微積分英文課件：Chapter2 Derivatives

1、Chapter2 Derivatives2.1 The Derivative as a function The Tangent Problem Let f be a function and let P(a, f(a) be a point on the graph of f. To find the slope m of the tangent line l at P(a, f(a) on the graph of f, we first choose another nearby point Q(x, f(x) on the graph (see Figure 1) and then c

2、ompute the slope mPQ of the secant line PQ.)()(axafxfmpQ0P(a,f(a)Q(x,f(x)f(x)0P(a,f(a)Q(x,f(x)f(x)l.)()(limaxafxfmaxLet Q get closer to P and QP. The slope m of the tangent line l is the limit of the slopes of the secant lines,i.eLet axhThen haxSo the slope of the second line PQ is .)()(hafhafmpQThe

3、 slope m of the tangent line l is.)()(lim0hafhafmhThe velocity problemSuppose an object moves along a straight line according to an equation of motion )(tfs )(tfis called the position function of the objectos)(af)(hafAverage velocity displacementtimehafhaf)()(Rates of changeIf x change from 1xto ,th

4、en then change , 2x)(xfy The corresponding change in y is)()(12xfxfyThe instantaneous velocity at t=a hafhafavh)()(lim)(0then the change in (increment of )isx12xxxxThe average rate of change of y with respect to xis1212)()(xxxfxfxyThe instantaneous rate of change of y with respect to x at isxxfxxfxx

5、xfxfxyxxxxx)()(lim)()(limlim110121212121xDefinition of DerivativeDefinition Let y=f(x) be a function defined on an open interval containing a number a.The derivative of f(x) at number a, denoted by f(a) , is if this limit exists. xafxafhafhafafxh)()(lim)()(lim)(00If we write x=a+h, then h=x-a and h

6、approaches 0 if and only if x approaches a.thenaxafxfhafhafafaxh)()(lim)()(lim)(02)2()(lim)2()2()2(lim)2()(202xfxffhfhffxxfxhhafhafafh)()(lim)(0Ixhxfhxfxfh)()(lim)(0 xhxhxhxfhxfxfhh2)(lim)()(lim)(2200The right-hand derivative of f at a ,is denoted by f +(a)The left-hand derivative of f at a , is den

7、oted by f -(a).00()( )()( )( )limlim .hxf ahf af axf afahx 00()( )()( )( )lim lim .hxf ahf af axf afahx f (a) exists if and only if both the right-hand derivative f +(a) and the left-hand derivative f -(a) exist and are equal. If exists, we say that f(x) is differentiable at a or that f(x) has a der

8、ivative at a. )(af Example Let f(x)=x.Show that f (x) is not differentiable at 0.Proof By Definition, we have Therefore the function f(x) is not differentiable at 0.lim)0()0(lim)0(00hhhfhffhhexists.not doeslim that have we , 1limlim and 1limlim Since00000-hhhhhhhhhhhhhhhInterpretation of the Derivat

9、ive as the Slope of a TangentThe tangent line to at is the line through whose slope is equal to ,the derivative of at a.)(xfy )(,(afa)(,(afa)(af f0P(a, f(a)Q(x, f(x) f(x)lAn equation of the tangent line to the curve at the point : )(,(afa)(xfy )()(axafafyThe geometric interpretation of a derivative.

10、Example Find an equation of the tangent lineto the parabola at the point (4,2). xxf)(SolutionThe slope of the tangent line at (4,2) is41)2)(4(4lim42lim4)4()(lim)4(444xxxxxxfxffxxxan equation of the tangent line is)4(412xyor141xyInterpretation of the Derivative as a Rate of ChangeThe derivative is th

11、e instantaneous rate of changeof with respect to when )(af )(xfy x. ax ( )cosc x is called the marginalt( ).R x is called the marginal revenue function( ).P x is called the marginal profit functionExample Find and given that Solution By definition Now lets calculate ( 3)f (1)f 2,1( )21,1xxf xxx02200

12、( 3)( 3)( 3)lim( 3)( 3)limlim( 6)6hhhfhffhhhh 0(1)(1)(1)lim.hfhffhSince f is not defined by the same formula on both sides of 1, we will evaluate this limit by taking one-sided limits. To the left of 1, Thus To the right of 1, ThusAnd 2( ).f xx2000(1)(1)(1)1(1)limlimlim(2)2.hhhfhfhfhhh( )21.f xx000(

13、1)(1)2(1) 1 1(1)limlimlim 22.hhhfhfhfhh(1)2.f hxfhxfxfIxh)()(lim)(0is a function of x, denoted by )()()()(xfDxDfxfdxddxdfdxdyyxfXaxaxaxdxdfdxdyyaf)( Leibniz notationA function f is differentiable on an open interval I if f (x) exists for every x in that interval I. ThenSoExample Find the derivative

14、of xxf)(xhxhxhxfhxfxfhh21lim)()(lim)(00Solutionxx21)(so A function f is differentiable on a closed interval a, b if f is differentiable on an open interval (a, b) and both the right-hand derivative f +(a) and the left-hand derivative f -(b) exist)()(fdomainfdomainTheorem If a function f is different

15、iable at a number, then it is continuous at a. Proof we have Hence Therefore This implies that f is continuous at a. .)()(lim)(axafxfafax. 0)()()(lim)()(limaxaxafxfafxfaxax)()(limafxfaxNOTE 1.The converse of theorem is false 2. f is not continuous at a, then f is notdifferentiable at a.Three ways fo

16、r f(x) not to be differentiable at a 1. a corner at a 2. discontinuous at a 3. have a vertical tangent line at aaaaTechniques of Calculating DerivativesEx1. Cxf)(C is a constant). Solution:yxCCx0lim0That is 0)(CEx 2. )N()(nxxfnsolution:axafxf)()(ax lim)(af axaxnnaxlim(limax1nx2nxa32nxa)1na1nanxxfxxf

17、)()(0limxFind ),(af Note：For any power functionxy )(R1)(xxeg，)(x)(21 x2121xx21x1)(1x11x21x)1(xx)(43x4743x（we will prove it later!）hxhxhsin)sin(lim0Ex 3. xxfsin)(. solution:)(xf hxfhxf)()(0limh0limh)2cos(2hx 2sinh)2cos(lim0hxh22sinhhxcosThat is, xxcos)(sinSimilarly:xxsin)(coshExponential Functionsxax

18、f)(10. 113lim)0(30hfaforhhh0.10.010.0010.00010.71770.69560.69340.69321.16121.10471.09921.0987hh12hh13 hfhffaforh)0()0(lim)0(2069. 012lim0hhh11lim)0(),32(0hefifeeaDenotehhDefinition of the Number ee is a number such that 11lim0hehhExponential Functionsxaxf)()0(1limlim)(00fahaahaaxfxhhxxhxh)0()(faaxxD

19、erivative of the natural exponential Functionxxeedxd)()0()(faaxxDifferential Laws )()( )()() 1 (xvxuxvxu)()()()( )()()2(xvxuxvxuxvxu)()()()()()()()3(2xvxvxuxvxuxvxu)0)(xv Suppose that u(x) and v(x) are differentiable, then their sum, difference, product, and quotient are also differentiable, andPf:

20、Let, thenvuvu )() 1 ()()()(xvxuxfhxfhxfxfh)()(lim)(0hxvxuhxvhxuh )()( )()(lim0hxuhxuh)()(lim0hxvhxvh)()(lim0)()(xvxuThe law can be generalized to more functions.(2)vuvuvu )(Pf: Let, )()()(xvxuxfthenhxfhxfxfh)()(lim)(0hxvxuhxvhxuh)()()()(lim0)()()()(xvxuxvxuhhxuh )(lim0)(xu)(hxvhxv)( )(xu)(hxvCorolla

21、ry: )() 1uC )()2wvuuC wvuwvuwvu )log()3xaaxlnlnaxln1( C is a constant )Ex 1. Solution:xsin41(21)1sin, )1sincos4(3xxxy.,1xyyFind y)(xx)1sincos4(213xxx23( xx)1xy1cos4)1sin43( 1cos21sin2727)1sincos4(3xx)1sincos4(3xx)()( lim0 xvhxvh)()()()()()(xvhxvhxvxuxvhxuh)()(xvxu(3)2vvuvuvuPf: Let)(xfthenhxfhxfxfh)

22、()(lim)(0hh lim0,)()(xvxu)()(hxvhxu)()(xvxuhhxu )( )(xu)(xvhhxv )( )(xu)(xv)()()()()(2xvxvxuxvxuCorollary:2vvCvC( C is a constant ) )(cscxxsin1x2sin)(sinxx2sinExample:,sec)(tan2xxPf: .cotcsc)(cscxxxxxxcossin)(tan x2cosxx cos)(sin)(cossinxx x2cosx2cosx2sinx2secxcosxxcotcscSimilarly:,csc)(cot2xx.tanse

23、c)(secxxx )( xfLaws for inverse function:Th2. is monotonic and differentiable, Pf:,)()(1yfxofinversetheisxfySuppose)(1yfand0 )(1yf)()(xfxxfy,0 xyyxxyxfx0lim)( lim0yyx 1 )(1yf1 )(1yf111Ex3. Derivatives for inverse trigonometric functions.1) Let,arcsin xy then,sin yx , )2,2(y)(arcsinx)(sinyycos1y2sin1

24、1211xSimilarly:?)(arccosx,11)(arctan2xx211)arccot(xx211xxxarcsin2arccosusing0cosy, thenyyeedydxylnxeexyyy11)(1)(lnEx4:Find the derivative of axaxaxln1)lnln()(logThe chain rule ( ( ),( ) ( ( )( ( )( ),( )( )If f and g areboth differentiable and Ffgis the composite function defined by Ff g xthen F is

25、differentiable and F is given by theproductF xf g xfg x g xIn Leibniz notation if yf u and ug x areboth differ,entiable function thendydydudxdudxyux,sinuy ,2xu dxdydudydxduucosx22cos2xx2sin)1(xy Find the derivative of the follow ingfunction2(2)sinyxWe letSolutionandthen2(2)sinyx2yusinuxdxdydudydxducosx2sin cossin2xxx2u( ( )( ( )( )dfg xfg x g xdxOutfunctionEvaluatedat innerfunctionDerivativeof outfunctionDerivativeof innerfunctiontanxFind the derivative of yetantanuxand ux then the deri