Section 2.1: the Derivative and the Tangent Line Problem
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Section 2.1: The Derivative and the Tangent Line Problem
Practice HW from Larson Textbook (not to hand in) p. 87 # 1, 5-23 odd, 59-62
Tangent Lines
Recall that a tangent line to a circle is a line that touches the circle only once.
If we magnify the circle around the point P
we see that slope of the slope of the tangent line very closely resembles the slope of the circle at P. For functions, we can define a similar interpretation of the tangent line slope.
Definition: The tangent line to a function at a point P is the line that best describes the slope of the graph of the function at a point P. We define the slope of the tangent line to be equal to the slope of the curve at the point P.
Examples of Tangent Lines: 2 3
Consider the following graph:
Slope of Secant line between the points = (x, f(x)) and (x+h, f(x+h))
As h→0, the slope of the secant lines approach that of the tangent line of f at x = a.
Slope of tangent line = m = of f at (x, f(x)) 4
Definition: Given a function y f (x) , the derivative of f, denoted by f , is the function defined by
f (x h) f (x) f (x) lim , h0 h provided the limit exists.
Facts 1. For a function y f (x) at a point x = a, f (a) gives the slope of the tangent line to the graph of f at the point (a, f (a)) . 2. f (a) represents the instantaneous rate of change of f at x = a, for example, instantaneous velocity.
Example 1: Use the definition of the derivative to find the derivative of the function f (x) 3x 2 .
Solution:
█ Example 2: Use the definition of the derivative to find the derivative of the function f (x) 2x2 x 1. 5
Solution:
█ Example 3: Find the equation of the tangent line to the curve f (x) x3 3x 2 4 at the point (1, 2).
Solution: 6
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Example 4: Use the definition of the derivative to find f (x) if f (x) 1 x .
Solution: Using the limit definition of the derivative, we see that
f (x h) f (x) f (x) lim h0 h
1 x h 1 x f (x) 1 x lim Note that h0 h f (x h) 1 x h
( 1 x h 1 x) ( 1 x h 1 x) Multiply numerator and denominator lim h0 h ( 1 x h 1 x) by the conjugate of the numerator
(1 x h) 1 x h 1 x 1 x h 1 x (1 x) FOIL the numerator and lim h0 h ( 1 x h 1 x) multiply denominator
1 x h 1 x lim (Simplify and distribute - sign to 1 x) h0 h ( 1 x h 1 x)
h lim (Simplify) h0 h ( 1 x h 1 x)
1 lim (Simplify by canceling h terms) h0 1 x h 1 x
1 (Substitute h 0 to evaluate the limit) 1 x 0 1 x
1 (Simplify) 1 x 1 x
1 (Simplify by combining like terms) 2 1 x
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Example 5: Find the derivative using the definition and use the result to find the equation of the 1 line tangent to the graph of f (x) at the point (0, -1). x 1
Solution: To find the equation of any line, including a tangent line, we need to know the line’s slope and a point on the line. Since we already have a point on line, we must find the tangent line’s slope, which is found using the derivative. Using the limit definition of the derivative, we see that
f (x h) f (x) f (x) lim h0 h
1 1 1 1 lim x h 1 x 1 Note that f (x) , f (x h) h0 h x 1 x h 1
1 [x 1] 1 [x h 1] (x h 1) [x 1] (x 1) [x h 1] lim The common demonimator is (x h 1)(x 1) h0 h
(x 1) (x h 1) (x h 1)(x 1) (x 1)(x h 1) lim Rewrite fractions in terms of common denominator h0 h
(x 1) (x h 1) (x h 1)(x 1) lim (Combine fractions over common denominator) h0 h
(x 1 x h 1) (x h 1)(x 1) lim (Distribute - sign) h0 h
h (x h 1)(x 1) lim (Simplify) h0 h h (x h 1)(x 1) h lim (Write h as ) h0 h 1 1 h 1 lim (Take reciprocal of denominator and multiply) h0 (x h 1)(x 1) h
1 lim (Cancel h terms) h0 (x h 1)(x 1) 1 1 1 (Simplify to get the derivative ) (x 0 1)(x 1) (x 0 1)(x 1) (x 1)2 Continued on next page 9
1 Using f (x) , we can now find the slope at the give point (0, -1). (x 1)2
Slope at the point 1 1 1 (0,1) m f (0) 1. 2 2 1 (x 0) (0 1) (1)
Using m 1, we see that from the slope intercept equation y mx b that y x b
To find b, use the fact that at the point (0, -1), x 0 and y 1. Thus
1 0 b giving b 1. Thus, the equation of the tangent line is:
y x 1. 10
Differentiability and Points where the Derivative Does Not Exist
Note: The derivative of a function may not exist a point.
Fact: If a function is not continuous at a point, its derivative does not exist at that point.
1 For, example, f (x) is not continuous at x = 1. This implies that f (1) will not x 1 1 exist. Note that f (x) , which computationally says f (1) does not exist. (x 1)2
Note! However, a function may still be continuous at a point but the derivative may still not exist. 11
Example 10: Use the definition of the derivative to demonstrate that f (0) does not exist for the function f (x) x .
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Fact: In general, a function that displays any of these characteristics at a point is not differentiable at that point
Cases
1. The function is not continuous at a point – it has a jump, break, or hole in the graph at that point.
2. The function has a sharp point or corner at a point.
3. The function has a vertical tangent at a point.
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Example 11: Determine the point)s on the following graph where the derivative does not exist. Give a short reason for your answer.
Solution:
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