Intro+to+the+derivative+notes.pdf

# Solution a by definition the instantaneous rate of

• Notes
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Solution: a. By definition the instantaneous rate of change is given by C 0 (100) = lim h 0 C (100 + h ) - C (100) h =

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3. Introduction to the derivative: Limits and Continuity 23 = lim h 0 5000 + 10(100 + h ) + 0 . 05(100 + h ) 2 - 5000 - 10 · 100 - 0 . 05 · 100 2 h = = lim h 0 10 h + 0 . 05 h 2 + 0 . 05 · 200 h h = lim h 0 ( 20 + 0 . 05 h ) = 20 \$/item b. It means that the cost is increasing when the 100 units are produced at an instantaneous rate of \$20 per item produced. Example 3.24. Let f ( x ) = x 2 . Use the definition of the derivative to compute f 0 (3) algebraically. Solution: By definition: f 0 (3) = lim h 0 f (3 + h ) - f (3) h = lim h 0 (3 + h ) 2 - 3 2 h = lim h 0 9 + h 2 + 6 h - 9 h = = lim h 0 h 2 + 6 h h = lim h 0 ( h + 6) = 6 Thus f 0 (3) = 6. Other Notations for the derivative Let y = f ( x ) be a function. Common notations for the derivative of f are f 0 ( x ) = y 0 = dy dx = df dx = d dx f ( x ) = Df ( x ) = D x f ( x ) The symbols D and d/dx are called differential operators because they indicate the operation of differentiation, which is the process of calculating a derivative. The symbol df/dx , which was introduced by Leibniz, is not a quotient of two numbers . It is just another way to write f 0 ( x ). If we want to indicate the value of a derivative df/dx in Leibniz notation at a specific number a , we use the notation df dx x = a
24 J. S´ anchez-Ortega 3.7 Geometric Interpretation: Secant and Tangent Lines Recall that the slope of the secant line through the points on the graph of f where x = a and x = a + h is given by the average rate of change, or difference quotient, m sec = slope of secant = f ( a + h ) - f ( a ) h The slope of the tangent line through the point on the graph of f where x = a and x = a + h is given by the instantaneous average rate of change, or derivative m tan = slope of tangent = f 0 ( a ) = lim h 0 f ( a + h ) - f ( a ) h assuming the limit exists. We define the tangent line to the graph of f at the point P ( a, f ( a )) as the straight line passing through P with slope f 0 ( a ). Its equation is y - f ( a ) = f 0 ( a )( x - a ) or y = f 0 ( a )( x - a ) + f ( a ) Note that the tangent line is the limiting position of the secant line PQ as Q approaches P . (See Figure below).

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3. Introduction to the derivative: Limits and Continuity 25 Example 3.25. Let f ( x ) = 3 /x . Find an equation of the tangent line to the curve y = 3 /x at the point (3 , 1) . Solution: The slope of the tangent line is f 0 (3), and by definition f 0 (3) = lim h 0 f (3 + h ) - f (3) h = lim h 0 3 3+ h - 1 h = lim h 0 3 - (3+ h ) 3+ h h = lim h 0 - h h (3 + h ) = = lim h 0 - 1 3 + h = - 1 3 Therefore an equation of the tangent at the point (3 , 1) is y - 1 = - 1 3 ( x - 3) or x + 3 y - 6 = 0 The curve y = 3 /x and the tangent at the point (3 , 1) are shown in the figure above. Exercise 3.26. Find an equation of the tangent line to the parabola y = x 2 - 8 x + 9 at the point (3 , - 6) .
26 J. S´ anchez-Ortega Example 3.27. A Function Not Differentiable at 0: f ( x ) = | x | Solution: Recall that the absolute function is the following piecewise defined function f ( x ) = - x if x < 0 x if x 0 Let us investigate the existence of the limit lim h 0 f (0 + h ) - f (0) h In order to do so, we compute the one sided limits: lim h 0 - f (0 + h ) - f (0) h = lim h 0 - (0 + h ) - 0 h = lim h 0 - h h = lim h 0 ( - 1) = - 1 lim h 0 + f (0 + h ) - f (0) h = lim h 0 (0 + h ) - 0 h = lim h 0 h h = lim h 0 1 = 1

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