Now that we understand limits and can compute them, we have established the foundation for studying calculus, the branch of mathematics involving derivatives and integrals. Calculus was independently developed by the Englishman Isaac Newton (1643-1727) and the German Gottfried Leibniz (1646-1716).

When we credit Newton and Leibniz with developing calculus, we refer to their understanding of the relationship between the derivative and the integral. Both benefitted from the work of predecessors like Barrow, Fermat, and Cavalieri. Initially, Newton and Leibniz had an amicable relationship, but a controversy later erupted over who developed calculus first. Although it appears Newton arrived at the ideas first, we are indebted to Leibniz for the notation that we commonly use today.

Tangent Lines

Let’s start by revisiting the notion of secant lines and tangent lines. The slope of a secant line to a function at a point [latex](a,f(a))[/latex] helps estimate the rate of change. We can find the slope of the secant by choosing a value of [latex]x[/latex] near [latex]a[/latex] and drawing a line through the points [latex](a,f(a))[/latex] and [latex](x,f(x))[/latex]. The slope of this line is given by the difference quotient:

We can also calculate the slope of a secant line to a function at a value [latex]a[/latex] by using this equation and replacing [latex]x[/latex] with [latex]a+h[/latex], where [latex]h[/latex] is a value close to [latex]0[/latex]. This gives us the slope of the secant line through the points [latex](a,f(a))[/latex] and [latex](a+h,f(a+h))[/latex]:

These two expressions for calculating the slope of a secant line are illustrated in Figure 2. Depending on the setting, we can choose either method based on ease of calculation.

In Figure 3(a), as the values of [latex]x[/latex] approach [latex]a[/latex], the slopes of the secant lines provide better estimates of the rate of change of the function at [latex]a[/latex]. The secant lines themselves approach the tangent line to the function at [latex]a[/latex], which represents the limit of the secant lines. Similarly, Figure 3(b) shows that as the values of [latex]h[/latex] get closer to [latex]0[/latex], the secant lines also approach the tangent line. The slope of the tangent line at [latex]a[/latex] is the rate of change of the function at [latex]a[/latex], as shown in Figure 3(c).

Figure 4 shows the graph of [latex]f(x)=\sqrt{x}[/latex] and its tangent line at [latex](1,1)[/latex] in a series of tighter intervals about [latex]x=1[/latex]. As the intervals narrow, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a goodapproximation of the function near [latex]x =1[/latex]. In fact, the graph of [latex]f(x)[/latex] appears locally linear close to [latex]x=1[/latex].

Formally we may define the tangent line to the graph of a function as follows.

Just as we used two expressions to define the slope of a secant line, we use two forms to define the slope of a tangent line. In this text, we use both definitions depending on the context.

Now that we have defined a tangent line to a function at a point, we can find equations of tangent lines. This requires recalling two algebraic techniques: evaluating a function with variable inputs and using point-slope form to write an equation of a line.

Find the equation of the line tangent to the graph of [latex]f(x)=x^2[/latex] at [latex]x=3[/latex].

Show Solution

First find the slope of the tangent line. In this example, use the first definition above.

[latex]\begin{array}{lllll}m_{\tan} & =\underset{x\to 3}{\lim}\frac{f(x)-f(3)}{x-3} & & & \text{Apply the definition.} \\ & =\underset{x\to 3}{\lim}\frac{x^2-9}{x-3} & & & \text{Substitute} \, f(x)=x^2 \, \text{and} \, f(3)=9. \\ & =\underset{x\to 3}{\lim}\frac{(x-3)(x+3)}{x-3}=\underset{x\to 3}{\lim}(x+3)=6 & & & \text{Factor the numerator to evaluate the limit.} \end{array}[/latex]

 

Next, find a point on the tangent line. Since the line is tangent to the graph of [latex]f(x)[/latex] at [latex]x=3[/latex], it passes through the point [latex](3,f(3))[/latex]. We have [latex]f(3)=9[/latex], so the tangent line passes through the point [latex](3,9)[/latex].

Using the point-slope equation of the line with the slope [latex]m=6[/latex] and the point [latex](3,9)[/latex], we obtain the line [latex]y-9=6(x-3)[/latex]. Simplifying, we have [latex]y=6x-9[/latex]. The graph of [latex]f(x)=x^2[/latex] and its tangent line at [latex]x=3[/latex] are shown in Figure 5.

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.1 Defining the Derivative” here (opens in new window).