CHAPTER 3: Derivatives 3.1: Derivatives, Tangent Lines, and Rates of Cange 3.2: Derivative Functions and Differentiability 3.3: Tecniques of Differentiation 3.4: Derivatives of Trigonometric Functions 3.5: Differentials and Linearization of Functions 3.6: Cain Rule 3.7: Implicit Differentiation 3.8: Related Rates Derivatives represent slopes of tangent lines and rates of cange (suc as velocity). In tis capter, we will define derivatives and derivative functions using limits. We will develop sort cut tecniques for finding derivatives. Tangent lines correspond to local linear approximations of functions. Implicit differentiation is a tecnique used in applied related rates problems.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.1 SECTION 3.1: DERIVATIVES, TANGENT LINES, AND RATES OF CHANGE LEARNING OBJECTIVES Relate difference quotients to slopes of secant lines and average rates of cange. Know, understand, and apply te Limit Definition of te Derivative at a Point. Relate derivatives to slopes of tangent lines and instantaneous rates of cange. Relate opposite reciprocals of derivatives to slopes of normal lines. PART A: SECANT LINES For now, assume tat f is a polynomial function of x. (We will relax tis assumption in Part B.) Assume tat a is a constant. Temporarily fix an arbitrary real value of x. (By arbitrary, we mean tat any real value will do). Later, instead of tinking of x as a fixed (or single) value, we will tink of it as a moving or varying variable tat can take on different values. Te secant line to te grap of f on te interval [ a, x], were a < x, is te line tat passes troug te points ( a, f ( a) ) and ( x, f ( x) ). secare is Latin for to cut. Te slope of tis secant line is given by: rise run = f ( x ) f ( a). x a difference of outputs We call tis a difference quotient, because it as te form: difference of inputs.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.2 PART B: TANGENT LINES and DERIVATIVES If we now treat x as a variable and let x a, te corresponding secant lines approac te red tangent line below. tangere is Latin for to touc. A secant line to te grap of f must intersect it in at least two distinct points. A tangent line only need intersect te grap in one point, were te line migt just touc te grap. (Tere could be oter intersection points). Tis limiting process makes te tangent line a creature of calculus, not just precalculus. Below, we let x approac a Below, we let x approac a from te rigt ( x a + ). from te left ( x a ). (See Footnote 1.) We define te slope of te tangent line to be te (two-sided) limit of te difference quotient as x a, if tat limit exists. We denote tis slope by f( a), read as f prime of (or at) a. f( a), te derivative of f at a, is te slope of te tangent line to te grap of f at te point ( a, f ( a) ), if tat slope exists (as a real number). \ f is differentiable at a f a exists. Polynomial functions are differentiable everywere on. (See Section 3.2.) Te statements of tis section apply to any function tat is differentiable at a.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.3 Limit Definition of te Derivative at a Point a (Version 1) f ( x) f ( a) f( a), if it exists x a x a If f is continuous at a, we ave te indeterminate Limit Form 0 0. Continuity involves limits of function values, wile differentiability involves limits of difference quotients. Version 1: Variable endpoint (x) Slope of secant line: f ( x) f ( a) x a a is constant; x is variable A second version, were x is replaced by a +, is more commonly used. Version 2: Variable run () Slope of secant line: f ( a+ ) f ( a) a is constant; is variable If we let te run 0, te corresponding secant lines approac te red tangent line below. Below, we let approac 0 Below, we let approac 0 from te rigt ( 0 + ). from te left ( 0 ). (Footnote 1.)
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.4 Limit Definition of te Derivative at a Point a (Version 2) f ( a+ ) f ( a) f( a), if it exists 0 Version 3: Two-Sided Approac Limit Definition of te Derivative at a Point a (Version 3) f ( a+ ) f ( a ) f( a), if it exists 0 2 Te reader is encouraged to draw a figure to understand tis approac. Principle of Local Linearity Te tangent line to te grap of f at te point ( a, f ( a) ), if it exists, represents te best local linear approximation to te function close to a. Te grap of f resembles tis line if we zoom in on te point ( a, f ( a) ). Te tangent line model linearizes te function locally around a. We will expand on tis in Section 3.5. (Te figure on te rigt is a zoom in on te box in te figure on te left.)
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.5 PART C: FINDING DERIVATIVES USING THE LIMIT DEFINITIONS Example 1 (Finding a Derivative at a Point Using Version 1 of te Limit Definition) Let f ( x)= x 3. Find f () 1 using Version 1 of te Limit Definition of te Derivative at a Point. Solution f ( x) f 1 f ()= 1 lim x 1 x 1 x 1 x 3 1 x 1 () () 3 ( Here, a = 1. ) TIP 1: Te brackets ere are unnecessary, but better safe tan sorry. x 1 x 3 1 x 1 Limit Form 0 0 We will factor te numerator using te Difference of Two Cubes template and ten simplify. Syntetic Division can also be used. (See Capter 2 in te Precalculus notes). () 1 x 1 ( x 2 + x + 1) x 1 ( x 1) () 1 x 1 x 2 + x + 1 () 2 + ()+ 1 1 = 1 = 3
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.6 Example 2 (Finding a Derivative at a Point Using Version 2 of te Limit Definition; Revisiting Example 1) Let f ( x)= x 3, as in Example 1. Find f () 1 using Version 2 of te Limit Definition of te Derivative at a Point. Solution f ( 1+ ) f 1 f ()= 1 lim 0 0 1 0 3 1+ () () 3 1 ( Here, a = 1. ) We will use te Binomial Teorem to expand ( 1+ ) 3. (See Capter 9 in te Precalculus notes.) () 3 + 31 () 2 + 31 1 + 3 + 3 2 + 3 1 0 3 + 3 2 + 3 0 () 1 0 ( 3+ 3 + ) 2 1 () 3+ 3 + 2 0 = 3+ 30 + ( 0) 2 = 3 () 2 + ( ) 3 1 We obtain te same result as in Example 1: f ()= 1 3.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.7 PART D: FINDING EQUATIONS OF TANGENT LINES Example 3 (Finding Equations of Tangent Lines; Revisiting Examples 1 and 2) Solution Find an equation of te tangent line to te grap of y = x 3 at te point were x = 1. (Review Section 0.14: Lines in te Precalculus notes.) Let f ( x)= x 3, as in Examples 1 and 2. Find f (), 1 te y-coordinate of te point of interest. f ()= 1 () 1 3 = 1 Te point of interest is ten: ( 1, f () 1 )= ( 1, 1). Find f (), 1 te slope (m) of te desired tangent line. In Part C, we sowed (twice) tat: f ()= 1 3. Find a Point-Slope Form for te equation of te tangent line. y y 1 = mx x 1 y 1 = 3 x 1 Find te Slope-Intercept Form for te equation of te tangent line. y 1 = 3x 3 y = 3x 2 Observe ow te red tangent line below is consistent wit te equation above.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.8 Te Slope-Intercept Form can also be obtained directly. Remember te Basic Principle of Graping: Te grap of an equation consists of all points (suc as 1, 1 ere) wose coordinates satisfy te equation. PART E: NORMAL LINES y = mx + b ()= 1 ( 3) ()+ 1 b ( Solve for b. ) b = 2 y = 3x 2 Assume tat P is a point on a grap were a tangent line exists. Te normal line to te grap at P is te line tat contains P and tat is perpendicular to te tangent line at P. Example 4 (Finding Equations of Normal Lines; Revisiting Example 3) Find an equation of te normal line to te grap of y = x 3 at P( 1, 1). Solution In Examples 1 and 2, we let f ( x)= x 3, and we found tat te slope of te tangent line at 1, 1 f 1 was given by: ()= 3. Te slope of te normal line at 1, 1 of te slope of te tangent line. is ten 1 3, te opposite reciprocal A Point-Slope Form for te equation of te normal line is given by: y y 1 = mx x 1 y 1 = 1 ( 3 x 1 ) Te Slope-Intercept Form is given by: y = 1 3 x + 4 3.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.9 WARNING 1: Te Slope-Intercept Form for te equation of te normal line at P cannot be obtained by taking te Slope-Intercept Form for te equation of te tangent line at P and replacing te slope wit its opposite reciprocal, unless P lies on te y-axis. In tis Example, te normal line is not given by: y = 1 3 x 2. PART F: NUMERICAL APPROXIMATION OF DERIVATIVES Te Principle of Local Linearity implies tat te slope of te tangent line at te point ( a, f ( a) ) can be well approximated by te slope of te secant line on a small interval containing a. Wen using Version 2 of te Limit Definition of te Derivative, tis implies tat: f( a) f ( a+ ) f ( a) wen 0. Example 5 (Numerically Approximating a Derivative; Revisiting Example 2) Let f ( x)= x 3, as in Example 2. We will find approximations of f 1 (See Example 8 in Part H.) f ( 1+ ) f () 1 0.1 3.31 0.01 3.0301 0.001 3.003001 0 3 0.001 2.997001 0.01 2.9701 0.1 2.71 ()., or 3+ 3 + 2 ( 0) ( See Example 2. ) If we only ave a table of values for a function f instead of a rule for f ( x), we may ave to resort to numerically approximating derivatives.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.10 PART G: AVERAGE RATE OF CHANGE Te average rate of cange of f on a, b is equal to te slope of te secant line on a, b, wic is given by: rise run = f ( b ) f ( a). (See Footnotes 2 and 3.) b a Example 6 (Average Velocity) Average velocity is a common example of an average rate of cange. Let s say a car is driven due nort 100 miles during a two-our trip. Wat is te average velocity of te car? Let t = te time (in ours) elapsed since te beginning of te trip. Let y = st (), were s is te position function for te car (in miles). s gives te signed distance of te car from te starting position. Te position (s) values would be negative if te car were sout of te starting position. Let s( 0)= 0, meaning tat y = 0 corresponds to te starting position. Terefore, s( 2)= 100 (miles). Te average velocity on te time-interval a, b is te average rate of cange of position wit respect to time. Tat is, cange in position cange in time = s t Here, te average velocity of te car on 0, 2 is: s( 2) s( 0) 100 0 = 2 0 2 were (uppercase delta) denotes cange in = sb sa, a difference quotient b a = 50 miles our or mi r or mp TIP 2: Te unit of velocity is te unit of slope given by: unit of s unit of t.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.11 Te average velocity is 50 mp on 0, 2 in te tree scenarios below. It is te slope of te orange secant line. (Axes are scaled differently.) Below, te velocity is constant (50 mp). (We are not requiring te car to slow down to a stop at te end.) Below, te velocity is increasing; te car is accelerating. Below, te car breaks te rules, backtracks, and goes sout. WARNING 2: Te car s velocity is negative in value wen it is backtracking; tis appens wen te grap falls. Note: Te Mean Value Teorem for Derivatives in Section 4.2 will imply tat te car must be going exactly 50 mp at some time value t in ( 0, 2). Te teorem applies in all tree scenarios above, because s is continuous on 0, 2. [ ] and is differentiable on 0, 2
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.12 PART H: INSTANTANEOUS RATE OF CHANGE Te instantaneous rate of cange of f at a is equal to f( a), if it exists. Example 7 (Instantaneous Velocity) Instantaneous velocity (or simply velocity) is a common example of an instantaneous rate of cange. Let s say a car is driven due nort for two ours, beginning at noon. How can we find te instantaneous velocity of te car at 1pm? (If tis is positive, tis can be tougt of as te speedometer reading at 1pm.) Let t = te time (in ours) elapsed since noon. Let y = st (), were s is te position function for te car (in miles). Consider average velocities on variable time intervals of te form a, a +, if > 0, or te form a +, a, if < 0, were is a variable run. (We can let = t.) Te average velocity on te time-interval a, a +, if > 0, or a +, a, if < 0, is given by: sa+ ( ) sa Tis equals te slope of te secant line to te grap of s on te interval. (See Footnote 1 on te < 0 case.)
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.13 Let s assume tere exists a non-vertical tangent line to te grap of s at te point ( a, sa ). Ten, as 0, te slopes of te secant lines will approac te slope of tis tangent line, wic is s( a). Likewise, as 0, te average velocities will approac te instantaneous velocity at a. Below, we let 0 +. Below, we let 0. Te instantaneous velocity (or simply velocity) at a is given by: sa+ ( ) sa s( a), or v( a), if it exists 0 In our Example, te instantaneous velocity of te car at 1pm is given by: s( 1+ ) s() 1 s(), 1 or v()= 1 lim 0 Let s say st ()= t 3. Example 2 ten implies tat s()= 1 v()= 1 3 mp.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.14 Example 8 (Numerically Approximating an Instantaneous Velocity; Revisiting Examples 5 and 7) Again, let s say te position function s is defined by: st ()= t 3 on 0, 2. We will approximate v(), 1 te instantaneous velocity of te car at 1pm. We will first compute average velocities on intervals of te form 1, 1 +. Here, we let 0 +. Interval Value of (in ours) 1, 2 1 1,1.1 0.1 1, 1.01 0.01 1, 1.001 0.001 0 + Average velocity, s ( 1+ ) s 1 s( 2) s() 1 () = 7 mp 1 s( 1.1) s() 1 = 3.31 mp 0.1 s( 1.01) s() 1 = 3.0301 mp 0.01 s( 1.001) s() 1 = 3.003001 mp 0.001 3 mp Tese average velocities approac 3 mp, wic is v(). 1 WARNING 3: Tables can sometimes be misleading. Te table ere does not represent a rigorous evaluation of v 1 (). Answers are not always integer-valued.
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.15 We could also consider tis approac: Interval Value of Average velocity, s ( 1+ ) s() 1 (rounded off to six significant digits) 1, 2 1 our 7.00000 mp 1, 1 1 60 1 minute 3.05028 mp 1, 1 1 3600 1 second 3.00083 mp 0 + 3 mp Here, we let 0. Interval Value of (in ours) 0,1 1 0.9,1 0.1 0.99, 1 0.01 0.999, 1 0.001 0 Average velocity, s ( 1+ ) s 1 s( 0) s() 1 = 1 mp 1 () () s( 0.9) s 1 = 2.71 mp 0.1 s( 0.99) s() 1 = 2.9701 mp 0.01 s( 0.999) s() 1 = 2.997001 mp 0.001 3 mp Because of te way we normally look at slopes, we may prefer to rewrite te first difference quotient s ( 0 ) s() 1 as s () 1 s( 0), 1 1 and so fort. (See Footnote 1.)
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.16 Example 9 (Rate of Cange of a Profit Function) Solution A company sells widgets. Assume tat all widgets produced are sold. P( x), te profit (in dollars) if x widgets are produced and sold, is modeled by: P( x)= x 2 + 200x 5000. Find te instantaneous rate of cange of profit at 60 widgets. (In economics, tis is referred to as marginal profit.) WARNING 4: We will treat te domain of P as 0, ), even toug one could argue tat te domain sould only consist of integers. Be aware of tis issue wit applications suc as tese. We want to find P( 60). P 60 P( 60 + ) P( 60) 0 ( 60 + ) 2 + 200( 60 + ) 5000 60 0 0 2 + 200 60 5000 WARNING 5: Grouping symbols are essential wen expanding P 60 ere, because we are subtracting an expression wit more tan one term. 2 ( 3600 + 120 + )+ 12,000 + 200 5000 3600 + 12,000 5000 TIP 3: Instead of simplifying witin te brackets immediately, we will take advantage of cancellations. 3600 120 2 + 12,000 + 200 5000 + 3600 12,000 + 5000 0 80 2 0
() 1 0 0 = 80 0 (Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.17 ( 80 ) 1 () ( 80 ) = 80 dollars widget Tis is te slope of te red tangent line below. If we produce and sell one more widget (from 60 to 61), we expect to make about $80 more in profit. Wat would be your business strategy if marginal profit is positive?
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.18 FOOTNOTES 1. Difference quotients wit negative denominators. Our forms of difference quotients allow negative denominators ( runs ), as well. Tey still represent slopes of secant lines. Left figure ( x < a): slope = rise run = f ( a ) f ( x) = f ( x ) f a a x ( x a) Rigt figure ( < 0): slope = rise run = f ( a ) f ( a+ ) f ( a+ ) f ( a) = a a + = f ( x ) f a x a = f ( a+ ) f a 2. Average rate of cange and assumptions made about a function. Wen defining te average rate of cange of a function f on an interval a, b, were a < b, sources typically do not state te assumptions made about f. Te formula f ( b ) f ( a) seems only to require b a te existence of f ( a) and f ( b), but we typically assume more tan just tat. Altoug te slope of te secant line on a, b can still be defined, we need more for te existence of derivatives (i.e., te differentiability of f ) and te existence of non-vertical tangent lines. We ordinarily assume tat f is continuous on a, b. Ten, tere are no oles, jumps, or vertical asymptotes on a, b wen f is graped. (See Section 2.8.) We may also assume tat f is differentiable on a, b. Ten, te grap of f makes no sarp turns and does not exibit infinite steepness (corresponding to vertical tangent lines). However, tis assumption may lead to circular reasoning, because te ideas of secant lines and average rate of cange are used to develop te ideas of derivatives, tangent lines, and instantaneous rate of cange. Differentiability is defined in terms of te existence of derivatives. We may also need to assume tat f is continuous on a, b. (See Footnote 3.)
(Section 3.1: Derivatives, Tangent Lines, and Rates of Cange) 3.1.19. 3. Average rate of cange of f as te average value of f. Assume tat a, b f is continuous on. Ten, te average rate of cange of f on a, b is equal to te average value of f on a, b. In Capter 5, we will assume tat a function (say, g) is continuous on a, b b g( x)dx a and ten define te average value of g on a, b to be ; te numerator is a b a definite integral, wic will be defined as a limit of sums. Ten, te average value of b f( x)dx a a, b is given by:, wic is equal to f ( b ) f ( a) by te Fundamental b a b a Teorem of Calculus. Te teorem assumes tat te integrand [function], on a, b. f on f, is continuous
(Section 3.2: Derivative Functions and Differentiability) 3.2.1 SECTION 3.2: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY LEARNING OBJECTIVES Know, understand, and apply te Limit Definition of te Derivative Function. Know sort cuts for differentiation, including te Power Rule. Evaluate derivative functions and relate teir values to slopes of tangent lines and rates of cange. Be able to find iger-order derivatives, and recognize notations for various orders. Understand te relationsips between position, velocity, and acceleration in rectilinear motion. Recognize differentiability of functions on open and closed intervals. Recognize possible beaviors of functions and teir graps were tey are not differentiable. PART A: DERIVATIVE FUNCTIONS Let f be a function. f may ave a derivative function, called be given by eiter of te following. f, wose rule may Limit Definition of te Derivative Function (Version 1) f ( x+ ) f ( x) f( x), if it exists 0 Limit Definition of te Derivative Function (Version 2; Two-Sided Approac) f ( x+ ) f ( x ) f( x), if it exists 0 2 We ave taken Limit Definitions from Section 3.1 and replaced te constant a wit te variable x. f ( x+ ) f ( x) x, are variable
(Section 3.2: Derivative Functions and Differentiability) 3.2.2 Take eiter version. Te domain of f consists of all real values of x for wic te indicated limit exists. We say tat f is differentiable at tose values. is called differentiation. We may say tat f( x) is te derivative of te expression wit respect to x. x is te variable of differentiation. Te process of finding f x derivative of te function f, or tat f x (See Footnote 1.) PART B: SHORT CUTS FOR DIFFERENTIATION We may tink of derivative functions as slope functions. Some Sort Cuts for Differentiation Assumptions: c, m, b, and n are real constants. f is a function tat is differentiable were we care. If g( x)= ten g( x)= Comments f is te 1. c 0 Te derivative of a constant is 0. 2. mx + b m Te derivative of a linear function is te slope. 3. x n nx n1 Power Rule 4. c f x f x c Constant Multiple Rule Proofs. Te Limit Definition of te Derivative can be used to prove tese sort cuts. (See Footnotes 2 and 3.)
(Section 3.2: Derivative Functions and Differentiability) 3.2.3 Example 1 (Rule 1: Differentiating a Constant [Function]) Let g( x)= 2. Ten, g( x)= 0 (for all real values of x; tis goes witout saying). For instance, g( 1)= 0 and g( )= 0. Observe tat, for eac real value of x, te corresponding point on te grap of g below as a orizontal tangent line, namely te grap itself. Example 2 (Rule 2: Differentiating a Linear Function) Let g( x)= 3x + 1. Ten, g( x)= 3 (for all real values of x). For instance, g( 0)= 3 and g()= 1 3. Observe tat, for eac real value of x, te corresponding point on te grap of g below as a tangent line of slope 3, namely te grap itself.
(Section 3.2: Derivative Functions and Differentiability) 3.2.4 Example 3 (Motivating Rule 3: Differentiating a Power Function; Evaluating Derivatives and Velocities) Let g( x)= x 3. Unlike in Examples 1 and 2, te derivative function g will not be a constant function. Different tangent lines to te grap of g can ave different slopes. We will use te Limit Definition of te Derivative to find g( x). Tis will parallel our work in Example 2 in Section 3.1. g( x+ ) g( x) g( x) 0 ( x + ) 3 x3 0 0 We will use te Binomial Teorem to expand ( x + ) 3. (See Capter 9 in te Precalculus notes.) ( x) 3 + 3 x 2 + 3 x ( ) 2 + ( ) 3 x 3 + 3x 2 + 3x 2 + 3 x 3 0 3x 2 + 3x 2 + 3 0 () 1 0 ( 3x 2 + 3x + ) 2 1 () x3
(Section 3.2: Derivative Functions and Differentiability) 3.2.5 3x 2 + 3x + 2 0 = 3x 2 + 3x( 0)+ ( 0) 2 = 3x 2 We now ave te derivative function rule g( x)= 3x 2 (for all real values of x). For instance, g()= 1 31 () 2 = 3. We already knew tis from Section 3.1, Part C, but we can now quickly find derivatives for oter values of x. For example, g( 0)= 30 2 = 0, and g( 1.5)= 3( 1.5) 2 = 6.75. WARNING 1: Do not confuse te original function s values, wic correspond to y-coordinates of points, wit derivative values, wic correspond to slopes of tangent lines. For example, g( 1.5)= 1.5 wic means tat te point 1.5, 3.375 = 3 1.5 g 1.5 3 = 3.375, lies on te grap of g. Also, 2 = 6.75, wic is te slope of te tangent line at tat point. In Section 3.1, we saw tat derivatives can be related to rates of cange, suc as velocity. In Section 3.1, Example 7, if te position function is given by st ()= t 3, ten te velocity function is given by v()= t s()= t 3t 2. For instance, v()= 1 3 mp, and v( 1.5)= 6.75 mp.
Example Set 4 (Rule 3: Power Rule) (Section 3.2: Derivative Functions and Differentiability) 3.2.6 Unless we are instructed to use te Limit Definition of te Derivative, as in Example 3, we will use te Power Rule of Differentiation as a sort cut to differentiate a power of x suc as x 3. We bring down te exponent (3) as a coefficient, and we ten subtract one to get te new exponent, resulting in 3x 2. TIP 1: Be prepared to rewrite a variety of expressions as powers of x so tat te Power Rule may be readily applied. WARNING 2: We will differentiate expressions suc as 3 x and x x in Capter 7. Tey do not represent power functions, and te Power Rule ere does not apply. If g( x)= ten g( x)= x 2 2x 1 = 2x x 3 3x 2 x 4 4x 3 x 17 1 x = x1 1 x = 2 x 2 3 17x 17 1 1x 2 = 1 x 2 2x 3 = 2 x 3 1 x = x 1/2 2 x1/2 = 1 2x, or 1 1/2 2 x 1 x = x 1/3 3 x 2/3 = 1 3x, or 1 2/3 3 3 x 2 1 = 1 4 x = 5/4 x 5/4 5 4 x 9/4 = x 5 In an algebra class, 4 x 9 may be rewritten as x 2 4 ( x ). We will discuss domain issues later in tis section. Te above table demonstrates te following: Te derivative of an even function is odd. Te derivative of an odd function is even. 5 4x, or 5 9/4 4 4 x 9
(Section 3.2: Derivative Functions and Differentiability) 3.2.7 Example 5 (Rule 4: Constant Multiple Rule) Informally, te Constant Multiple Rule states tat te derivative of a constant multiple equals te constant multiple of te derivative. For example, te derivative of twice x 3 is twice te derivative of x 3. Tat is, if g( x)= 2x 3, ten g( x)= 2( 3x 2 )= 6x 2. TIP 2: Basically, we multiply te coefficient by te exponent, and we ten subtract one from te old exponent to get te new exponent. For instance, g()= 1 61 () 2 = 6.
PART C: HIGHER-ORDER DERIVATIVES (Section 3.2: Derivative Functions and Differentiability) 3.2.8 Indeed, we can take te derivative of te derivative of, and so on. Higer-Order Derivatives f ( x), read as f double prime of (or at) x, is te second derivative (or te second-order derivative) of f wit respect to x. It is te [first] derivative of f( x) wit respect to x. f ( x), read as f triple prime of (or at) x, is te tird derivative (or te tird-order derivative) of f wit respect to x. It is te [first] derivative of f ( x) wit respect to x. Higer-order derivatives are denoted by f ( 4) ( x), f ( 5) ( x), etc. Roman numerals migt also be used: f IV ( x), f V ( x), etc. (See Footnote 4. See Capter 4 for grapical interpretations of f.) Example 6 (Higer-Order Derivatives) Let f ( x)= x 3. Ten, f( x)= 3x 2, f ( x)= 6x, f ( x)= 6, and f ( 4) ( x)= 0. PART D: RECTILINEAR MOTION: POSITION, VELOCITY, and ACCELERATION In Section 3.1, Parts G and H, we discussed te motion of a car being driven due nort. Tis is an example of rectilinear motion, or motion along a coordinate line. Te starting position of te car corresponds to 0 on te line. Te real numbers on te line correspond to position values, wic are signed distances of te car from te starting position. For instance, 3 corresponds to te position tree miles due nort of te starting point. Positive directions are typically associated wit nort, up, east, rigt, or forward. Also, 3 corresponds to te position tree miles due sout of te starting point. Negative directions are typically associated wit sout, down, west, left, or backward. + 0 0 +
(Section 3.2: Derivative Functions and Differentiability) 3.2.9 In our car examples, we let s be te position function for te car. st () gives te position value of te car (in miles) t ours after te trip begins. Te independent variable t represents time or time elapsed. It will be our variable of differentiation. ()= Let v be te velocity function for te car. Ten, vt s (), t because velocity is te rate of cange of position wit respect to time. Te unit of velocity ere is miles per our, or mi, or mp. r ()= ()= Let a be te acceleration function for te car. Ten, at v t s (), t because acceleration is te rate of cange of velocity wit respect to time. Te unit of acceleration ere is miles per our per our, or mi r. 2 Example 7 (Average Acceleration) A commercial says tat a car can go from 0 [mp] to 60 [mp] in 5 seconds. Te average acceleration of te car on tat five-second interval 0, 5 [ ]is given by: v( 5) v 0 5 0 = 60 0 5 = 12 mp sec We can convert to a more internally consistent unit: 12 mp sec mi / r = 12 sec 3600 sec 1r = 43,200 mi r 2 Example 8 (Position, Velocity, and Acceleration) Let st ()= t 3, as in Example 3. Ten, v()= t s()= t 3t 2, and at ()= v()= t 6t. For instance, a()= 1 61 ()= 6 mi. Tis means tat te 2 r car s acceleration is 6 miles per our per our wen one our as elapsed.
PART E: NOTATIONS FOR DERIVATIVES (Section 3.2: Derivative Functions and Differentiability) 3.2.10 Let y = f ( x), were f ( x)= x 3, say. Te various orders of derivatives can be denoted in a variety of ways. First derivative Second derivative nt derivative f( x)= 3x 2 f ( x)= 6x f ( n) x (Lagrange s notation) y = 3x 2 (See Warning 3.) dy dx = 3x2 (Leibniz s notation; see note on next page) d dx y = 3x2, or d ( dx x3 )= 3x 2 (See Differential operators note.) D x ( x 3 )= 3x 2 (Euler s notation; see Differential operators note.) (Lagrange s notation) y = 6x (Lagrange s notation) (See Warning 3.) (See Warning 4.) d 2 y dx = 6x d n y 2 dx n (Leibniz s notation; (Leibniz s notation; see Differential see Differential operators note) operators note) d 2 y = 6x, or 2 dx d 2 ( x 3 )= 6x dx 2 (See Differential operators note.) 2 D x ( x 3 )= 6x, or D xx ( x 3 )= 6x (Euler s notation; see Differential operators note.) d n y, or n dx d n ( x 3 ) dx n (See Differential operators note.) n D x ( x 3 ) (Euler s notation; see Differential operators note.) WARNING 3: Te y notation suffers te critical drawback of not indicating te variable of differentiation (ere, x). In tis work, we will assume tat y = dy dx denoted by y. dy. Oter derivatives suc as dt, dy, etc. will not be d WARNING 4: It is not recommended to use y n, since it is easily confused wit y n, te nt power of y. (See Footnote 4.)
(Section 3.2: Derivative Functions and Differentiability) 3.2.11 Leibniz s notation. Te notation dy evokes te idea of slope. It may be dx tougt of as a quotient of differentials (dy and dx), wic represent infinitesimal (arbitrarily small) canges in y and x. (See Section 3.5.) Separating te differentials is frequently done in practice, altoug many tink of dy as an inseparable entity rater tan a quotient in rigorous work. dx Differential operators. d dx and D x operate on te following expression by differentiating it wit respect to x. Some sources simply use D. Generically, if f is te cubing function, ten Df is tree times te squaring function. d 2 indicates repeated (or iterated ) differentiation. 2 dx For example, d 2 dx y = d d 2 dx dx y. Newton s notation (obsolete). Sir Isaac Newton referred to fluxions, were derivatives were taken wit respect to time: x i = dx dt.
(Section 3.2: Derivative Functions and Differentiability) 3.2.12 PART F: DIFFERENTIABILITY ON INTERVALS; RIGHT-HAND and LEFT-HAND DERIVATIVES and TANGENT LINES Assume tat f is a function and a and b are real constants suc tat a < b. Differentiability on an Open Interval f is differentiable on te open interval a, b f is differentiable at all real numbers in a, b Tis extends to unbounded open intervals of te form ( a, ), (, b), or (, ). Rigt-Hand Derivative at a Point a; Rigt-Hand Tangent Lines Te rigt-and derivative at a is defined as: f ( a+ ) f a f + ( a) 0 +, if it exists We define te rigt-and tangent line at te point a, f a passing troug tis point wose slope is equal to f + ( a). ( ) to be te line If te above limit can be said to be or, and if f is continuous from te rigt at a, ten te rigt-and tangent line is vertical. Informally, a vertical tangent line indicates were a grap is becoming infinitely steep. (See Footnote 5 on notation.) Left-Hand Derivative at a Point b; Left-Hand Tangent Lines Te left-and derivative at b is defined as: f ( b+ ) f b f ( b) 0, if it exists We define te left-and tangent line at te point b, f b passing troug tis point wose slope is equal to f ( b). ( ) to be te line If te above limit can be said to be or, and if f is continuous from te left at a, ten te left-and tangent line is vertical. (See Section 3.1, Footnote 1 on sign issues.)
(Section 3.2: Derivative Functions and Differentiability) 3.2.13 Relating One-Sided and Two-Sided Derivatives As wit limits, f( a) exists f + ( a) and f ( a) bot exist, and f + ( a)= f ( a). If c is a real constant, ten f( a)= c ( a)= c and ( a)= c. Differentiability on a Closed Interval f is differentiable on te closed interval [ a, b] 1) f is defined on a, b, 2) f is differentiable on ( a, b), 3) f + ( a) exists, and 4) f ( b) exists 3) and 4) weaken te differentiability requirements at te endpoints, a and b. Imagine taking limits as we pus outwards towards te endpoints. Observe te similarity wit te idea of continuity on a closed interval. Differentiability on alf-open, alf-closed intervals suc as a, b similarly defined. In te case of [ a, b), we would replace a, b in 1), and we would delete 4). f + f [ ) can be [ ] wit [ a, b) [ ] or [ a, b) WARNING 5: Differentiability of f on an interval suc as a, b does not imply differentiability [in a two-sided sense] at a. Tat is, a migt not be in Dom( f ). (Many sources avoid tis issue.)
(Section 3.2: Derivative Functions and Differentiability) 3.2.14 Example 9 (Differentiability on a Closed Interval) Let f ( x)= 1 x 2 on te restricted x-interval [ 0.8, 0.8]. Ten, f is differentiable on tat interval. Parts of te one-sided tangent lines at te endpoints of te grap of f are drawn in red below. Metods from Section 3.6 will allow us to find f( x). It turns out tat f + ( 0.8)= 4 3 and f ( 0.8 )= 4 3. Example 10 (Vertical Tangent Lines) Let g( x)= 1 x 2 on te implied domain, [ 1, 1]. Ten, g is differentiable on te open interval 1, 1 on te closed interval [ 1, 1]. but is not differentiable Tere is a rigt-and vertical tangent line (in red) at te point ( 1, 0), because te limit of te slopes of te secant lines (in orange) coming in from te rigt can be said to be. Informally, we will write g + ( 1):, g ( 1+ ) g ( 1) because lim =. (It is not, because te secant 0 + lines rise from left to rigt, and we still look at slopes left-to-rigt. ) Also, g is continuous from te rigt at 1.
(Section 3.2: Derivative Functions and Differentiability) 3.2.15 Tere is a left-and vertical tangent line (in red) at te point ( 1, 0), because te limit of te slopes of te secant lines (in orange) coming in from te left can be said to be. Informally, we will write g (): 1, because lim 0 at 1. g ( 1+ ) g 1 () PART G: NON-DIFFERENTIABILITY =. Also, g is continuous from te left We will examine a variety of situations in wic a function f is not differentiable at a real constant a. Tat is, f a does not exist (DNE). Differentiability Implies (and terefore Requires) Continuity 1) If f is differentiable at a, ten f is continuous at a. 2) If f is not continuous at a, ten f is not differentiable at a. 1) and 2) form a pair of contrapositive statements. Terefore, tey are logically equivalent. Since 1) is true, 2) must also old true. Footnote 6 as a proof. Example 11 (Differentiability Requires Continuity) Let f ( x)= 1, x > 1. Ten, f is not continuous at x = 1. 1, x 1 Terefore, f is not differentiable at x = 1. Te slopes of te secant lines coming in from te rigt at x = 1 f ( 1+ ) f () 1 approac, so lim =, and f () 1 does not exist 0 + (DNE). However, because f is not continuous from te rigt at x = 1, its grap does not ave a rigt-and vertical tangent line at te point ( 1, 1).
(Section 3.2: Derivative Functions and Differentiability) 3.2.16 We now consider situations were a function is continuous at a, but it is not differentiable tere. Tis typically means tat its grap makes a sarp turn at x = a (indicating two tangent lines) or it as a vertical tangent line tere. Losing Differentiability at a Corner Assume tat f is continuous at a. f is not differentiable at a, and its grap as a corner at te point ( a, f ( a) ) 1), 2), or 3) below olds: 1) f + ( a) and f ( a) bot exist, but f + ( a) f ( a), 2) f + ( a) exists and f ( a): ± (left-and tangent line is vertical), or 3) f + ( a): ± (rigt-and tangent line is vertical) and f ( a) exists A point on a grap is a corner tere are two distinct tangent lines tere, one from eac side. Example 12 (Losing Differentiability at a Corner; Derivatives of Piecewise-Defined Functions) Let f ( x)= x = x, x 0. x, x < 0 on We can use our basic rules to differentiate te different rules for f x teir different subdomains (indicated by x 0 and x < 0 ), altoug we must investigate values of x were te rule for f ( x) canges (ere, at x = 0 ). f( x)= 1, x > 0 1, x < 0 Altoug f is continuous at x = 0, we can see from te grap of f below tat f is not differentiable tere, and tere is a corner at te origin. Tis is because f + ( 0)= 1, wile f ( 0)= 1. Grap of f Grap of f
(Section 3.2: Derivative Functions and Differentiability) 3.2.17 Here is te proof tat f + ( 0) 0 + Here is te proof tat f ( 0) 0 f + ( 0)= 1: f ( 0 + ) f ( 0) 0 + f ( 0)= 1: f ( 0 + ) f ( 0) 0 f ( ) 0 f ( ) 0 0 + = 1 0 = 1 Example 13 (Losing Differentiability at a Corner wit a Vertical Tangent Line) Let f ( x)= x2, x < 0 x, x 0. Ten, f 2x, x < 0 ( x)= 1 2 x, x > 0. Altoug f is continuous at x = 0, we can see from te grap of f below tat f is not differentiable tere, and tere is a corner at te origin. Tis is because f + ( 0):, wile f ( 0)= 0. Grap of f Grap of f Here is te proof tat f + ( 0) 0 + 0 + 1 f + ( 0): : f ( 0 + ) f ( 0) 0 + Limit Form: 1 0 + = f ( ) 0 0 +
(Section 3.2: Derivative Functions and Differentiability) 3.2.18 Here is te proof tat f ( 0) 0 0 = 0 f ( 0)= 0 : f ( 0 + ) f ( 0) 0 f ( ) 0 2 0 An informal sort cut. If f is a simple function tat is continuous at a, a one-sided derivative (even informally as ± ) can be guessed at by taking te corresponding one-sided limit of te derivative as x a. Tat is, we tentatively guess tat f + ( a) f( x), and f x a + ( a) f( x), x a were and are possible informal results. Instead of taking te limit of slopes of secant lines, we are taking te limit of slopes of tangent lines. Here: Guess: lim f( x) x 0 + 2 x wic suggests tat f + ( 0):, and x 0 + 1 Limit Form: 1 0 + =, Guess: lim f x 0 wic suggests tat ( x) x 0 f ( 0)= 0. ( 2x )= 0, However, tis trick does not work for more complicated functions! (See Footnote 7.)
(Section 3.2: Derivative Functions and Differentiability) 3.2.19 Losing Differentiability at a Cusp Assume tat f is continuous at a. f is not differentiable at a, and its grap as a cusp at te point ( a, f ( a) ) 1) or 2) below olds: 1) f + ( a): and f ( a):, or 2) f + ( a): and f ( a): Te rigt-and and left-and tangent lines at a cusp are bot vertical, but te secant lines coming in from one side fall, wile te secant lines coming in from te oter side rise. Example 14 (Losing Differentiability at a Cusp) Let f ( x)= x 2/3, or 3 x 2. For graping purposes, observe tat f is an even, nonnegative function wit domain. Altoug f is continuous at x = 0, we can see from te grap of f below tat f is not differentiable tere, and tere is a cusp at te origin. Tis is because f + ( 0):, wile f ( 0):. Grap of f Wit secant lines (in orange) and tangent line (in red) Here is te proof tat f + ( 0) 0 + 0 + 1 f + ( 0): : f ( 0 + ) f ( 0) 0 + 1/3 0 1 + 3 f ( ) 0 2/3 0 + Limit Form: 1 0 + =
Here is te proof tat (Section 3.2: Derivative Functions and Differentiability) 3.2.20 f ( 0) 0 0 1 f ( 0): : f ( 0 + ) f 0 1/3 0 1 3 Using te informal sort cut. Guess: lim x 0 + wic suggests tat Guess: lim x 0 wic suggests tat f( x) 0 f ( ) 0 Limit Form: 1 0 f( x)= 2 3 x1/3 = 2 x 0 + 3 3 x f + ( 0):, and f( x) 2 x 0 3 3 x f ( 0):. 2 3 3 x. 2/3 0 = Limit Form: 2 0 + Limit Form: 2 0 =, =, Example 15 (Losing Differentiability at a Point wit a Vertical Tangent Line) Let f ( x)= x 1/3 3, or x. For graping purposes, observe tat f is an odd function wit domain. Altoug f is continuous at x = 0, we can see from te grap of f below tat f is not differentiable tere, and tere is a vertical tangent line at te origin, even toug tere is neiter a corner nor a cusp tere. Tis is because f + ( 0): and f ( 0):. Grap of f Wit secant lines (in orange) and tangent line (in red)
(Section 3.2: Derivative Functions and Differentiability) 3.2.21 FOOTNOTES 1. Functions tat are nowere differentiable. Functions tat are nowere continuous are also nowere differentiable. See Footnote 3 in Section 2.8. 2. Proof of te Power Rule of Differentiation. An elegant proof of te Power Rule for all real constants n will be found in te Footnotes for Section 7.5; it will employ Logaritmic Differentiation. Some sources use te Binomial Teorem to first prove it for positive integers n. n = 0 corresponds to te special case of differentiating 1; 0 0 is often defined to be 1 for tis purpose. For positive rational values of n, let n = p, were p and q are positive q integers. Let y = x n. Ten, y = x p/q, and y q = x p. Te Implicit Differentiation tecnique from Section 3.7 can be used to prove te Power Rule in tis case. For negative rational values of n, let m = n. Ten, x n = x m = 1, and te Reciprocal (or Quotient) Rule of m x Differentiation in Section 3.3 can be applied. In tese last two cases, te domain of te derivative migt not be. 3. Proof of te Constant Multiple Rule of Differentiation. Assume tat f is a function tat is differentiable were we care, and c is a real constant. Let g( x)= c f( x). g( x+ ) g( x) c f( x+ ) c f( x) c g( x) 0 0 0 ( We will exploit te Constant Multiple Rule of Limits. ) f( x+ ) f( x) = c lim = c f 0 ( x) f( x+ ) f x 4. f n notation. We use f ( n) to denote an nt-order derivative, as opposed to f n. Tis is because n often represents an exponent in te notation f n, except wen n = 1 (in wic case we ave a function inverse). For example, f 2 is often taken to mean ff ; tat is, f 2 ( x)= f ( x) f ( x). For example, we will accept tat sin2 x = ( sin x) ( sin x), wic is te standard interpretation. Note: f x f x f x. is typically not equivalent to On te oter and (and tis compounds te confusion), some sources use n to indicate te number of applications of f in compositions of f wit itself; te result is called an iterated function. For example, tey would let f 2 = f f, and tey would use te rule: f 2 ( x)= f ( f ( x) ). Tis is typically different from te rule f 2 ( x)= f ( x) f ( x). However, our use of te notation f 1 for f inverse is more consistent wit tis second interpretation, since f 1 f is an identity function, wic could be construed as f 0 in tis context. Note: f( x) is typically not equivalent to f ( f( x) ).
(Section 3.2: Derivative Functions and Differentiability) 3.2.22 5. Notation for rigt-and and left-and derivatives. Tere appears to be no standard notation for rigt-and and left-and derivatives. In fact, f + ( a) sometimes denotes te upper rigt Dini derivative at a, wic is a bit different from wat we are calling a rigt-and derivative. If te upper rigt Dini derivative at a and te lower rigt Dini derivative at a exist and are equal, ten teir common value is te rigt-and (or rigt-and Dini) derivative at a. Likewise, if te upper left and lower left Dini derivatives at a are equal, ten teir common value is te left-and derivative at a. See T.P. Lukasenko, Dini derivative, SpringerLink, Encyclopedia of Matematics, Web, 4 July 2011, <ttp://eom.springer.de/>. 6. Proving tat differentiability implies (and requires) continuity. f is differentiable at a f( x) f( a) lim exists; see Version 1 of te Limit Definition of f xa ( a)in Section 3.1. x a A rigorous approac: Assume tat f is differentiable at a. Tis implies tat f( a) exists; in fact, it implies tat f is defined on an open interval containing a. lim f( x) f( a)+ f( a) f( x) f( a) ( x a) f x xa xa xa x a + f a f( x) f( a) = lim xa x a lim x a xa + f ( a )= f ( a) 0 + f( a)= f( a). Terefore, lim f x xa = f( a), wic defines continuity of f at a. An intuitive approac: As x a, x a 0. Te only way te limit can exist as a real number is if it as te Limit Form 0 DNE. (Te Limit Form cannot yield a real number c 0 0 as a limit, basically because c 0 = 0.) Tis requires tat f( x) f( a) 0 as x a. Tis, in turn, requires tat f( x) f( a) as x a, or lim f x xa = f( a), wic defines continuity of f at a. 7. A function wit a derivative tat is defined but discontinuous at 0; failure of te limit of te derivative sort cut for one-sided derivatives. See Gelbaum and Olmsted, Counterexamples in Analysis (Dover), p.36. Let f( x)= x2 sin 1 x, x 0. 0, x = 0 Ten, f( x)= 2xsin 1 x cos 1 x, x 0. 0, x = 0 Te metods of Sections 3.2-3.6 can be used to work out te top rule. Wy is f ( 0)= 0? f ( 0 + ) f ( 0) 2 sin 1 0 f ( 0) sin 1 0 0 0 = 0 by te Squeeze (Sandwic) Teorem from Section 2.6. f is discontinuous at 0, because lim f x x0
(Section 3.2: Derivative Functions and Differentiability) 3.2.23. does not exist (DNE). Also, because lim x0 + f ( x) does not exist (DNE) and lim x0 f ( x) does not exist (DNE), te limit of te derivative sort cut for guessing one-sided derivatives (even informally as ± ), as described in Example 13, fails for tis example. (See also Section 3.6, Footnote 4.) Grap of f Grap of f (Axes are scaled differently.)
(Section 3.3: Tecniques of Differentiation) 3.3.1 SECTION 3.3: TECHNIQUES OF DIFFERENTIATION LEARNING OBJECTIVES Learn ow to differentiate using sort cuts, including: te Linearity Properties, te Product Rule, te Quotient Rule, and (peraps) te Reciprocal Rule. PART A: BASIC RULES OF DIFFERENTIATION In Section 3.2, we discussed Rules 1 troug 4 below. Basic Sort Cuts for Differentiation Assumptions: c, m, b, and n are real constants. f and g are functions tat are differentiable were we care. If ( x)= ten ( x)= Comments 1. c 0 Te derivative of a constant is 0. 2. mx + b m Te derivative of a linear function is te slope. 3. x n nx n1 Power Rule 4. c f ( x) c f( x) Constant Multiple Rule (Linearity) 5. f ( x)+ g( x) f( x)+ g( x) Sum Rule (Linearity) 6. f ( x) g( x) f( x) g( x) Difference Rule (Linearity) Linearity. Because of Rules 4, 5, and 6, te differentiation operator D x is called a linear operator. (Te operations of taking limits (C.2) and integrating (C.5) are also linear.) Te Sum Rule, for instance, may be tougt of as te derivative of a sum equals te sum of te derivatives, if tey exist. Linearity allows us to take derivatives term-by-term and ten to pop out constant factors. Proofs. Te Limit Definition of te Derivative can be used to prove tese sort cuts. Te Linearity Properties of Limits are crucial to proving te Linearity Properties of Derivatives. (See Footnote 1.)
(Section 3.3: Tecniques of Differentiation) 3.3.2 Armed wit tese sort cuts, we may now differentiate all polynomial functions. Example 1 (Differentiating a Polynomial Using Sort Cuts) Solution Let f ( x)= 4x 3 + 6x 5. Find f( x). f( x)= D x 4x 3 + 6x 5 = D x ( 4x 3 )+ D x ( 6x) D x 5 = 4 D x ( x 3 )+ D x ( 6x) D x 5 ( Sum and Difference Rules) ( Constant Multiple Rule) TIP 1: Students get used to applying te Linearity Properties, skip all of tis work, and give te answer only. = 4( 3x 2 )+ 6 0 = 12x 2 + 6 Callenge to te Reader: Observe tat te 5 term as no impact on te derivative. Wy does tis make sense grapically? Hint: How would te graps of y = 4x 3 + 6x and y = 4x 3 + 6x 5 be different? Consider te slopes of corresponding tangent lines to tose graps.
(Section 3.3: Tecniques of Differentiation) 3.3.3 Example 2 (Equation of a Tangent Line; Revisiting Example 1) Solution Find an equation of te tangent line to te grap of y = 4x 3 + 6x 5 at te point ( 1, 3). Let f ( x)= 4x 3 + 6x 5, as in Example 1. lies on te grap by ()= 3. (Remember tat function values correspond to Just to be safe, we can verify tat te point 1, 3 verifying tat f 1 y-coordinates ere.) Find m, te slope of te tangent line at te point were x = 1. Tis is given by f (), 1 te value of te derivative function at x = 1. m = f () 1 From Example 1, remember tat f( x)= 12x 2 + 6. = 12x 2 + 6 x=1 = 12() 1 2 + 6 = 6 We can find a Point-Slope Form for te equation of te desired tangent line. Te line contains te point: ( x 1, y 1 )= ( 1, 3). It as slope: m = 6. y y 1 = m x x 1 y ( 3)= 6 x 1 If we wis, we can rewrite te equation in Slope-Intercept Form. y + 3 = 6x + 6 y = 6x + 3
(Section 3.3: Tecniques of Differentiation) 3.3.4 We can also obtain te Slope-Intercept Form directly. y = mx + b ( 3)= ( 6) ()+ 1 b b = 3 y = 6x + 3 Observe ow te red tangent line below is consistent wit te equation above.
(Section 3.3: Tecniques of Differentiation) 3.3.5 Example 3 (Finding Horizontal Tangent Lines; Revisiting Example 1) Solution Find te x-coordinates of all points on te grap of y = 4x 3 + 6x 5 were te tangent line is orizontal. Let f ( x)= 4x 3 + 6x 5, as in Example 1. We must find were te slope of te tangent line to te grap is 0. We must solve te equation: f( x)= 0 12x 2 + 6 = 0 See Example 1. 12x 2 = 6 x 2 = 1 2 x =± 1 2 x =± 2 2 Te desired x-coordinates are 2 2 and 2 2. Te corresponding points on te grap are: 2 2, f 2 2, wic is 2 2, 2 2 5, and 2 2, f 2 2 2, wic is, 2 2 5 2.
Te red tangent lines below are truncated. (Section 3.3: Tecniques of Differentiation) 3.3.6
PART B: PRODUCT RULE OF DIFFERENTIATION (Section 3.3: Tecniques of Differentiation) 3.3.7 WARNING 1: Te derivative of a product is typically not te product of te derivatives. Product Rule of Differentiation Assumptions: f and g are functions tat are differentiable were we care. If ( x)= f ( x)g( x), ten ( x)= f( x)g( x)+ f x g ( x). Footnote 2 uses te Limit Definition of te Derivative to prove tis. Many sources switc terms and write: ( x)= f ( x) g( x)+ f( x)g( x), but our form is easier to extend to tree or more factors. Example 4 (Differentiating a Product) Solution Find D x x 4 + 1 ( x 2 + 4x 5). TIP 2: Clearly break te product up into factors, as as already been done ere. Te number of factors (ere, two) will equal te number of terms in te derivative wen we use te Product Rule to expand it out. TIP 3: Pointer metod. Imagine a pointer being moved from factor to factor as we write te derivative term-by-term. Te pointer indicates wic factor we differentiate, and ten we copy te oter factors to form te corresponding term in te derivative. ( x 4 + 1) ( x 2 + 4x 5) ( x 2 + 4x 5) D x x 4 + 1 copy + D x copy D x + = D x x 4 + 1 x2 + 4x 5 ( x 4 + 1) D x ( x 2 + 4x 5) + = 4x 3 x 2 + 4x 5 ( x 4 + 1)[ 2x + 4]
(Section 3.3: Tecniques of Differentiation) 3.3.8 Te Product Rule is especially convenient ere if we do not ave to simplify our result. Here, we will simplify. = 6x 5 + 20x 4 20x 3 + 2x + 4 Callenge to te Reader: Find te derivative by first multiplying out te product and ten differentiating term-by-term. Te Product Rule can be extended to tree or more factors. Te Exercises include a related proof. Example 5 (Differentiating a Product of Tree Factors) Solution t t 2 + 2 Find d 3 t + 4 t dt. Te result does not ave to be simplified, and negative exponents are acceptable ere. (Your instructor may object!) d dt 3 t ( t + 4) t 2 + 2 ( t 1/3 t) ( t + 4) t 2 + 2 ( D t ) copy copy + copy ( D t ) copy + copy copy D t ( t) = D t ( t + 4) ( t 2 + 2) 3 t t ( t + 4) D t ( t 2 + 2) ( t + 4)( t 2 + 2) D t ( t 1/3 t) + 3 ( t t) + 3 t t = [] 1 t 2 + 2 ( t + 4) 2t [ ] t ( t + 4)( t 2 + 2) + 3 ( t) + 1 3 t 2/3 1 TIP 4: Apply te Constant Multiple Rule, not te Product Rule, to someting like D x 2x 3. Wile te Product Rule would work, it would be inefficient ere.
(Section 3.3: Tecniques of Differentiation) 3.3.9 PART C: QUOTIENT RULE (and RECIPROCAL RULE) OF DIFFERENTIATION WARNING 2: Te derivative of a quotient is typically not te quotient of te derivatives. Quotient Rule of Differentiation Assumptions: f and g are functions tat are differentiable were we care. g is nonzero were we care., If ( x)= f x g x ten ( x)= g x f ( x) f ( x) g( x). g x 2 Footnote 3 proves tis using te Limit Definition of te Derivative. Footnote 4 more elegantly proves tis using te Product Rule. TIP 5: Memorizing. Te Quotient Rule can be memorized as: D Hi Lo = Lo D( Hi) Hi D( Lo) ( Lo) 2, te square of wat's below Observe tat te numerator and te denominator on te rigt-and side ryme. At tis point, we can differentiate all rational functions.
Reciprocal Rule of Differentiation If ( x)= 1 g x ten, ( x)= g x g x. 2 (Section 3.3: Tecniques of Differentiation) 3.3.10 Tis is a special case of te Quotient Rule were f ( x)= 1. Tink: DLo Lo 2 TIP 6: Wile te Reciprocal Rule is useful, it is not all tat necessary to memorize if te Quotient Rule as been memorized. Example 6 (Differentiating a Quotient) Solution 7x 3 Find D x 3x 2 + 1. 7x 3 Lo D( Hi) Hi D( Lo) D x 3x 2 + 1 = ( Lo) 2, te square of wat's below ( = 3x2 + 1) D x ( 7x 3) ( 7x 3 ) D x ( 3x 2 + 1) ( 3x 2 + 1) 2 ( = 3x2 + 1)[ 7] ( 7x 3)[ 6x] 3x 2 + 1 2 = 21x2 + 18x + 7, or ( 3x 2 + 1) 2 7 21x 2 + 18x 3x 2 + 1 2 18x 7, or 21x 2 ( 3x 2 + 1) 2
(Section 3.3: Tecniques of Differentiation) 3.3.11 TIP 7: Rewriting. Instead of running wit te first tecnique tat comes to mind, examine te problem, tink, and see if rewriting or simplifying first can elp. Example 7 (Rewriting Before Differentiating) Solution Let sw = 6w2 w 3w. Find s ( w). Rewriting sw by splitting te fraction yields a simpler solution tan applying te Quotient Rule directly would ave. sw = 6w2 3w w 3w = 2w 1 3 w1/2 s ( w)= 2 + 1 6 w 3/2 = 2 + 1 6w, or 12w3/2 + 1, or 12w2 + w 3/2 6w 3/2 6w 2
(Section 3.3: Tecniques of Differentiation) 3.3.12 FOOTNOTES 1. Proof of te Sum Rule of Differentiation. Trougout te Footnotes, we assume tat f and g are functions tat are differentiable were we care. Let p = f + g. (We will use for run in te Limit Definition of te Derivative.) p( x+ ) p( x) p( x) f( x+ )+ g( x+ ) f( x)+ g( x) 0 0 f( x+ )+ g( x+ ) f( x) g( x) f( x+ ) f( x) + g( x+ ) g( x) 0 0 f( x+ ) f( x) + g( x+ ) g( x) 0 f( x+ ) f( x) g( x+ ) g x + lim 0 0 ( Observe tat we ave exploited te Sum Rule (linearity) of Limits. ) = f( x)+ g( x) Te Difference Rule can be similarly proven, or, if we accept te Constant Multiple Rule, we can use: f g = f + ( g). Sec. 2.2, Footnote 1 extends to derivatives of linear combinations. 2. Proof of te Product Rule of Differentiation. Let p = fg. p( x+ ) p( x) p( x) 0 f( x+ )g( x+ ) f( x)g( x) 0 f( x+ )g( x+ ) f( x+ )g( x)+ f( x+ )g( x) f( x)g( x) 0 f( x+ )g( x+ ) f( x+ )g( x) + f( x+ )g( x) f( x)g( x) 0 f( x+ )g( x+ ) f( x+ )g( x) f( x+ )g( x) f( x)g( x) + lim 0 0 f( x+ ) g( x+ ) g( x) + lim f( x+ ) f( x) g( x) 0 0 f x+ 0 g( x+ ) g( x) + lim f( x+ ) f( x) g( x) 0 = lim f x+ 0 lim g( x+ ) g( x) 0 + lim f( x+ ) f( x) 0 lim g( x) 0 = f( x) g ( x) + f( x) g( x), or f( x)g( x)+ f( x) g( x) Note: We ave: lim f x+ 0 = f x continuity. We ave someting similar for g in Footnote 3. by continuity, because differentiability implies
(Section 3.3: Tecniques of Differentiation) 3.3.13 3. Proof of te Quotient Rule of Differentiation, I. Let p = f / g, were g( x) 0. p( x+ ) p( x) p( x) 0 f( x+ ) g( x+ ) f( x) g( x) 0 f( x+ ) 0 g( x+ ) f( x) 1 g( x) f( x+ )g( x) f( x)g( x+ ) 0 1 g( x+ )g( x) f( x+ )g( x) f( x)g( x+ ) 1 0 g( x+ )g( x) f( x+ )g( x) f( x)g( x)+ f( x)g( x) f( x)g( x+ ) 1 0 g( x+ )g( x) f( x+ )g( x) f( x)g( x) + f( x)g( x) f ( x )g ( x+ ) 1 0 g( x+ )g( x) f( x+ ) f( x) g( x) + f( x) g ( x ) g ( x+ ) 1 0 g( x+ )g( x) f( x+ ) f( x) g( x) f ( x ) g ( x+ ) g( x) 1 0 g( x+ )g( x) f( x+ ) f( x) g( x) f( x) lim g( x+ ) g( x) 1 0 0 lim 0 g( x+ )g( x) f( x+ ) f( x) = g( x) lim 0 f( x) g( x+ ) g( x) 1 lim 0 g( x)g( x) ( See Footnote 2, Note. ) = g( x) f ( x) f( x) g ( x) 1 ( ) g( x) = g( x) f( x) f( x) g( x) 2 g( x) 2
(Section 3.3: Tecniques of Differentiation) 3.3.14. 4. Proof of te Quotient Rule of Differentiation, II, using te Product Rule. Let ( x)= f( x) g( x), were g( x) 0. Ten, g( x)( x)= f( x). Differentiate bot sides wit respect to x. Apply te Product Rule to te left-and side. We obtain: g( x)( x)+ g( x) ( x)= f( x). Solving for ( x), we obtain: ( x)= f ( x ) g ( x) x g x ( x)= f( x) g x g( x) f x g x. Remember tat ( x)= f x g x f x f( x) g( x) g x g x = g( x) g( x) = g( x) f( x) f( x) g( x) g x 2. Ten, Tis approac is attributed to Marie Agnessi (1748); see Te AMATYC Review, Fall 2002 (Vol. 24, No. 1), p.2, Letter to te Editor by Joe Browne. See also Quotient Rule Quibbles by Eugene Boman in te Fall 2001 edition (vol.23, No.1) of Te AMATYC Review, pp.55-58. Te article suggests tat te Reciprocal Rule for 1 D x can be proven directly by using te Limit Definition of te Derivative, and ten g( x) te Product Rule can be used in conjunction wit te Reciprocal Rule to differentiate 1 f( x) ; te Spivak and Apostol calculus texts take tis approac. Te article g( x) presents anoter proof, as well.
(Section 3.4: Derivatives of Trigonometric Functions) 3.4.1 SECTION 3.4: DERIVATIVES OF TRIGONOMETRIC FUNCTIONS LEARNING OBJECTIVES Use te Limit Definition of te Derivative to find te derivatives of te basic sine and cosine functions. Ten, apply differentiation rules to obtain te derivatives of te oter four basic trigonometric functions. Memorize te derivatives of te six basic trigonometric functions and be able to apply tem in conjunction wit oter differentiation rules. PART A: CONJECTURING THE DERIVATIVE OF THE BASIC SINE FUNCTION Let f ( x)= sin x. Te sine function is periodic wit period 2. One cycle of its grap is in bold below. Selected [truncated] tangent lines and teir slopes (m) are indicated in red. (Te leftmost tangent line and slope will be discussed in Part C.) Remember tat slopes of tangent lines correspond to derivative values (tat is, values of f ). Te grap of f must ten contain te five indicated points below, since teir y-coordinates correspond to values of f. Do you know of a basic periodic function wose grap contains tese points?