Defining the Derivative: Learn It 1

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Defining the Derivative: Learn It 1

Calculate the slope of a tangent line to a curve and find its equation
Find the derivative of a function at a given point
Explain how velocity measures speed over time, and compare average velocity over a period with the exact speed at a specific moment

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 $(a,f(a))$ helps estimate the rate of change. We can find the slope of the secant by choosing a value of $x$ near $a$ and drawing a line through the points $(a,f(a))$ and $(x,f(x))$ . The slope of this line is given by the difference quotient:

m_{\sec} = \frac{f (x) - f (a)}{x - a}

$m_{\sec}=\dfrac{f(x)-f(a)}{x-a}$

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

m_{\sec} = \frac{f (a + h) - f (a)}{a + h - a} = \frac{f (a + h) - f (a)}{h}

$m_{\sec}=\dfrac{f(a+h)-f(a)}{a+h-a}=\dfrac{f(a+h)-f(a)}{h}$

difference quotient

Let $f$ be a function defined on an interval $I$ containing $a$ . If $x\ne a$ is in $I$ , then

Q = \frac{f (x) - f (a)}{x - a}

$Q=\dfrac{f(x)-f(a)}{x-a}$

is a difference quotient.

Also, if $h\ne 0$ is chosen so that $a+h$ is in $I$ , then

Q = \frac{f (a + h) - f (a)}{h}

$Q=\dfrac{f(a+h)-f(a)}{h}$

is a difference quotient with increment $h$ .

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.

This figure consists of two graphs labeled a and b. Figure a shows the Cartesian coordinate plane with 0, a, and x marked on the x-axis. There is a curve labeled y = f(x) with points marked (a, f(a)) and (x, f(x)). There is also a straight line that crosses these two points (a, f(a)) and (x, f(x)). At the bottom of the graph, the equation msec = (f(x) - f(a))/(x - a) is given. Figure b shows a similar graph, but this time a + h is marked on the x-axis instead of x. Consequently, the curve labeled y = f(x) passes through (a, f(a)) and (a + h, f(a + h)) as does the straight line. At the bottom of the graph, the equation msec = (f(a + h) - f(a))/h is given. — Figure 2. We can calculate the slope of a secant line in either of two ways.

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

This figure consists of three graphs labeled a, b, and c. Figure a shows the Cartesian coordinate plane with 0, a, x2, and x1 marked in order on the x-axis. There is a curve labeled y = f(x) with points marked (a, f(a)), (x2, f(x2)), and (x1, f(x1)). There are three straight lines: the first crosses (a, f(a)) and (x1, f(x1)); the second crosses (a, f(a)) and (x2, f(x2)); and the third only touches (a, f(a)), making it the tangent. At the bottom of the graph, the equation mtan = limx → a (f(x) - f(a))/(x - a) is given. Figure b shows a similar graph, but this time a + h2 and a + h1 are marked on the x-axis instead of x2 and x1. Consequently, the curve labeled y = f(x) passes through (a, f(a)), (a + h2, f(a + h2)), and (a + h1, f(a + h1)) and the straight lines similarly cross the graph as in Figure a. At the bottom of the graph, the equation mtan = limh → 0 (f(a + h) - f(a))/h is given. Figure c shows only the curve labeled y = f(x) and its tangent at point (a, f(a)). — Figure 3. The secant lines approach the tangent line (shown in green) as the second point approaches the first.

Figure 4 shows the graph of $f(x)=\sqrt{x}$ and its tangent line at $(1,1)$ in a series of tighter intervals about $x=1$ . 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 $x =1$ . In fact, the graph of $f(x)$ appears locally linear close to $x=1$ .

This figure consists of four graphs labeled a, b, c, and d. Figure a shows the graphs of the square root of x and the equation y = (x + 1)/2 with the x-axis going from 0 to 4 and the y-axis going from 0 to 2.5. The graphs of these two functions look very close near 1; there is a box around where these graphs look close. Figure b shows a close up of these same two functions in the area of the box from Figure a, specifically x going from 0 to 2 and y going from 0 to 1.4. Figure c is the same graph as Figure b, but this one has a box from 0 to 1.1 in the x coordinate and 0.8 and 1 on the y coordinate. There is an arrow indicating that this is blown up in Figure d. Figure d shows a very close picture of the box from Figure c, and the two functions appear to be touching for almost the entire length of the graph. — Figure 4. For values of $x$ close to 1, the graph of $f(x)=\sqrt{x}$ and its tangent line appear to coincide.

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

tangent line

Let $f(x)$ be a function defined in an open interval containing $a$ . The tangent line to $f(x)$ at $a$ is the line passing through the point $(a,f(a))$ having slope

m_{\tan} = lim_{x \to a} \frac{f (x) - f (a)}{x - a}

$m_{\tan}=\underset{x\to a}{\lim}\dfrac{f(x)-f(a)}{x-a}$

provided this limit exists.

Equivalently, we may define the tangent line to $f(x)$ at $a$ to be the line passing through the point $(a,f(a))$ having slope

m_{\tan} = lim_{h \to 0} \frac{f (a + h) - f (a)}{h}

$m_{\tan}=\underset{h\to 0}{\lim}\dfrac{f(a+h)-f(a)}{h}$

provided this limit exists.

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.

Evaluating Functions: Functions can be evaluated for inputs that are variables or expressions. The process is the same as evaluating with a constant, but the simplified answer will contain a variable.
Point-Slope Form: The point-slope form of a linear equation takes the form:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$

where $m$ is the slope, and $(x_1,y_1)$ are the coordinates of a specific point through which the line passes.

Find the equation of the line tangent to the graph of $f(x)=x^2$ at $x=3$ .

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

\begin{array}{lllll} m_{\tan} & = lim_{x \to 3} \frac{f (x) - f (3)}{x - 3} & Apply the definition. \\ = lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & Substitute f (x) = x^{2} and f (3) = 9. \\ = lim_{x \to 3} \frac{(x - 3) (x + 3)}{x - 3} = lim_{x \to 3} (x + 3) = 6 & Factor the numerator to evaluate the limit. \end{array}

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

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

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

This figure consists of the graphs of f(x) = x squared and y = 6x - 9. The graphs of these functions appear to touch at x = 3. — Figure 5. The tangent line to $f(x)$ at $x=3$ .

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).

Use the second definition to find the slope of the line tangent to the graph of $f(x)=x^2$ at $x=3$ .

The steps are very similar to the previous example.

\begin{array}{lllll} m_{\tan} & = lim_{h \to 0} \frac{f (3 + h) - f (3)}{h} & Apply the definition. \\ = lim_{h \to 0} \frac{(3 + h)^{2} - 9}{h} & Substitute f (3 + h) = (3 + h)^{2} and f (3) = 9. \\ = lim_{h \to 0} \frac{9 + 6 h + h^{2} - 9}{h} & Expand and simplify to evaluate the limit. \\ = lim_{h \to 0} \frac{h (6 + h)}{h} = lim_{h \to 0} (6 + h) = 6 \end{array}

$\begin{array}{lllll}m_{\tan} & =\underset{h\to 0}{\lim}\frac{f(3+h)-f(3)}{h} & & & \text{Apply the definition.} \\ & =\underset{h\to 0}{\lim}\frac{(3+h)^2-9}{h} & & & \text{Substitute} \, f(3+h)=(3+h)^2 \, \text{and} \, f(3)=9. \\ & =\underset{h\to 0}{\lim}\frac{9+6h+h^2-9}{h} & & & \text{Expand and simplify to evaluate the limit.} \\ & =\underset{h\to 0}{\lim}\frac{h(6+h)}{h}=\underset{h\to 0}{\lim}(6+h)=6 \end{array}$

We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.