Introduction to Derivatives

Transcription

1 Introduction to Derivatives 5-Minute Review: Instantaneous Rates and Tangent Slope Recall te analogy tat we developed earlier First we saw tat te secant slope of te line troug te two points (a, f (a)) and (x, f (x)) on a curve f was given by te difference quotient difference quotient = m sec = (2) x a Tis is illustrated in te left part of Figure 2 x a f (a) Figure 2: Rigt: Te secant lines troug (a, f (a)) and (x, f (x)) as x a better approximate te slope of te curve rigt at (a, f (a)) Rigt: Te tangent line tat we seek a x a Interpretation: If f represents position (distance) and x represents time, ten difference quotient = m sec = x a = position time = Average Velocity (22) Remember if f represented some oter variable quantity, ten te difference quotient would simply be te average rate of cange in tat quantity Of course we were interested in te te slope of f rigt at te point (a, f (a)) or te instantaneous rate of cange rigt at te time x = a We needed to let x approac a in te formula for te difference quotient As te interval got sorter, te secant slope approaced te tangent slope or te average velocity approaced te instantaneous velocity We could not simply let x = a in te formula for te difference quotient in equation (2) because tat would produce te meaningless expression 0 0 It was tis indeterminate expression tat motivated our discussion of its We found tat m sec = (23) x a x a x a

2 mat 30 intro to derivatives: day 2 2 Or using te velocity interpretation Inst Rate of Cange = Ave Rate of Cange = (24) x a x a x a In sort, m tan is te same as te instantaneous rate of cange DEFINITION 2 (Te Tangent Line) If f is a function defined at point a, ten te tangent slope of f at (a, f (a)) is, x a x a if te it exists Te equation of te tangent line is given by y f (a) = m tan (x a) EXAMPLE 22 Let p(x) = 2x 2 + 5x Find te slope of te curve at te point (, p()) = (, 7) and ten find te equation of te tangent line tere Solution We could grap te function, but we do not need to to carry out te calculations We simply use Definition 2 p(x) p(a) 2x = 2 + 5x 7 x a x a x x (2x + 7)(x ) = = 2x + 7 Poly = 9 x x x Tat was easy Now to determine te equation of te tangent we again use Definition 2 Te formula is y p(a) = m tan (x a) In our case a = and p() = 7 and we just found tat m tan = 9 So y 7 = 9(x ) y = 9x 2 Again, very straigtforward if we know te definitions of te terms An Alternative Formula for Tangent Slope In many of te questions tat we will sortly face, it is simpler to carry out te calculation of te tangent slope using sligtly different notation It can often greatly simplify te algebra of te difference quotient We now let (a, f (a)) and (a +, f (a + )) be te two points on te grap of f Ten te difference of te x-coordinates is (a + ) a = and te difference in te y-coordinates is f (a + ) f (a) (See Figure 22 and compare it to Figure 2) So te difference quotient becomes f (a + ) f (a) difference quotient = m sec = Te denominator is simpler tan x a Te penalty is tat one must be careful in computing f (a + ) Consequently, we can rewrite Definition 2 as a a + Figure 22: Black: Te secant lines troug (a, f (a)) and (, f (a + )) as 0 Red: Te tangent line at (a, f (a)) Lecture09tex Version: Mitcell-206/09/226:07:48

3 3 DEFINITION 23 (Te Tangent Line, Alternative Form) If f is a function defined at point a, ten te tangent slope or instantaneous rate of cange of f at (a, f (a)) is if te it exists f (a + ) f (a), 0 EXAMPLE 24 (Equation of a Tangent) Use te alternative form of te definition of tangent slope to find te equation of te tangent line to y = f (x) = x at a = 4 Ten find m tan at a = 8 Solution First for te te point at x = 4 we need to find m tan Using Definition 23, we find f (a + ) f (a) = = = 0 ( ) = 0 ( ) = = 4 By definition te equation of te tangent line is y f (a) = m tan (x a) In tis case a = 4, f (a) = 2, and m tan = 4, so y 2 = 4 (x 4) y = x 4 + Take a second to look back at te calculation of m tan Te algebra was simplified especially in te denominator and tat fact tat is approacing 0 Now for te te point at x = 8 we need to find m tan Using Definition 23 again, we find = = 0 ( ) = 0 ( ) = = 2 8 = 4 2 Notice ow similar te calculation was to te one for te point x = 4 YOU TRY IT 2 Determine te tangent slope for y = f (x) = x at a = 9 Notice ow similar te calculation is to tat in Example 24 Answer to you try it 2 : m tan = 6

4 mat 30 intro to derivatives: day 2 4 We could continue to find tangent slopes for f (x) = x at several oter points Te calculation is always te same in format, but te numbers are a bit different eac time Is tere a way to discern a pattern ere so tat we can do several of tese calculations more quickly? Te answers is yes and it will lead us to define a new function associated wit te original function f Te Derivative Function We saw in Example 24 tat te slope of te curve f (x) = x canged at te as te point canged Eac time we cange te point we ad to redo a similar slope calculation Te important idea is tat for eac point on te curve we could compute te slope of te curve at tat point Tat means tat we ave a new function At eac number a in te domain of te original function f, we can ask wat is te slope of te tangent line rigt at (a, f (a))? Te input to te function is a point a in te domain of f Te output is te corresponding tangent slope at (a, f (a)) Tis new function is called te derivative of f because it is derived from te original function Te derivative function as a special notation We let f (read it as f prime ) denote te derivative of f In Example 24 were f (x) = x, we found te f (4) = 4 because m tan = 4 at te point (4, f (4)) Similarly, f (8) = 4 In you 2 try it 2 f (9) = 6 Back in Example 22 were p(x) = 2x2 + 5x, we found tat p () = 9, tat is, te tangent slope at (, p()) was 9 More generally, te derivative function f (x), wen it exists, represents te slope m tan of te tangent line (or te instantaneous rate of cange) at a variable point (x, f (x)) Often we can determine a formula for f (x) by replacing a by te variable x in te calculation for m tan We call f (x) te derivative function Specifically Figure 23: If f (x) = x, ten f (x) is te piecewise function {, if x < 0 f (x) =, if x > because te derivative is te slope of te function DEFINITION 25 (Te Derivative) Te derivative of f is te function defined by f f (x + ) f (x) (x) =, 0 provided tat te it exists Wen f (x) does exist, we say tat f is differentiable at x If f is differentiable at every point of an open interval I, we say tat f is differentiable on I EXAMPLE 26 (A Derivative Function) Let f (x) = 4x x 2 Find f (x), if it exists Solution Using Definition 25, te derivative is found by calculating f (x+) {}}{{}}{ f f (x + ) f (x) 4(x + ) (x + ) 2 (4x x 2 ) (x) = = 0 0 Expand 4x + 4 x = 2 2x 2 4x + x 2 0 Simplify 4 2x = 2 0 Cancel = 0 4 2x Poly = 4 2x f (x) Lecture09tex Version: Mitcell-206/09/226:07:48

5 Te derivative is f (x) = 4 2x and it exists for all x Tis function gives us te slope at every point on te parabola Now we can get te slope of te original function f (x) = 4x x 2 at any point we coose witout doing a furter it calculation For example, te slope at is f () = 4 2 = 2, te slope at 2 is f (2) = 4 4 = 0, and te slope at 4 is f (4) = 4 8 = 4 Tese points and te corresponding tangent lines are indicated in Figure 25 EXAMPLE 27 (Instantaneous Velocity) Let s(t) = 4t t 2 represent te position of an object (in meters) at time t (in seconds) Find te instantaneous velocity of te object at time t =, t = 2, and t = 35 Solution Te instantaneous velocity is te same as te derivative We just found tis derivative in te previous problem (wit different notation): s (t) = 4 2t So te instantaneous velocity at t = is s () = 2 m/s, and similarly s (2) = 0 m/s, and s (35) = 3 m/s Te sign of te velocity tells us about te direction of motion: positive is forward or up wile negative is backward or down A velocity of 0 indicates tat te object is at rest for te moment (at te top of its fligt or canging direction, say) Figure 24: Tangent lines to te curve f (x) = 4x x 2 Te slope of tese tangents is given by te derivative f (x) = 2 4x Notice tat some slopes are positive, oters negative, and at te top of te curve te slope is 0 YOU TRY IT 22 Return to you try it 2 were f (x) = x Determine f (x) Wat is its domain? Use f (x) to find te tangent slope at x = 64 Wat is te equation of te tangent line at te tis same point? Answers to you try it 22 : f (x) = 2 x f (64) = 6 y = 6 x + 4 EXAMPLE 28 (Anoter Derivative Function) Let f (x) = x 2, (x = 0) Find f (x), if it exists Solution Using Definition 25, te derivative is found by calculating f f (x + ) f (x) (x) = = 0 0 f (x+) {}}{ (x + ) 2 f (x) {}}{ x 2 x 2 (x+) 2 Com Den x = 2 (x+) 2 0 Expand x = 2 x 2 2x 2 0 x 2 (x + ) 2 Simplify 2x = 2 0 x 2 (x + ) 2 Cancel 2x = 0 x 2 (x + ) 2 Rat l = 2x x 2 (x 2 ) Simplify = 2 x 3 So f (x) = 2 for x = 0 x3 YOU TRY IT 23 Let f (x) = x Determine f (x) Wat is its domain? Use f (x) to find te tangent slope at x = 2 Wat is te equation of te tangent line at te tis same point? Answers to you try it 22 : f (x) = x 2 f (2) = 4 y = x

6 mat 30 intro to derivatives: day 2 6 Derivatives from Graps Often in te real world we ave graps of functions (from data tat s been collected) but we do not ave a formula for te function sometimes no simple formula exists Noneteless from te grap of te original function we can often obtain a grap of te derivative by interpreting te derivative as te slope of te te original function Let s start wit a relatively easy example tat will also introduce te idea points were a function is not differentiable EXAMPLE 29 (Grapical Differentiation) Let f (x) be te piecewise function sown in black Figure Figure 25: Black: Te grap of te piecewise function f (x) Red: Te derivative f (x) Note tat tere are tree points were te derivative does not exist Tese points correspond to corners in te grap of f were tere is not unique tangent slope Solution Remember tat te geometric meaning of te derivative is as te tangent slope Using Figure 25, te slope of te section of te grap for x < 3 is 2, wic means tat f (x) = 2 for x < 2 Similarly te slope is f (x) = 2 for 3 < x < 3 and f (x) = for x > 3 Note: Te slope of te tangent line canges abruptly at bot x = 3 and x = 3 Tere are corners in te grap at tese two values As a result, tere is no single value of te slope tat makes sense at eac point In oter words, f ( 3) and f (3) do not exist Te derivative is not continuous at tese points EXAMPLE 20 (Grapical Differentiation) Sketc te grap of f (x) if f (x) is te function sown in te upper alf of Figure 26 Solution Te grap of f in black is repeated in te upper alf of Figure 26 Witout an equation for f, we cannot calculate f exactly Remember tat te geometric meaning of te derivative is as te tangent slope Te best we can do is estimate te slope of f at various points and use tese estimates to plot f We proceed in steps to construct te grap of f in red in te lower alf of te figure Te steps are listed to te left Lecture09tex Version: Mitcell-206/09/226:07:48

7 7 f (x) = 0 at, 0, and so f (x) = 0 at, 0, and Te tangent slopes indicate te sign of f f < 0 f > 0 f < 0 f > 0 A roug grap of f Figure 26: Te process to construct f On te left side, we see tat f as orizontal tangents at x =,, 3 Horizontal tangents mean te tangent slope (te derivative) is 0 So f ( ) = 0, f () = 0, and f (3) = 0 Tese tree points are marked wit on te lower left grap for f 2 Now use te grap of f on te upper rigt (a) For x < : Te tangent slopes to te grap f are negative So f < 0 wen x < Tese slopes increases f reaces 0 rigt at x = (b) For < x < : Te slopes of te tangents are positive Te slope of f starts off small near and increases and ten decreases until it gets back to 0 at x = We must make te grap of f do te same (c) For < x < 3: Te slope of a tangent (and so f ) is negative f decreases at first near and ten increases until it gets back to 0 at x = 3 (d) For x > 3: Te tangent slope and ence f are positive and increasing