Instantaneous Rate of Change Lecture 8. The Derivative.
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1 Instantaneous Rate of Cange Lecture 8 Te Derivative Recall tat te average rate of cange of a function y = f(x) on an interval from x 1 to x 2 is just te ratio of te cange in y to te cange in x: x = f(x 2) f(x 1 ) x 2 x 1 For example, if f measures distance traveled wit respect to time x, ten tis average rate of cange is te average velocity over tat interval But tat leaves us wit te question of wat is te instantaneous velocity at some moment x 0, te velocity tat te speedometer in a car is claimed to give us? Te answer is in some sense quite easy to give: Te instantaneous rate of cange of te function y = f(x) at te point x 0 in its domain is: x x 0 x f(x 0 ) f(x) x x 0 x 0 x provided tis it exists Example 1 Let f(x) = 1/x and let s find te instantaneous rate of cange of f at x 0 = 2 Te first step is to compute te average rate of cange over some interval x 0 = 2 to x; and in order for tis to make sense we need x 2 So tat average rate of cange is ( 1 f(2) f(x) 2 1 ) x = = = x 2 x 2 x 2 x 2x(2 x) = 1 2x Tus, te instantaneous rate of cange at x 0 = 2 is x 2 x 1 x 2 2x = 1 4
2 Te instantaneous rate of cange at some point x 0 = a involves first te average rate of cange from a to some oter value x So if we set = a x, ten 0 and te average rate of cange from x = a + to x = a is f(x) f(a) f(a + ) f(a) = = x x a Eiter of tese last two ratios is known as a difference quotient, a term we sall us repeatedly Wit tis notation te instantaneous rate of cange of f at x = a is te it, if it exists, f(a + ) f(a) Tis as a brief official name: Te derivative of f at x = a, denoted by f (a) is f f(a + ) f(a) (a), te instantaneous rate of cange of f at a, if it exists Example 2 Let s calculate te derivative of f(x) = x 2 at x = 3 From te above definition we ave f f(3 + ) f(3) (3) (3 + ) (6 + ) =
3 It is definitely important to understand te geometric significance of te derivative or instantaneous rate of cange Te key is to remember tat te average rate of cange of a function y = f(x) from some value a to some oter value a + is just te cange in f(x) divided by te (non-zero!!!) cange in x, and tis is just te slope of te line L between te two points (a, f(a)) and (a +, f(a + )) Now let s assume te te grap of y = f(x) near x = 1 is smoot and not too wiggly Ten te smaller we coose te closer te point a + is to a, and te closer to te line L troug tese points is to a line tat just touces te grap at te point (a, f(a) on te grap Untitled L (tangent line) 15 (a, f(a)) Tis iting line L is called te tangent line to te grap at 3
4 te point (a, f(a)), and te punc line is Te slope of te line tangent to te grap y = f(x) at te point (a, f(a)) is te derivative f f(a + ) f(a) (a), of y = f(x) at (a, f(a)), if it exists Example 3 We saw in Example 2 tat te derivative of f(x) = x 2 at x = 3 is f (3) = 6, so te line tangent to te parabola y = x 2 at te point (3, f(3)) = (3, 9) is y = 6(x 3) + 9 = 6x 9 Example 4 Next let s look at a case were tere is no derivative and no tangent line Consider te function f(x) = x and let s see wat, if anyting, its derivative is at x = 0 Remember { x, if x 0; x = x, if x < 0 So te appropriate difference quotient at x = 0 is x = = = { 1, if > 0; 1, if < 0 But simply does not exist! So at x = 0 te function f(x) = x as no derivative and te grap of y = x as no tangent at x = 0 someting tat is quite clear from a glance at te grap! 4
5 Most of te functions we will encounter will ave derivatives at most points in teir domains Generally speaking, derivatives will exist at points on te grap were te function is continuous and tere is no sarp corner (as in te absolute value function) Tere are some pretty nasty functions out tere wit quite bizarre beavior, but fortunately, we won t ave to deal wit tem ere! So, let y = f(x) be one of our reasonably nice functions wit derivatives at most points in its domain Ten at eac suc point tere is a derivative, and ence tere is a new function tat assigns to eac suc nice point x in te domain of f a value, f (x) Not surprisingly, we call tis new function te derivative of f(x) Tus, Te derivative of a function y = f(x) is te function defined by f f(x + ) f(x) (x) So te derivative f (x) of a function y = f(x) spews out te slope of te tangent to te grap y = f(x) at eac x in te domain of f were tere is a tangent line One ting we will ave to deal wit is tat tere is quite a variety of notational versions of te derivative of a function y = f(x) Here are te ones we are most likely to meet: f (x), d dx f, d dx f(x), df dx, dy dx 5
6 Example 4 Let s find te derivative of te function f(x) = x 2 Well, from te definition above f (x) f(x + ) f(x) (x + ) 2 x 2 x 2 + 2x + 2 x 2 2x + 2 = 2x (2x + ) Tus at eac point x in R, te domain of f, te tangent to te parabola y = x 2 as slope f (x) = 2x Example 5 Given y = 1/x, let s find te equation of te line tangent to te grap of tis equation at te point (4, 1/4) First step is to calculate te slope function, te derivative dy dx Well, dy dx f(x + ) f(x) 1/(x + ) 1/x x (x + ) x(x + ) 1 x 2 + x = 1 x 2 So wen x = 4, te derivative is dy = 1 Terefore, te dx x=4 16 tangent line at (4, 1/4) is y = 1/16(x 4) + 1/4 = 1/16x + 1/2 6
7 Practice Problems 1 Find te derivative f (x) of te function f(x) = 5x Find te derivative dy/dx of te constant function y = 4 3 Find te tangent line to te grap y = x at te point (4, 2) 4 Find all points on te grap of f(x) = 3x 2 +1 were te tangent line as slope 1 5 Find te derivative of te function y = f(x) = x 2 at te point x = 2 6 A car starting from a dead stop is s(t) = t 2 feet from te starting point t seconds after it begins to move Wat is te velocity of te car 20 seconds after it begins its journey? How long does it take for te car to reac a speed of 60 mp? Of 80 mp? 7
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