Finding the Equation of a Tangent Line using the First Derivative, Study Guides, Projects, Research of Calculus

How to find the equation of a tangent line to a curve at a specific point using the first derivative. It covers the concept of a tangent line, the slope-intercept and point-slope formula for a line, and the process of finding the first derivative using the power rule. A worked-out example is provided to illustrate the steps.

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2021/2022

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Finding the Equation of a Tangent Line
Using the First Derivative
Certain problems in Calculus I call for using the first derivative to find the equation of the
tangent line to a curve at a specific point.
The following diagram illustrates these problems.
There are certain things you must remember from College Algebra (or similar classes)
when solving for the equation of a tangent line.
Recall :
A Tangent Line is a line which locally touches a curve at one and only one point.
The slope-intercept formula for a line is y = mx + b, where m is the slope of the
line and b is the y-intercept.
The point-slope formula for a line is y – y1 = m (x – x1). This formula uses a
point on the line, denoted by (x1, y1), and the slope of the line, denoted by m, to
calculate the slope-intercept formula for the line.
Also, there is some information from Calculus you must use:
Recall:
The first derivative is an equation for the slope of a tangent
line to a curve at an indicated point.
The first derivative may be found using:
()()
h
xf+
hxf
lim
:derivative a of definition The A)
0h
B) Methods already known to you for derivation, such as:
Power Rule
Product Rule
Quotient Rule
Chain Rule
(For a complete list and description of these rules see your text)
pf3

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Finding the Equation of a Tangent Line

Using the First Derivative

Certain problems in Calculus I call for using the first derivative to find the equation of the tangent line to a curve at a specific point.

The following diagram illustrates these problems.

There are certain things you must remember from College Algebra (or similar classes) when solving for the equation of a tangent line.

Recall :

  • A Tangent Line is a line which locally touches a curve at one and only one point.
  • The slope-intercept formula for a line is y = mx + b , where m is the slope of the line and b is the y-intercept.
  • The point-slope formula for a line is y – y 1 = m (x – x 1 ). This formula uses a point on the line, denoted by (x 1 , y 1 ), and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.

Also, there is some information from Calculus you must use:

Recall :

  • The first derivative is an equation for the slope of a tangent line to a curve at an indicated point.
  • The first derivative may be found using:

h

  • f x

f x h lim

A)Thedefinitionofaderivative:

h 0

B) Methods already known to you for derivation, such as:

  • Power Rule
  • Product Rule
  • Quotient Rule
  • Chain Rule (For a complete list and description of these rules see your text)

With these formulas and definitions in mind you can find the equation of a tangent line.

Consider the following problem:

Find theequationofthelinetangentto f ( x ) = x^2 atx = 2.

Having a graph is helpful when trying to visualize the tangent line. Therefore, consider the following graph of the problem:

8

6

4

2

-3 -2 -1 1 2 3

The equation for the slope of the tangent line to f(x) = x^2 is f '(x), the derivative of f(x). Using the power rule yields the following:

f(x) = x^2 f '(x) = 2x (1)

Therefore, at x = 2, the slope of the tangent line is f '(2).

f '(2) = 2(2) = 4 (2)

Now , you know the slope of the tangent line, which is 4. All that you need now is a point on the tangent line to be able to formulate the equation.

You know that the tangent line shares at least one point with the original equation, f(x) = x^2. Since the line you are looking for is tangent to f(x) = x^2 at x = 2, you know the x coordinate for one of the points on the tangent line. By plugging the x coordinate of the shared point into the original equation you have:

4 or y 4 (3)

f(x) 22 = =

Therefore, you have found the coordinates, (2, 4), for the point shared by f(x) and the line tangent to f(x) at x = 2. Now you have a point on the tangent line and the slope of the tangent line from step (1).

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