Tangent lines and derivatives

Section 2.2 Tangent lines and derivatives

In the previous section we introduced the concept of derivative through kinematics. Here we see that the derivative also has a deep interpretation in geometry, as calculating the slope of the tangent line to the graph of a function. We study our first “tangent line problem”, which leads us to the question of how to evaluate (and define) limits.

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Objectives

You should be able to:

Describe and illustrate the connection between the velocity and tangent problems.
Explain the difference between the slope of a secant line connecting two points on a curve and the slope of the tangent line to a curve at a point.
Describe the limit process that arises in the calculation of the slope of a tangent line.
Use correct notation for the limit process, to represent the slope of a tangent line as an appropriate limit of the slope of a secant line.
Calculate the equation of the tangent line for simple functions.

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Subsection 2.2.1 Instructional videos

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Subsection 2.2.2 Key concepts

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Concept 2.2.1. Secant line.

Given two points $P_1(c,f(c))$ and $P_2(x_2,f(x_2))$ on the graph of a function $f\text{,}$ the slope of the secant line through $P_1$ and $P_2$ is given by

m_{P_{1} P_{2}} = \frac{f (x_{2}) - f (c)}{x_{2} - c} .

$\begin{equation*} m_{P_1 P_2} = \frac{f(x_2) - f(c)}{x_2 - c}. \end{equation*}$

Equivalently, defining $x_2 = c+h\text{,}$ it can be written as

m_{P_{1} P_{2}} = \frac{f (c + h) - f (c)}{h} .

$\begin{equation*} m_{P_1 P_2} = \frac{f(c+h) - f(c)}{h}. \end{equation*}$

This is known as the difference quotient of the function $f$ at $x=c\text{.}$

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Concept 2.2.2. Tangent line.

The slope of the tangent line to the curve $y=f(x)$ at $P_1(c,f(c))$ is obtained by taking the limit $x_2 \to c$ or, equivalently, $h \to 0\text{:}$

\begin{matrix} m = & lim x_{2} \to c \frac{f (x_{2}) - f (c)}{x_{2} - c} = & lim h \to 0 \frac{f (c + h) - f (c)}{h} . \end{matrix}

$\begin{align*} m =\amp \lim_{x_2 \to c} \frac{f(x_2) - f(c)}{x_2 - c}\\ =\amp \lim_{h \to 0} \frac{f(c+h)- f(c)}{h}. \end{align*}$

Thus, $m$ is precisely equal to the derivative of $f$ at $x=c\text{;}$ that is, $m = f'(c)\text{.}$

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MATH 144: Calculus for the Physical Sciences I