College of Arts and Sciences



§3.1 Exponential FunctionsForm y=ax, a>0 constant, x variableWarning! Do not confuse these with power functions of form y=xa.Laws of Exponents:1.ax+y=axay2. ax-y=axay3. axy=axy4.abx=axbxExamples. From 2. with x=0a-y=1ayFrom 4.ab=ab12=a12b12=abab=ab12definition of square root=ab-112by first example=a12b-112by 4.=a12b-12by 3.=a12b12-1by 3.=a12b12by first example=abdefinition of square rootGraph y=2x and y=2-xTable x/ -2, -1 0, 1, 2 2x/ … (?)x/… -2…0…2…x-, 0…4…y-, y=2x, y=12xFor a>0 and a≠1fx=ax has domain R={real numbers} and range (0,∞).fx=1x=1 has domain R and range {1}For any base a>0, the graph of y=ax increases for a>1 and decreases for 0<a<1 and passes through the point (0,1).Limits at Infinityfor a>1limx→-∞ax=0 and limx→∞ax=∞for 0<a<1limx→-∞ax=∞ and limx→∞ax=0for a>0 and a≠1y=0 is a horizontal asymptote of y=axExample.limx→-∞3+4x=limx→-∞3+limx→-∞4x property of exponential functions=3+0=3■ Study of DerivativesBy definitionf'x=limh→0fx+h-f(x)happly to the exponential functionfx=axsimplify the difference quotientfx+h-f(x)h=ax+h-axhfactor ax out of numerator=ax(ah-1)h=axah-1hThusf'x=limh→0axah-1hconst. multiple rule ! f'(x)=axlimh→0ah-1hLet x=0 and use a0=1f'0=limh→0ah-1hThus f'x=f'0axthe derivative of ax is proportional to ax !0…-1…0…1…x-, 0…1…2…3…y- , y=2x, y=3x, y=x+1ddx2xx=0<1ddx3xx=0>1There is a number e≈2.7 such thatddxexx=0=1Let fx=ex. Then f'0=1.Since ex is an exponential functionf'x=f'0ex=exddxex=exExampleLet gx=x2+2exg'x =ddxx2+ddx 2ex=2x+2ddxex=2x + 2 ex ■?? Find the line tangent to g(x) at x=0.One-to-one functionsA function fis called one-to-one (1-1) if it never takes the same value twice.fx1≠fx2 whenever x1≠x2Example. fx=x2, -∞<x<∞ is not 1-1…-2…0…2…x- , 0…1…4…y-, y=x2,y=1f-1=f1although -1≠1Horizontal Line TestA function is one-to-one if and only if no horizontal line intersects its graph more than once.Example. gx=x2, domain x≥0 , is a 1-1 function.0…2…x-, 0…1…4…y-, y=x2, y=1§3.2 Inverse Functions and LogarithmsInverse FunctionsLet f be a one-to-one function with domain A and range B. Then its inverse function f-1 has domain B and range A and is defined byx=f-1y?y=f(x) for all y∈BFunction Machine Picturex→f→yA=domain of f B=range of f x∈A y∈B y→f-1→xB=domain of f-1 A=range of f-1 y∈B x∈A Cancellation Equationsf-1fx=x for each x∈Aff-1y=y for each y∈BExample. gx=x2, domain x≥0.The inverse function is g-1x=xg-1g3=32=3 gg-19=92=9 ■ Warning! f(x) is a number but f is a function. f-1(x) and fx-1 do not mean the same thing. Example. Let fx=x2Then f-1x=x=x12 but fx-1=1fx=x-2 ■ Graphs of f and f-1If b=f(a) then (a,b) is on the graph of fIf a=f-1(b) then (b,a) is on the graph of f-1Example. gx=x2, domain x≥0.g-1x=xpoints 12,14, 1,1, (2,4) are on the graph of g 14,12, 1,1, 4,2 g-1 0…1…4…x-, 0…1…4…y-, y=x2, y=xThe graphs of g and g-1 are symmetric about y=x.In general, the graphs of functions f and f-1 are symmetric about the line y=x. ■How to find the inverse of a one-to-one function y=f(x) algebraically?1. write y=f(x)2.solve for x in terms of y (if possible) to get x=f-1(y)3.exchange roles of x and yy=f-1(x)Example gx=x2, domain x≥01.y=x22.Take the positive square rootx=y3.exchange roles of x and yy=x ■Example hx=2x-73x+41.y=2x-73x+42.3x+4y=2x-73xy-2x=-4y-7x3y-2=-4y-7x=-4y-73y-23.y=-4x-73x-2=h-1(x) ■Derivatives and Tangent Functions0…a…x-, 0…y-, y=f(x), ∠θθ is the angle that the tangent line to f at x=a makes with the x-axis.f'a=tan?(θ)Derivatives of Inverse Function by Geometry0…x-, 0…y-, (b,a), y=fx, (a,b), y=f-1(x), y=xadd ∠θ, complementary ?, ∠?, complementary θf'b=tan?(θ) f-1'a=tan?(?) θ+?=π/2 tan?=tanπ2-θ =sinπ2-θcosπ2-θ =cos?(θ)sin?(θ) =1tan?(θ) =1f'(b) Thusf-1'a=1f'(b)Derivatives of Inverse Functions by CalculusCancellation Equationff-1x=x has the formfgx=x where g=f-1differentiate wrt x using the chain rulef'gxg'x= 1 solve for g'(x)g'x=1f'gx orf-1'x=1f'f-1x suppose x=a and b=f-1(a)f-1'a=1f'(b)suppose y=f-1xf-1'x=1f'(y)In Leibniz NotationLet y=f-1x, then x=fydydx=f-1'x dxdy=f'(y) the relationship between derivatives looks likedydx=1dxdyExample. y=gx=x2 with domain x≥0x=g-1y=y=y12 dydx=1dxdy=112y-12=2y12=2x ■LogarithmsRecall exponential functionsy=ax , a>0, these are 1-1 if a≠1The logarithm base a is the inverse of this functionx=logay ? y=axreverse the role of x and yy=logax ? x=ayExample. 2=log10100 ? 100=102 ■Laws of Logarithms1.logaxy=logax+logay2.logaxy=logax-logay3.logaxr=rlogaxExample. Find the exact value of log1050+log102.By law #1log1050+log102=log10(50?2) =log10(100) =2 ■Natural LogarithmsIt is simplest to use base e≈2.7 in calculus. The logarithm base e is the natural logarithm.Notationlnx=logexRecall the exponential function base ey=exThe logarithm base e is its inversex=lny ? y=exreverse roles of x and yy=lnx ? x=eyin particular0=ln1 ? 1=e01=lne ? e=e1Cancellation Equationselnx=xlnex=xExample. Solve for x: e3x-4=2lne3x-4=ln2 3x-4=ln2 3x=4+ln2 x=43+13ln2 ■Compare the graphs of y=ex and y=lnx.xex-1e-1≈0.37011e≈2.7xlnxe-1-110e1-1…3…x-, -1…3…y-, y=ex, y=lnx, y=xChange of base formularecall y=logax ? x=aytake natural logarithm of equation on rightlnx=y ln?(a) y=lna/ln?(x) or logax=ln(x)ln(a)Example. Evaluate log3(5) correct to 4 significant figures.log35=ln(5)ln(3)=1.465■ §3.3 Derivatives of Logarithmic and Exponential Functionsddxex=ex?? Example. Let Fx=exx . Find F'(x).F'x= =exx12-ex12x-12x■?? Example Let gx=e3x. Find g'(x)Write as a composition.Apply the chain rule.dydx= =3 e3x■Combine the natural exponential function with the chain rule.Let y=efx. Find dydx.Write as a composition.y=eu, u=f(x)Apply the chain rule.dydx = dydududx=eududx we haveddxefx=efxf'(x)Examples.ddxe3x= ddue-u= ddtesint= ?? ddxex2-3x=■Example. Suppose y=cosex. Find dydx.Write as a composition.Apply the chain rule.dydx= = -exsin(ex)■Derivative of exponentials with base a.y=ex, dydx=ex y=ax, dydx= ? Strategy: transform ax into esomething. Then use the chain rule to differentiate esomething.lny=lnax=xlna y=elny=exlna thendydx=ddxexlna =exlnaddx(xlna) =exlnaln?(a) = axln(a) ddxax=axlnaExample. ddx10x= ■Example. Let Fx=2x2-3x. Find F'(x).Write F as a compositionApply the chain rule.dydx= =2x2-3x(ln2) (2x-3)■?? Class PracticeDerivatives of natural logarithmic functionsThe natural logarithm is the inverse of the natural exponential function.y=lnx ?x=eyDifferentiate the right hand wrt x, using the chain rule.1=eydydx dydx= 1ey ddxln(x)=1xExample. Let y=lnx2. Find dydx.Generalized power rule:ddx gxn=n gxn-1 g'(x) ddxlnx2=2 (lnx)1x ■Combine the natural logarithm and the chain ruleLet y=lnfxWrite as a composition.y=lnu, u=f(x) Apply the chain rule.dydx=dydududx ddxlnu=1ududx ddxlnfx=f'(x)f(x)Example. Let y=ln(x2-1). Find dydx.fx=x2-1 f'x=2x ddxlnx2-1=2xx2-1 ■?? Class PracticeVery Important Example.ddxlnx= ? two cases:(1) x>0ddxlnx=ddxlnx=1x(2)x<0x=-x write y=ln|x| as a compositiony=lnu, u=-x chain ruledydx=dydududx=1ududx=1-x (-1)=1xwe have for bothx>0 and x<0:ddxlnx=1/x■Derivative of Logarithm with base a.Recall the change of base formulalogax=ln(x)ln(a) ddxlogax=1ln(a)ddxln(x) Example. Let y=log5(2x+3). Find dydx.Write as a compositiony=log5(u), u=2x+3 Apply the chain ruledydx=dydududx = 1ln5u 2 = 2ln5(2x+3) ■?? Class PracticeSimplification before differentiationRecall laws of logarithmslnab=lna+lnblnab=lna-lnblnab=blnaExample. Let y=ln(x 1-x2sinx ) . Find dydx .First simplify y using the laws of logarithmsy=lnx+ln1-x212+lnsin(x) =lnx+12ln1-x2+lnsin(x) Compute dydx using the chain rule: ddxlnfx=f'(x)f(x)dydx=ddxlnx+12ddxln1-x2+ddxlnsin(x) =1x+12-2x1-x2+ cos(x)sin(x)■?? Class PracticeLogarithmic DifferentiationRecall ddxlnf(x)= f'(x)f(x)Problem is to differentiate y=f(x), where f is a complicated expression involving products, quotients and/or powers.Take logarithms of both sides and simplify f(x) using the laws of logarithms.Differentiate to get y'y=…Solve for y'=f'(x).Example. Let y=x3+14sin2(x)3x. Find dydx.Step 1.lny=lnx3+14+lnsin2x-ln3x =4lnx3+1+2lnsinx-13ln?(x) Step 2.y'y=4 3 x2x3+1+2 cosxsinx- 13x Step 3.y'= y 12x2x3+1+2cotx- 13x dydx= x3+14sin2(x)3x12x2x3+1+2cotx- 13x ■?? Example. Let y=x2+1x2-1 . Find dydx.Differentiating expressions with variable base and exponentUse logarithmic differentiationExample. Let y=xsinx. Find dydx.§3.6 Hyperbolic FunctionsCombinations of exponential functions that are analogous to trig functionssinhx=12ex-e-xcoshx=12ex+e-xtanhx=sinh(x)cosh(x)cothx=cosh?(x)sinh(x)sechx=1cosh?(x)cschx=1sinh(x)Graph cosh(x)…0…x-, …0…y-, 12ex, 12e-x, cosh(x), evenGraph sinh(x)…0…x-, …0…y-, 12ex, -12e-x, sinh(x), oddGraph tanh(x)?? tanh0=?? tanh-x= symmetry?? limx→∞tanhx=By odd symmetry limx→-∞tanhx=…0…x-, …0…y-, y=1, y=-1, tanh(x)Most important hyperbolic identitycosh2x-sinh2x=1analogous tocos2x+sin2(x)=1ProofNotice thatcosh2x=14ex2+2exe-x+e-x2=14e2x+2+e-2x sinh2x=14e2x-2+e-2x thencosh2x-sinh2x=14e2x+2+e-2x -14e2x-2+e-2x =1■Derivatives of Hyperbolic FunctionsHyperbolicTrigonometricddxsinhx=cosh?(x)ddxsinx=cos?(x)ddxcoshx=sinh?(x)ddxcosx=-sin?(x)ddxtanhx=sech2(x)ddxtanx=sec2(x)Example. Show that ddxcoshx=sinh?(x)ddxcoshx=ddx12ex+e-x =12ddxex+ddxe-x =12ex-e-x =sinh?(x) Example. Let y=lnsinhx Find dydx.=coth?(x)■?? Example. Let y=tanh?(e2t). Find dydx.=sech2e2t 2e2t■§3.5 Inverse Trigonometric FunctionsLet f be a one-to-one function with domain A and range B. Then its inverse f-1 has domain B and range A and is defined byx=f-1y ? y=f(x)Inverse siney=sin?(x) is not 1-1. It fails the horizontal line test.-π…0…π…x-, -1…0…1…y-, sin(x)The restricted sine function is 1-1 domain [-π2,π2 ]range -1,1 emphasizeIts inverse is y=sin-1(x) or y=arcsin?(x)domain -1,1range [-π2,π2 ] addFor the restricted sine functiony=sin-1x? x=sin?(y)Example. Evaluate sin-1-12Recall 30°-60°-90° triangle sketchsinπ6=12 since sin?(x) is an odd functionsin-π6=-12 it followssin-1-12=-π6 ■Example. Evaluate sec?sin-115If θ=sin-115 then sinθ=15 triangleTo find the derivative of y=sin-1(x) consider y=sin-1x ? x=sin?(y)differentiate the equation on the right wrt x.1=cosydydx dydx=1cosy Triangle for x=sin?(y) trianglecosy=1-x2 dydx= 11-x2 we have obtainedddxsin-1x=11-x2Example. Let y=ln?(sin-1(x)). Find dydx.Recall ddxlnfx=f'(x)fxdydx=ddxln?(sin-1(x)) =11-x2sin-1(x) =11-x2sin-1(x)■Example. Let y=sin-1(x3). Find dydx.write as a compositiony=sin-1u, u=x3 apply the chain ruledydx = dydududx = 11-u2dudx = 3x21-x6■ Inverse Tangent-π2…3π2…x-,…-1…1…y-,vert. asymptotes, y=tan?(x)y=tan?(x) is not 1-1. It fails the horizontal line test.The restricted tangent function is 1-1. emphasizedomain -π2,π2 range (-∞,∞)Its inverse is y=tan-1(x) or y=arctan(x) adddomain (-∞,∞) range -π2,π2y=-π/2 and y=π/2 are horizontal asymptotes of y=tan-1(x)limx→∞tan-1x=π2limx→-∞tan-1x=-π2For the restricted tangenty=tan-1x ? x=tan(y)Example. Evaluate tan-1(-1)45°-45°-90° triangletanπ4=1 tan(x) is oddtan-π4=-1 tan-1-1=-π4 ■Example. Evaluate cos?(tan-1(x)) Triangle for y=tan-1(x) trianglecos?(tan-1x) =cosy=11+x2 ■To find the derivative of y=tan-1(x) considery=tan-1x ? x=tan(y)differentiate the equation on the right wrt x1=sec2ydydx dydx=cos2y=11+x2 ddxtan-1x=11+x2Example. Let y=etan-1(x). Find dydx.Recall ddxefx=efxf'(x)dydx=etan-1(x)11+x2■?? Example. Let y=ex tan-1(x). Find dydx.■Example. Let y=tan-12t. Find dydt .Write as a compositiony=tan-1(u), u=2/tApply the chain ruledydt= dydududt =11+u2dudt =11+4t2-2t-2 =t2t2+4-2t2 =-24+t2 ■?? Example. Let y=tan-1(x3). Find dydx.■§3.7 Indeterminate forms and L’Hospital’s Rulerecall the quotient law for limitsSuppose limx→af(x) and limx→ag(x) exist and limx→agx≠0. Thenlimx→afxgx=limx→af(x)limx→ag(x) .Sometimes we cannot apply this ruleExample. Find limx→0sin?(x)sinh(x)both sinx→0 and sinhx→0 as x→a.An indeterminate form of type 0/0 is a limit of form limx→af(x)g(x) (1)where both fx→0 and gx→0 as x→a. L’Hospital’s Rule. Suppose (1) is of form 0/0. Thenlimx→af(x)g(x)=limx→af'(x)g'(x) if the limit on the right exists or is +∞ or -∞.Notes:A. L’Hospital’s rule usually gives the wrong answer if eq. (1) is not indeterminate!B. L’Hospital’s rule is valid for one-sided limits or limits at infinity.Example. Evaluate L=limx→1x2+x-2x2-1 .This is a type 0/0 indeterminate form!Method 1. Divide out a common factorx2+x-2x2-1= x-1(x+2)x-1(x+1)= x+2x+1if x≠1.L=limx→1x+2x+1=32 Method 2. Use L’Hospital’s rule.L=limx→12x+12x=32.Warning! We cannot apply L’Hospital’s rule again!L=limx→12x+12x≠limx→122=1 this is because the middle expression is not an indeterminate form. ■Example. Evaluate L=limx→0sin?(x)sinh?(x) .This is a type 0/0 indeterminate form.L=limx→0sinx'sinhx' =limx→0cosxcoshx = cos0cosh0 =1 ■Example. Evaluate L=limx→0cosx-1x2This is a type 0/0 indeterminate form.L=limx→0-sinx2x Still type 0/0.L=limx→0-cosx2= -12 ■Example (One-sided Limit) Evaluate L=limx→0+ln(x+1)xThis is a type 0/0 indeterminate form.L=limx→0+1/(x+1)12x-12 simplify =limx→0+2x12x+1 =0 ■An indeterminate form of type ∞/∞ is a limit of form limx→af(x)g(x) (2)where both fx→∞ (or -∞) and gx→∞ (or -∞) as x→a.L’Hospital’s Rule. Suppose (2) is of form ∞/∞. Then limx→af(x)g(x)=limx→af'(x)g'(x)if the limit on the right exists (or is +∞ or -∞).Notes:A. L’Hospital’s rule usually gives the wrong answer if eq. (1) is not indeterminate!B. L’Hospital’s rule is valid for one-sided limits or limits at infinity.Example. (Limit at infinity)Evaluate L=limx→∞ln?(1+ex)5xThis is a type ∞/∞ indeterminate form.L=limx→∞ex1+ex5 =15limx→∞ex1+exWe still have an ∞/∞ indeterminate form.L=15limx→∞exex=15. ■ STOPAn indeterminate form of type 0?∞ is a limit of formlimx→afxg(x) where fx→0 and gx→∞ (or -∞) as x→a.Method. Put product in quotient formfg=f1/g and then apply L’Hospital’s rule.Example. Find L=limx→-∞x exL is of form 0?∞. Write product as a quotient.L=limx→-∞xe-x We kept x in the numerator so that differentiation simplifies it.L is now of form ∞/∞. Apply L’Hospital’s rule.L=limx→∞1e-x(-1) =limx→-∞-ex =0. ■Indeterminate powers have formlimx→afxgx Three cases type1. fx→0, gx→0 as x→a. 002. fx→∞, gx→0 as x→a. ∞03. fx→1, gx→±∞ as x→a. 1∞ Method. Take the logarithm f(x)g(x), use the previous techniques, and then exponentiate your result. Example. FindL=limx→∞1+1xx L is of type 1∞.Let y=1+1xx.lny=ln1+1xx = xln1+1xConsiderlimx→∞y=limx→∞xln1+1x type 0?∞ =limx→∞ln1+1xx-1 type 0/0 =limx→∞1+1x-1-x-2-x-2 simplifyWe have limx→∞y=limx→∞1+1x-1 =1ThenL=limx→∞y =limx→∞explny where ez=exp?(z) =*exp(limx→∞lny) =exp(1)=e*Since the exponential function is continuous, the limit symbol passes into the argument of the exponential (see theorem 7 in section 1.5)This is an important result. ■ ................
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