§2.1 Derivatives and Rates of Change



§2.1 Derivatives and Rates of ChangeTangent Linesaxes, curve CConsider a smooth curve C.A line tangent to C at a point P both intersects C at P and has the same slope as C at P. add line tThe Tangent Line ProblemGiven point P on curve C, how do you find the tangent line?Example. Consider y=fx=x20…2…x-, 0…4…y-, f(x)What is the equation of the line tangent to the curve at x=1? add P, linePoint-slope form for a straight line passing through P(1,1)y-1=m(x-1) What is the slope m?What is the slope of the secant passing through P(1,1) and Q12,4? add Q1, PQ1mPQ1=ΔyΔx =f2-f(1)2-1=4-12-1=3 What is the slope of the secant passing through P(1,1) and Q21.5,f1.5? add Q2, PQ2mPQ2=f1.5-f(1)1.5-1 =2.25-11.5-1 =1.250.5 =2.5What is the slope of the secant line passing through P(1,1) and Q(x,x2)?mPQ=f(x)-f(1)x-1 this ratio is called a difference quotient =x2-1x-1 =(x-1)(x+1)x-1 = x+1, x≠1undefined, x=1As long as x≠1, mPQ=x+1 xmPQ231.52.51.12.11.012.01The slope of the tangent line is the limit of the difference quotient as x→1.m=limx→1x+1=2 The equation of the tangent line isy-1=2(x-1) ■Example. Find an equation of the line tangent to the curve gx=x+1 at P(3,2).Point slope form of the tangent liney-2=m(x-3) where m=limx→3gx-g(3)x-3 simplify the difference quotientgx-g(3)x-3=x+1-2x-3 multiply by 1 to rationalize the numerator =x+1-2x-3 x+1+2x+1+2 =x+1-4x-3(x+1+2)cancel factors of x-3 =1x+1+2true if x≠3Thusm=limx→31x+1+2=12+2=14 Equation of tangent liney-2=14(x-3) ■The Velocity ProblemDrive to Spokane airport (~85 miles)Start at noonDrive slowly through ColfaxHave lunch at HarvesterArrive 2pmAverage velocity=distance traveledtime elapsed=85 miles2 hours=42.5 mph The speedometer 5 miles north of Colfax reads 65 mph. This is the “instantaneous velocity.”Mathematical definition of instantaneous velocity?Galileo drops a ball off the leaning Tower of Pisasketch ground, tower, coordinate s with origin at topball falls distance s at time t after release.s(t)=5t2 s meters, t seconds0…2…t-, 0…5…20…s-, curveWhat is the average velocity between t=1 and t=2?Average velocity =distance traveledtime elapsed =s2-s(1)2-1 =5?22-5?121 =15meterssecond What is the average velocity between t=1 and a variable t?Average velocity =st-s(1)t-1 =5?t2-5t-1 =5?(t2-1)t-1 =5t-1(t+1)t-1 =5(t+1) as long as t≠1.Tabletaverage velocity (meters per sec.)2151.15?2.1=10.51.0015?2.001=10.005Define instantaneous velocity at t=1 as the limit of average velocities over shorter and shorter time intervals around t=1.Denote instantaneous velocity v(t).v1=limt→1st-s(1)t-1 =limt→15(t+1) =10meterssecond DerivativesDefine the derivative of a function f at a number a, denoted f'(a)f'a=limx→afx-f(a)x-a [1]From the example above v1=s'1=limt→1st-s(1)t-1 .Alternatively, introduceh=x-a x=a+h and insert in equation [1] to getf'a=limh→0fa+h-f(a)h [2]§2.2 The Derivative as a FunctionReplace the symbol a in [2] by x. Regard x as a variable.f'x=limh→0fx+h-f(x)h Regard f' as a new function.Example. Let fx=x2. Find f'(x)Simplify the difference quotientfx+h-f(x)h=x+h2-x2h =x2+2xh+h2-x2h =2xh+h2h =2x+h this step assumes h≠0Thenf'x=limh→02x+h=2x Graph and compare f(x) and f'(x)-2…0…2…x-, -4…0…4…y-, f(x)-2…0…2…x-, -4…0…4…y-, f'(x)add tangent segments at x=-1, 0, 1 to graph of fadd dots at x=-1, 0 , 1 to graph of f' ■Example. Let gx=x+1. Find g'(x).Simplify the difference quotientgx+h-g(x)h=x+h+1-x+1h rationalize numerator by multiplying by 1 =x+h+1-x+1h x+h+1+x+1x+h+1+x+1multiply terms in numerator =x+h+1-(x+1)hx+h+1+x+1divide through by h (assumes h≠0) =1x+h+1+x+1Theng'x=limh→01x+h+1+x+1 =12x+1Show transparency comparing g and g'?? Transparency: match f and f'Notations for Derivativeoriginal function: y=f(x)derivative function: f'x=dydxprime notation emphasizes idea of derivative as a new functionf'(x) the prime means differentiate with respect to function argumentevaluate at no. af'(a) Leibniz notation emphasizes idea of derivative as the limit of a ratio…x…x+h…,…, f, h=Δx, fx+h-fx=Δylimh→0fx+h-f(x)h=limh→0ΔyΔx=dydx evaluate at no. a dydxa Operator notation for derivativeSometimes we writef'x=ddxf orf'x=Df view ddx and D as operators: machines that convert the functions they operate on into other functionsDifferentiabilityf differentiable at a means f'(a) existsf differentiable on an open interval (a,b) means f is differentiable at every point in (a,b)Example (a function not differentiable at a point)fx=|x| Is f differentiable at x=0?if sof'0=limh→0f0+h-f(0)f =limh→00+h-|0|h =limh→0|h|hdoes this limit exist? find the limit from the rightlimh→0+|h|h=limh→0hh=1 find the limit from the leftlimh→0-|h|h=limh→0--hh=-1 the right and left hand limit do not agree.conclude f'(0) does not exist Geometrical Ideaaxes, graph of |x|, no tangent line here (at origin) To be differentiable at point, the graph must have a unique tangent line at that point. ■Three ways that a function can fail to be differentiable(a)at any discontinuity …a…,…, function with dcty at a limh→0fa+h-f(a)h DNE(b)at any corner or kink …a…,…, function with a kink at a limh→0fa+h-f(a)h DNE(c) at a vertical tangent …a…,…, function w/ a vertical tangent at a limx→afx=∞, f'(a) DNERelationship between differentiability and continuityWe have shown: if f is not continuous then f is not differentiableLet A be the statement f is continuous at a no. xLet B be the statement f is differentiable at xWe have shownIf (not A) then (not B)This is logically equivalent toIf B then A If f is differentiable at x then it is continuous at xHigher DerivativesConsider y=f(x) First derivative of fdydx=f'(x) Regarded as a function, f may itself be differentiable.Second derivative of fd2ydx2=f''(x) If f''(x) is differentiable, form the third derivatived3ydx3=f'''(x) If f'''(x) is differentiable, form the fourth derivatived4ydx4=f4(x) Notation for the nTh derivative, with n≥4:dnydxn=fn(x) Application of higher derivativesLet s(t) be the position of an object at time t.s't=dsdt=v(t) is the velocity of the objects''t=v't=a(t) is the acceleration of the objectFirst and second derivatives are the most important in applicationsExample. Mechanicsmomentum = mass ? velocityforce=mass ? acceleration ■2.3 Basic Differentiation RulesWe first consider those rules that will enable us to differentiate polynomials.Derivative of a Constant Function…x-,…0…c…y-, line y=cslope of tangent line?ddxc=0 Derivative of fx=x0…x-, 0…y-, y=xslope of tangent line?ddxx=1 Derivative of gx=x2We have seen thatg'x=2x Derivative of Fx=x3F'x=limh→0Fx+h-F(x)h simplify the difference quotientFx+h-F(x)h=x+h3-x3h =x3+3x2h+3xh2+h3-x3h =3x2h+3xh2+h3h =3x2+3xh+h3assumes h≠0thusF'x=limh→03x2+3xh+h2 =3x2The Power Rule.Let n be a positive integerddxxn=nxn-1 Example. ddxx7=7x6. ■Proof. Preliminary fact: x-a? (xn-1+xn-2a+xn-3a2+…+xan-2+an-1) =xn+xn-1a+xn-2a2+…+xan-1 -xn-1a-xn-2a2-…-xan-1-an________________________________________ xn - anin other wordsxn-anx-a=xn-1+xn-2a+…xan-2+an-1 notice there are n terms on the right hand sideLet fx=xnf'a=limx→axn-anx-a =limx→a(xn-1+xn-2a+…xan-2+an-1 ) =an-1+an-2a+…aan-2+an-1 =nan-1Regard a as a variable. Replace a by x.f'x=nxn-1 ■The Power Rule (general version)Let n be any real number.ddxxn=nxn-1 Examples. ddxx-9=-9x-10 ddtt-2=-2t-3ddxx=12x-12 recall x=x12?? ddt1t= ?? dduu2= ■ The Constant Multiple RuleLet c be a constant and f a differentiable functionddxcfx=cddxf(x) a constant passes through the limit symbolExamples. ddx5x2=5ddxx2=5?2x=10x ?? ddt 7 10t =?? ddw 5 wπ=■The Sum RuleIf f and g are both differentiableddxfx+gx=ddxfx+ddxg(x) In words: “the derivative of a sum is the sum of the derivatives”prime notationfx+gx'=f'x+g'(x) shorthandf+g'=f'+g' the sum rule applies to the sum of any number of functionsf+g+h'=f'+g'+h' Example ddxx3+5x2+π=3x2+10x+0■The Difference RuleIf f and g are both differentiableddxfx-gx=ddxfx-ddxg(x) In words: “the derivative of a difference is the difference of the derivatives”prime notationfx-gx'=f'x-g'(x) shorthandf-g'=f'-g' Example.?? ddxx3-5x2= ■We can now differentiate any polynomialExample. Letgx=5x8-2x5+6 theng'x=ddx5x8-ddx2x5+ddx6 =5ddxx8-2ddxx5+ 0 =58x7-25x4+0 =40x7-10x4 ■We can differentiate other functions too. Example. Letu=3t2+2t3 Find dudtu=t23+2t32 dudt=23t-13+232t12 =23t-13+3t12 ■Example. Find an equation of the line tangent to the curvey=3x2+4x at the point (1,7)point slope form for straight line y-y0=m(x-x0) where x0,y0=(1,7)what is m?dydx=32x+4-x-2 =6x-4x-2m=dydxx=1=6-4=2 answery-7=2(x-1) ■?? Example. A ball is thrown straight up from the ground at 20 meters/second. Its height is given byy=20 t-5 t2 (a) Find the velocity at time t.(b) Find the velocity at t=1sec.(c) When is the ball at rest?(d) What is the average velocity between t=1 and t=2? ■Economics – Marginal costCx= cost to produce x widgetsaverage rate of change of cost =Cx2-C(x1)x2-x1=ΔCΔxmarginal cost =limΔx→0ΔCΔx=C'(x)Example. Jeans manufacture.Let Cx= cost of producing x pairs of jeans. =2000+3x+0.01x2where $2000 capital costs (sewing machines) 3x+0.01x2 cost of labor, materials, rentCost of producing 100 pairs of jeansC100=2000+300+0.011002 2400What is the cost of producing one additional pair of jeans?Cx+1=2000+3x+1+0.01x+12 Cx=2000+3x+0.01x2 ___________________________________________Cx+1-Cx=3+0.01x+12-x2 =3+0.01(2x+1)Cost of producing the 101ST pairC101-C100=3+0.01(201) =$5.01Compare with the marginal cost at 100th pairC'x=3+0.02x C'100=$5 C'(x) is often a very good approximation to the cost of producing one additional widget. ■Derivatives of Sine and CosineRecall the limitslimx→0sin?(x)x=1 limx→0cosx-1x=0 Recall the addition formula for cosinecosa+b=cosacosb-sinasin?(b) Now use the limit definition of derivativeddxcosx=limh→0cosx+h-cos?(x)h addition formula for cosine =limh→0cosxcosh-sinxsinh-cos?(x)h difference law for limits =limh→0cosxcosh-1h-limh→0sin?(x)sin?(h)h constant multiple law of limits =cosx?limh→0cosh-1h-sinx?limh→0sin?(h)h recalling the limits above = cosx?0-sinx?1 =-sin?(x) ■The derivative of sin?(x) may be found using a similar argument (see our text). In summary:ddxsinx=cosxddxcosx=-sin?(x)?? Differentiate the following 1. ft=sint+π cos?(t) 2. gy=Ay10+ Bcos(y)§2.4 The Product and Quotient Rules Product RuleIf f and g are both differentiableddxfxgx=fxddxgx+gxddxf(x) or alternately (as I personally prefer)ddxfxgx=ddxfxgx+fxddxgx prime notationfxgx'=f'xgx+fxg'(x) shorthandfg'=f'g+fg' WARNING: fg'≠f'g' The derivative of a product is not the product of derivativesThis is a common mistake!Example. By the power ruleddxx3=3x2 Now let f=x2 and g=x then f'=2x and g'=1ddxx3=(fg)' =f'g+fg' =2x?x+x2?1 =3x2 ■Proof of the Product RuleSuppose f and g are both differentiable functions.Let fxgx=F(x)then ddxfxgx=ddxF(x) =limh→0Fx+h-F(x)h =limh→0fx+hgx+h-fxg(x)h subtract and add the same term in the numerator=limh→0fx+hgx+h-fx+hgx+fx+hgx-fxg(x)h algebra=limh→0[ fx+h?gx+h-g(x)h+gxfx+h-f(x)h ] sum and product laws for limits=limh→0fx+h?limh→0gx+h-g(x)h +limh→0gx?limh→0fx+h-f(x)h continuity of f and definition of derivative =fx?g'x+gx?f'(x) ■Extension to a Product of Three FunctionsIf f, g and h are all differentiablef g h'=f'g h+f g'h+f g h' Example. Let fx=gx=hx=x then f' = g' = h' =1ddxx3= f'g h + f g'h + f g h' =1?x?x+x?1?x+x?x?1 = 3x2 ■Example. Differentiate Fx=sin?(x)x.Law of exponents: 1x=x-1ThenFx=x-1sin?(x) Product ruleF'x=x-1'sinx+x-1sinx' =-x-2sinx+x-1cos(x) =-sin?(x)x2+cos?(x)x ■Quotient RuleIf f and g are differentiable at a point x where g'x≠0 thenddxf(x)g(x)= ddxfxgx-fxddxg(x)gx2 shorthandfg'=f'g-fg'g2 terms in numerator in same order as my product rule (but take difference)Proof of quotient rule.Let fg=hThen f=g hBy product rulef'=g'h+g h' Solve for h'g h'=f'-g'h h'=f'-g'hg =f'g-g'fg2 ■Example. Differentiate Gx=x+1x-1.G'x=x+1'x-1-x+1x-1'(x-1)2 where x+1' is shorthand for ddx(x+1) =x-1-(x+1)(x-1)2 =-2x-12 ■Example. Find the equations of the tangent lines to the curveGx=(x+1)(x-1) that are parallel to the linex+2y=2 .Solution. Parallel means same slope. Slope of line?2y-2-x y=1-12x slope is m=-12where does G have slope -12 ?G'=-2x-12=-12 solve for x 1x-12=14 x-12=4 x=-1 or x=3form of equation for tangent liney-y0=m(x-x0) Consider x0=-1. y0=G-1=-1+1-1-1=0y=-12x+1Consider x0=3. y0=G3=3+13-1=2y-2=-12(x-3)■?? Class practice product and quotient rules ................
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