this direction using an angle [latex]\theta[/latex], measured counterclockwise in the [latex]xy[/latex]-plane, starting from the positive [latex]x[/latex]-axis (Figure 1).

We travel a distance [latex]h[/latex] in the direction given by the unit vector [latex]{\bf{u}}=(\cos\theta){\bf{i}}+(\sin\theta){\bf{j}}[/latex]. The [latex]z[/latex]-coordinate of our destination point on the graph is [latex]z=f(a+h\cos\theta, b+h\sin\theta)[/latex].

To find the slope of the secant line, we divide the difference in [latex]z[/latex]-values by the length of the line segment connecting the two points in the domain. Since the length is [latex]h[/latex], the slope of the secant line is:

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

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

This definition provides a formal way to calculate directional derivatives in many cases.

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.)

Computing Directional Derivatives Using Partial Derivatives

There’s another approach to calculating directional derivatives that uses partial derivatives directly. This method is often more efficient than using the limit definition.