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Table of Trig Functionssinxcosxtanx=sinxcosxcotx=cosxsin?(x)secx=1cos?(x)cscx=1sin?(x)Obtain co-functions in right column from left column by replacing sinx→cos?(x) and cosx→sin?(x).Two Famous Trianglessketch 45°-45°-90° triangle, label angles and sidesPythagorean theorem122+122=1 or 12+12=1 sinπ4=cosπ4=12, tanπ4=1 sketch 30°-60°-90° triangle, label angles and sidesPythagorean theorem322+122=1 or 34+12=1 sinπ6=12, cosπ6=32, tanπ6=13 sinπ3=32, cosπ3=12, tanπ3=3 Derivatives of Trig Functionsddxtanx=ddxsinxcosx =sinx'(cosx)-(sinx)cosx'cos2x =cosx?cosx-sinx?-sinxcos2x =cos2x+sin2xcos2x =1cos2x =sec2(x) where secx=1/cos?(x).ddxsecx=ddx1cosx =1'?cosx-1?cosx'cos2x =-(-sinx)cos2x =1cosx?sinxcosx =secx?tanx similarly, obtain the following table (to memorize)ddxsinx=cosxddxcosx=-sinxddxtanx=sec2xddxcotx=-csc2xddxsecx=secx?tanxddxcscx=-cscx?cot(x)to obtain the right hand column from the left hand column, replace functions by co-functions and multiply by -1.?? ddx xsec(x)?? ddθ tan?(θ)θ2+1 Example. Find an equation of the line tangent to the curvey=xtan(x) at the point π4,π4.Solution.Point-slope form of tangent liney-y0=m(x-x0) Need slopedydx=tanx+xsec2(x) dydxπ4=tanπ4+π4sec2π4 where secπ2=1cosπ4=2dydxπ4=1+π4?2=1+π2 equation for tangent liney-π4=1+π2x-π4 ■§2.5 The Chain RuleCompositionExample. Consider Fx=x2+1Let fu=u, gx=x2+1F is the composition of f and gFunction machines:x→g→x2+1=u→f→ u=x2+1■Notation for compositionFx=fgx f is the outer function, g is the inner functionDerivatives of CompositionsLet y=Fx=fgxor y=f(u), u=g(x).The Chain RuleIf g is differentiable at x and f is differentiable at u, thenF'x=f'ug'(x) The derivative of a composition of two functions is the product of their derivativesWrite another wayF'x=f'gxg'(x)Leibniz notationLet y=f(u) and u=g(x) and y=Fx=fgxthenF'x=f'u g'(x) dydx = dydu dudxEasy to remember: looks as though a term “du” cancels on the right hand side!Example. Let Fx=x2+1. Find F'(x).Write as a compositionu=gx=x2+1 y=fu=u where u=u12apply the chain rule dydx=dydu dudx =12u-12?2x =x2+1-12 x =xx2+1 ■Example. Let y=cos3x. Find dydx.Write as a compositiony=cos?(u) where u=3xapply the chain ruledydx= dydu dudx = -sinu?3 = -3sin(3x) ■?? Class PracticeLet y=x2+110. Find dydx.Let y=sinx. Find dydx.Let y=tancosxPlausibility Argument for the Chain RuleRecall the limit definition of derivativedydx=f'x=limh→0fx+h-f(x)h Change notationLet Δx=hΔy=fx+h-f(x) thendydx=limΔx→0ΔyΔx Statement of Chain RuleLet y=Fx=fgx or y=f(u) with u=g(x)If g is differentiable at x, and f is differentiable at u, then F'x=f'ug'(x) or dydx =dydu dudxPlausibility ArgumentLet Δu be the change in u corresponding to Δx.Δu=gx+Δx-g(x) The corresponding change in y isΔy=fu+Δu-f(u) Thendydx = limΔx→0ΔyΔx =limΔx→0ΔyΔuΔuΔxas long as Δu≠0=limΔx→0ΔyΔu?limΔx→0ΔuΔxproduct rule for limitsbut Δu→0 as Δx→0, sodydx=limΔu→0ΔyΔu?limΔx→0ΔuΔx =dydu?dudx The only problem is that we may have Δu=0. ■Generalized Power RuleCombine the power rule and the chain ruleConsiderFx=gxn write as a compositionu=g(x) y=fu=un apply the chain ruledydx=dydududx =n un-1dudx this is frequently writtenddxgxn=n gxn-1g'(x)Example. ConsiderFx=x3+4x10 this has the form given above, with gx=x3+4x n=10 thenF'x=10 x3+10x9ddx(x3+4x) =10x3+10x9 (3x2+4) ■Example. Fx=1sin?(x)=sinx-12 F'x=-12sinx-32cos?(x) ■WARNING It is a common mistake to forget the factor g'(x)!Compositions of Three FunctionsConsiderFt=fght write as a compositiony=fu, u=g(x), x=h(t) chain ruledydt=dydu dudx dxdtit looks as though ‘du’ and ‘dx’ factors cancelExampleFt=sin2(4t) write as a compositiony=u2, u=sinx, x=4t chain ruledydt=dydu dudx dxdt =2u?cosx?4 =8 u cos?(x) =8sinxcos?(x) =8sin4tcos?(4t) ■ExampleFind the line tangent to the graph of y=sin24t at t=π16 .Solution. Point slope form of a tangent liney-y0=mt-t0 t0=π/16 , y0=sin2π4=122=12 slope? m=dydxt=π16 =8sin4tcos4tt=π16 =8sinπ4cosπ4 =8 1212 =4 gives the tangent liney-12=4t-π16 ■?? Class practice differentiating!Find dydx1. y=x4-3x2+632. y=2xx2+13. y=x+1x274. y=t1-t25. y=sin?(sec(x)) 6. y=costan1+t2§2.6 Implicit Differentiationcircle of radius 1…0…x-, …0…y-, circlex2+y2=1 defines y implicitly as a function of xsolve for y: y2=1-x2 two possibilitiesy=1-x212=1-x2[1a]y=-1-x212=-1-x2[1b]derivativesdydx=121-x2-12-2x=-x1-x2[2a][2a]dydx=-121-x2-12-2x= x1-x2[2b]Using Implicit Differentiation is easier in many cases!x2+y2=1 [3][3]regard y as a function of xy=f(x) or y2=fx2 recall the generalized power ruleddx fx2=2 fxf'(x) write this asddxy2=2 y y' must remember y'=dydx ! differentiate equation [3] implicitly with respect to x2x+2ydydx=0 solve for dy/dxydydx=-x dydx=-xy from [1], two possibilitiesdydx=-x1-x2 dydx=x1-x2these match [2]!Example. Find dydx if x-1+y-1=1.[1]Solution. Regard y as a function of xy=f(x) Use the generalized power ruleddx fx-1=-fx-2f'(x) ddxy-1=-y-2dydx differentiate [1] implicitly with respect to x-x-2-y-2dydx=0 solve for dydx-y-2dydx= 1x2 dydx= - y2x2 ■Example. Find the equation of the line tangent to the curve2cosxsiny=1[1]at x0,y0=π4,π4.Solution. Recall sinπ4=cosπ4=12.think of y as a function of xddxsiny=cosydydx differentiate [1] implicitly-2sinxsiny+ 2cosxcosydydx=0 solve for dydx2cosxcosydydx=2sinxsin(y) dydx=tanxtan(y) slope of curve at x=y=π/4dydxx=y=π4=tanπ4tanπ4=1?1=1 point-slope form of tangent liney-y0=m(x-x0) y-π4=x-π4 y=x ■?? Class practice with implicit differentiation1. Let x3+x2y+4 y2=c, where c is a constant. Find dydx.Solution dydx=-(3x2+2xy)x2+8y2. Let sinx+cosy=sinxcos(y). Find dydx. Solution dydx=cos?(x)sinx-1cosy-1sin?(y)Find an equation of the line tangent to the following curve at the given point.2x2+y22=25(x2-y2) at the point (3,1)Solution. Differentiate implicitly with respect to x4x2+y22x+2yy'=25(2x-2yy')Solve for y'4x2+y2 2yy'+50yy'=-8x2+y2x+50xy'(8yx2+y2)+50y=-8x3-8xy2+50xy'=-8x3-8xy2+50x8yx2+y2+50y=2x(-4x2-4y2+25)2y(4x2+4y2+25)evaluate at x=3 and y=1y'=6(-36-4+25)2(36+4+25)= 6(-15)2(65)=-4565=-913point slope form of tangent line y-y0=mx-x0y-1=-913x-3in slope intercept formy=-913x+2713+1=-913x+4013show Maple image of curve and tangent line ■§2.7 Related RatesExample. A descending balloonist lets helium escape at a rate of 10 cubic feet / minute. Assume the balloon is spherical. How fast is the radius decreasing when the radius is 10 feet?sketch balloon, basket, escaping gasGiven V= volume of balloon dVdt= -10 feet3minute r= radius of balloon =10 feetFind drdt when r=10Equation relating these quantitiesV=43πr3 [1]differentiate with respect to time using the chain rulegeneralized power rule: ddt gtn=n gtn-1consider radius r a function of time tddtr3=3r2drdt Differentiate [1] with respect to time to getdVdt=43π 3 r2drdt=4πr2drdtSolve for dr/dtdrdt= 14πr2dVdt drdtr=10=14 π 1001feet2-10feet3min = -140 πfeetmin≈-0.008 feetmin≈-0.1 inchesmin ■Problem Solving StrategyRead problem carefullyDraw picture if possibleAssign notation to given information and unknown rateWrite down an equation relating the given information and the function whose rate is unknownDifferentiate with respect to timeSolve for the unknown rateExample. The length of a rectangle is increasing at a rate of 7 cm/s and its width is increasing at a rate of 7 cm/s. When the length is 6cm and the width is 4 cm, how fast is the area of the rectangle increasing?2. picture3. Given informationL(t)=6cm, W(t)=4cmL't=7 cm/s, W't=7 cm/s Unknown rate is A'(t), where A=LW.4. At=LtW(t)5. A't=L'tWt+ LtW'(t)6. =7cms?4 cm+6 cm?7cms =28cm2s+ 42cm2s=70cm2s ■Example. One end of a 13 foot ladder is on the ground and the other end rests on a vertical wall. If the bottom end is drawn from the wall at 3 feet/second, how fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?…x…x-, …y…y-, ladder betw. x & y, l= 13 ftgiven informationl=13 ft, x= 5ft, dxdt=3 ft./sec.unknown ratedydt= ?Pythagorean theoremx2+y2=l2differentiate with respect to time2 xdxdt+ 2 ydydt=0solve for ydydt= -xydxdtto evaluate dy/dt we need yy2=l2-x2=169-25=144y=144=12thendydtx=5 ft= -512?3 feetsec= - 54feetsec■Example. Pat walks 5 feet/second towards a street light whose lamp is 20 feet above the ground. If Pat is 6 feet tall, find how rapidly Pat’s shadow changes in length.ground, lamp, Pat, x, s, y, hgiven information y=20 feet, h= 6 feet, dxdt= -5 feetsecunknown ratedsdt= ? by similar trianglesyx+s= hsy s=h (x+s)y-hs=hxs= hy-h xdifferentiate wrt timedsdt= hy-hdxdt= 620-6-5 feetsec= - 157feetsec ■Example. The beacon of a lighthouse 1 mile from the shore makes 5 rotations per minute. Assuming that the shoreline is straight, calculate the speed at which the spotlight sweeps along the shoreline as it lights up sand 2 miles from the lighthouse.0…x…x-, 0…y…y-, lighthouse at (0,y), beam to (x,0), length L, angle θgiven informationy=1 mile L= 2 milesdθdt=5 revolutionsminute =5 revolutionsminute 2π radiansrevolution =10π radiansminute unknown rate dxdt=?from the definition of tangenttanθ=xy differentiate wrt timesec2θdθdt= 1ydxdtusing that y is a constantdxdt= ysec2θdθdt we need sec2θcosθ=yL secθ=Ly=2thendxdt=1 mile?4?10π radiansminute =40πmilesminute = 40πmilesminute 60 minuteshour =4060π mileshour ≈7539 mileshour ■ STOP§2.8 Linear Approximations and DifferentialsLinear or Tangent Line Approximation0…a…x-, 0…y-, y=f(x), Pa,faPoint-slope form of the tangent liney=y0=m(x-x0) Point x0,y0=a,faSlope m=f'aTheny-fa=f'ax-a ory=Lx=fa+f'a(x-a)L(x) approximates f(x) near x=aExample. Find the linear approximation to fx=1+x near x=0.fx=1+x12 f'x=121+x-12 f'0=12 Lx=f0+f'0x-0 =1+12xthis straight line approximates 1+x near x=0.-1…0…1…x-, 0…1…y-, f(x), L(x)■Example (continued). Use linear approximation to estimate 1.1.1.1= 1+0.1=f(0.1)L0.1=1+120.1=1.05Exact value 1.1=10.488…■Example (continued). For what values of x is the linear approximation1+x≈1+12xaccurate to within 0.05? This meansLx-fx<0.05or Lx-0.05<fx<Lx+0.05show Maple output From the plot, the linear approximation is accurate to within 0.05 for-0.54<x<0.74■Realistic Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat 0.01 inches thick.Exact calculation: volume of sphereVr=43 πr3volume between two concentric spheres of radii r and aΔV=Vr-Va=43 π r3-43 π a3a=20feet=240 inches r=240.01 inches ΔV≈7238.53 cubic inches ≈31 gallons1 gallon = 231 cubic inchesApproximate calculation:Linear approximation, usual notationfx≈fa+ f'(a)(x-a)Use V(r) instead of f(x)Vr≈Va+ V'a(x-a)ΔV=Vr-Va≈V'a(r-a)V'r=ddr43 π r3 =43 π 3 r2 =4 π r2andV'a=4πa2 thenΔV=4 πa2r-a = surface area of sphere × thickness of paint =4 π 240 inches2 (0.01 inches) ≈7237.23 cubic inchesthe error is only 0.3 cubic inches! ■STOP Differentials0…x-, 0…y-, f, P, L, x, x+Δxtangent line approximationLx+Δx=fx+f'xΔxwritedx=Δxdy=Lx+Δx-f(x)=f'xdxadd dx=Δx, Δy, dy=f'xdxΔy=fx+Δx-f(x)is approximated by dy=f'x dxNote that riserun=dydx=f'(x). This is clever notation.Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat of paint 0.01 inches thick.Volume of sphereVr=43 π r3approximate amount of paintdV=V'rdr=4 π r2 drif dr=0.01 inches, thendV= 4 π 240 inches2 (0.01 inches)the same result as above. This is a shorthand for linear approximation. ■Newton’s Method – An Application of Linear Approximation0…r…x-, 0…y-, fA root of f is a number r such that fr=0.How to find roots geometrically?guess x1addlet L1 be the line tangent to f at x1addfollow L1 to the x axis, get an intersection at x2addrepeat this processif the initial guess was good, the sequence x1, x2,… rapidly converges to rHow to find roots algebraically?Make an initial guess x1.Find the line tangent to f at x=x1.L1x=fx1+f'(x1)(x-x1)by the linear or tangent line approximation!L1 intersects the x axis at x2. Find x2.L1x2=0=fx1+f'(x1)(x2-x1)-fx1=f'(x1)(x2-x1)-f(x1)f'(x1)=x2-x1x2=x1-f(x1)f'(x1)Find the line tangent to f at x=x2.L2x=fx2+f'(x2)(x-x2)L2 intersects the x axis at x3. Find x3.L2x3=0=fx2+f'(x2)(x3-x2)-fx2=f'(x2)(x3-x2)-f(x2)f'(x2)=x3-x2x3=x2-f(x2)f'(x2)repeat this procedure to obtain a general formula for n=1, 2, …xn+1=xn-f(xn)f'(xn)If the initial guess x1 was good, the iterates x2, x3,… will quickly approach r.Example. fx=x2-2-2…-1…0…1…2…x-, -2…0…2…y-, f, r=2f'x=2xfrom the boxed formula abovexn+1=xn-xn2-22 xnguess x1=2x2=2-4-24=2-12=32x3=32-94-23=32-143=32-112=18-112=1712=1.4167…Exact value of positive rootr=2=1.4142…Transparency – How Newton’s Method can go wrong! ................
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