C3.5DifferentiationandIntegration1

Boardworks Ltd 20061 of 53These icons indicate that teachers notes or useful web addresses are available in the Notes Page.This icon indicates the slide contains activities created in Flash.These activities are not editable.For more detailed instructions,see the Getting Started presentation.Boardworks Ltd 20061 of 53A-Level Maths:Core 3for OCRC3.5 Differentiation and Integration 1 Boardworks Ltd 20062 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 20062 of 53The chain rule Boardworks Ltd 20063 of 53Review of differentiationSo far,we have used differentiation to find the gradients of functions made up of a sum of multiples of powers of x.We found that:and when xn is preceded by a constant multiplier k we have:1If=then=nndyykxknxdxfgfgIf=()()then=()()dyyxxxxdxAlso:1If=then=nndyyxnxdx Boardworks Ltd 20064 of 53Review of differentiationWe will now look at how to differentiate exponential,logarithmic and trigonometric functions.We will also look at techniques that can be used to differentiate:Compound functions of the form f(g(x).For example:21x xe32x3sin()Products of the form f(x)g(x),such as:xxlnxxe2xx23cosxx 23+11xex2sinxx2ln Quotients of the form ,such as:fg()()xx Boardworks Ltd 20065 of 53The chain ruleThe chain rule is used to differentiate composite functions.For instance,suppose we want to differentiate y=(2x+1)3 with respect to x.One way to do this is to expand(2x+1)3 and differentiate it term by term.Using the binomial theorem:xxxx332(2+1)=8+3(4)+3(2)+1xxx32=8+12+6+1dyxxdx2=24+24+6xx2=6(4+4+1)x2=6(2+1)Differentiating with respect to x:Boardworks Ltd 20066 of 53The chain ruleAnother approach is to use the substitution u=2x+1 so that we can write y=(2x+1)3 as y=u3.The chain rule states that:If y is a function of u and u is a function of x,thendydydudxdudx=So if y=u3 whereu=2x+1,dyudu2=3dudx=2Using the chain rule:dydydudxdudx=u2=32u2=6x2=6(2+1)Boardworks Ltd 20067 of 53The chain rule1212=dyududuxdx=6Using the chain rule,dydydudxdudx=1212=6ux122=3(35)xxUse the chain rule to differentiate y=with respect to x.x 235Let where u=3x2 5yu12=23=35xx xu12=3 Boardworks Ltd 20068 of 53The chain ruledyudu5=8duxdx2=3Using the chain rule:dydydudxdudx=ux52=8 3xx235=24(7)Let yu42=xx23 524=(7)x u25=24Find given that .yx342=(7)dydxu4=2where u=7 x3 Boardworks Ltd 20069 of 53The chain rule using function notationWith practice some of the steps in the chain rule can be done mentally.Suppose we have a composite functiony=g(f(x)If we lety=g(u)whereu=f(x)thenandg=()dyuduf=()duxdxUsing the chain rule:dydydudxdudx=gf=()()uxBut u=f(x)soIf y=g(f(x)theng ff=()()dyxxdx Boardworks Ltd 200610 of 53The chain ruleAll of the composite functions we have looked at so far have been of the form y=(f(x)n.In general,using the chain rule,If y=(f(x)n thenff1=()()ndynxxdxIf we use to represent f(x)and to represent f(x)we can write this rule more visually as:1=nndyyndyFind the equation of the tangent to the curve y=(x4 3)3 at the point(1,8).Boardworks Ltd 200611 of 53The chain ruleWhen x=1,dydx2=12(1 3)=48Using y y1=m(x x1)the equation of the tangent at the point(1,8)is:y+8=48(x 1)y=48x 48 8y=48x 56y=8(6x 7)dydxxdxdx424=3(3)(3)xx423=3(3)4xx 342=12(3)y=(x4 3)3 Boardworks Ltd 200612 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200612 of 53Connected rates of change Boardworks Ltd 200613 of 53Rates of changeWhen we talk about rates of change we are usually talking about the rate at which a variable changes with respect to time.Suppose,for example,that a spherical balloon is slowly being inflated.There are several changes we could measure,such as:We can connect these rates of change using the chain rule.The rate at which the radius r changes,drdtThe rate at which the surface area A changes,dAdtThe rate at which the volume V changes,dVdtFor example,the rate of change of the surface area is connected to the rate of change of the radius bydAdAdrdtdrdt=Boardworks Ltd 200614 of 53Rates of changeSo,the rate of change of the surface area is connected to the rate of change of the radius byIf the surface area of the balloon is increasing at a rate of 15 cm2 s1,find the rate at which the radius of the balloon is increasing at the moment when the radius is 4 cm.We can find by differentiating the formula for the surfacearea of a sphere with respect to the radius.dAdrdArdr=8A=4r2sodAdrrdtdt=8 Boardworks Ltd 200615 of 53Connected rates of changeWe are given that =15 and we want to find when r=4.dAdtdrdtUsing we have:dAdrrdtdt=8drdt15=32 drdt15=32=0.149 So,the radius is increasing at a rate of 0.149 cm s1(to 3 s.f.)at the moment when the radius of the balloon is 4 cm.Boardworks Ltd 200616 of 53Connected rates of changeSo If the air in the fully-inflated balloon is released at a rate of 30 cm3 s1,find the rate at which the surface area is decreasing at the moment when the radius is 5 cm.Using the chain rule,the rate of change of the volume is connected to the rate of change of the radius by:dVdVdrdtdrdt=We are given that =30 and we want to find when r=5.dVdtdAdt343=VrdVrdr2=4dVdrrdtdt2=4 Boardworks Ltd 200617 of 53Connected rates of changeHere =30 and r=5:dVdtdrdt30=100 We can now find using with r=5:dAdtdAdrrdtdt=8drdt30=1003=10dAdt3=40 104=12So,the surface area is decreasing at a rate of 12 cm2 s1 at the moment when the radius of the balloon is 5 cm.Boardworks Ltd 200618 of 53Connected rates of change Boardworks Ltd 200619 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200619 of 53The relationship betweendydxdxdyand Boardworks Ltd 200620 of 53The relationship betweenSuppose we are given x as a function of y instead of y as a function of x.For instance,dydxdxdyandx=4y2We can find by differentiating with respect to y:dxdy=8dxydyUsing the chain rule we can write=1,dydxdxdy1=dxdydydxSo by the above result,if 8y then=dxdy1=8dydxyfrom which we get:Boardworks Ltd 200621 of 53The relationship betweenFind the gradient of the curve with equation x=2y3 3y 7 at the point(3,2).dydxdxdyandx=2y3 3y 7dxydy2=63At the point(3,2),y=2:dxdy2=6(2)3=21We can now find the gradient using the fact that 1=dxdydydxdydx1=21 Boardworks Ltd 200622 of 53Differentiating inverse functions=cosdxydyFind ,writing your answer in terms of x.1(sin)dxdyLet y=sin1 x soThe result is particularly useful for differentiatinginverse functions.For example:1=dxdydydxx=sin yUsing the identity cos2y=1 sin2yBut sin y=x so1=cosdydxy21=1 sin y121(sin)=1dxdyx Boardworks Ltd 200623 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200623 of 53Differentiating ex and related functions Boardworks Ltd 200624 of 53The derivative of exFrom this,it follows thatxxdyyeedxIf=then=xxdyykekedxIf=then=A special property of the exponential function ex is thatwhere k is a constant.For example,if y=4ex x32=43xdyexdx Boardworks Ltd 200625 of 53Functions of the form ekxSuppose we are asked to differentiate a function of the form ekx,where k is a constant.For example,Differentiate y=e5x with respect to x.udyedu=5dudxUsing the chain rule:dydydudxdudx=5ue5=5xeLet where u=5xuye=5ueIn practice,we wouldnt need to include this much working.Boardworks Ltd 200626 of 53Functions of the form ekxWe would just remember that in general,For example,We can use the chain rule to extend this to any function of the form ef(x).If=then=kxkxdyyekedx22()=2xxdeedx77()=7xxdeedx33()=3xxdeedx Boardworks Ltd 200627 of 53Functions of the form ef(x)If y=ef(x)then we can letUsing the chain rule:dydydudxdudx=f=()uexLet where u=f(x)uye=ff()=()xx ef=()duxdxudyedu=thenSo in general,fff()()If=then=()xxdyyex edxIn words,to differentiate an expression of the form y=ef(x)we multiply it by the derivative of the exponent.Boardworks Ltd 200628 of 53Functions of the form ef(x)For example,Using to represent f(x)and to represent f(x):=dyyeedx3()=xdyedx3xe54()=xdyedx545xe29(5)=xdyedx2952=xxe2910 xxe Boardworks Ltd 200629 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200629 of 53Differentiating ln x and related functions Boardworks Ltd 200630 of 53The derivative of ln xRemember,ln x is the inverse of ex.So,if y=ln xthen x=eyDifferentiating with respect to y gives:ydxedy=11=ydxdydydxedydxx1=But ey=x so,dyyxdxx1If=ln then=Boardworks Ltd 200631 of 53Functions of the form ln kxSuppose we want to differentiate a function of the form ln kx,where k is a constant.For example:Differentiate y=ln 3x with respect to x.1=dyduu=3dudxUsing the chain rule:dydydudxdudx=3=u1=xLet where u=3x=lnyu3=3x Boardworks Ltd 200632 of 53Functions of the form ln kxWhen functions of the form ln kx are differentiated,the ks will always cancel out,so in general,We can use the chain rule to extend to functions of the more general form y=ln f(x).1If=ln then=dyykxdxxUsing the chain rule:dydydudxdudx=f()=xuLet where u=f(x)=lnyuff()=()xxf=()duxdx1=dyduuthen Boardworks Ltd 200633 of 53Functions of the form ln(f(x)fff()If=ln()then=()dyxyxdxxIn general,using the chain ruleUsing to represent f(x)and to represent f(x):For example,ln(74)=dxdx774x =ln=dyydy3ln(3+8)=dxdx2393+8xx Boardworks Ltd 200634 of 53Functions of the form ln(f(x)In some cases we can use the laws of logarithms to simplify a logarithmic function before differentiating it.Remember that,ln(ab)=ln a+ln b ln=lnlnaabbln an=n ln a Differentiate with respect to x.=ln2xy=ln2xy=lnln2x 12=lnln2x 12=lnln2x Boardworks Ltd 200635 of 53Functions of the form ln(f(x)ln 2 is a constant and so it disappears when we differentiate.12=lnln2=dyyxdx11=2x1=2xIf we had tried to differentiate without simplifying it first,we would have had:=ln2xy1212=ln=dyyxdx12121412xx11221=2x x1=2xThe derivative is the same,but the algebra is more difficult.Boardworks Ltd 200636 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200636 of 53The product rule Boardworks Ltd 200637 of 53The product ruleThe product rule allows us to differentiate the product of two functions.It states that if y=uv,where u and v are functions of x,thendydvduuvdxdxdx=+So3=4duxdx121=(32)22dvxdx4Find given that=32.dyyxxdxLetu=x4 andv 12=(32)x12=(32)x Boardworks Ltd 200638 of 53The product ruleUsing the product rule:112243=(32)+4(32)dyxxxxdx11221122434(32)(32)=+(32)(32)xxxxxx1243+4(32)=(32)xxxx12434+128=(32)xxxx1234129=(32)xxx343(43)=32xxx Boardworks Ltd 200639 of 53The product ruleGive the coordinates of any stationary points on the curve y=x2e2x.So=2duxdx2=2xdvedxLetu=x2 andv 2=xeUsing the product rule:222=2+2xxdyx exedx2=2(+1)xxex2=0 when 2=0 or+1=0 xdyxexdx22=0=0 xxex+1=0=1xx Boardworks Ltd 200640 of 53The product ruleWhen x=0,y=(0)2e0=0The point(0,0)is a stationary point on the curve y=x2e2x.When x=1,y=(1)2e2=e2The point(1,e2)is also a stationary point on the curve y=x2e2x.Boardworks Ltd 200641 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200641 of 53The quotient rule Boardworks Ltd 200642 of 53The quotient ruleThe quotient rule allows us to differentiate the quotient of two functions.2=dudvdxdxvudydxvIt states that if y=,where u and v are functions of x,thenuvFind given that .xyx22+1=5dydxLetu=2x+1andv=5x2Sodudx=2dvxdx=10 Boardworks Ltd 200643 of 53The quotient ruledyxxxdxx24(5)(2)(2+1)(10)=25xxxx224102010=25xxx3242=5xx322=5xx32(+1)=5 Boardworks Ltd 200644 of 53The quotient ruleLetu=ln x4 andv=x2So344=duxdxx=2dvxdxUsing the quotient rule:21444ln2=dyxxxxdxx448 ln=xxxxFind the equation of the tangent to thecurve y=at the point(1,0).42lnxx1=4x34(1 2ln)=xxUsing ln x4=4 ln x Boardworks Ltd 200645 of 53The quotient ruleWhen x=1,The gradient of the tangent at the point(1,0)is 4.Use y y1=m(x x1)to find the equation of the tangent at the point(1,0).y 0=4(x 1)y=4x 44(1 2 ln1)=1dydx=4Remember that ln 1=0 Boardworks Ltd 200646 of 53The chain ruleConnected rates of changeThe relationship betweenDifferentiating ex and related functionsDifferentiating ln x and related functionsThe product ruleThe quotient ruleExamination-style questionsdydxdxdyandContents Boardworks Ltd 200646 of 53Examination-style questions Boardworks Ltd 200647 of 53Examination-style question 1The height of a conical container is twice its radius,r cm,as shown in the diagram.2rrLiquid is poured into the container at a rate of 5 litres per minute.If x cm is the depth of the liquid at time t minutes,write an expression for the rate at which the depth is increasing when x=2 cm.(The volume of a cone of radius r and height h is given by ).213r hThe volume of the liquid in the container at time t is given by21=3vr x Boardworks Ltd 200648 of 53Examination-style question 1The liquid enters the container at a rate of 5 litres per minute so21=32xvx3=12xv2=4dvxdxSince the radius of the container is half its depth this can be written in term of x as=5dvdt Boardworks Ltd 200649 of 53Examination-style question 1So=dvdvdxdtdxdtUsing the chain rule25=4xdxdt220=dxdtxAt the instant when x=2 cm the rate at which the liquid is entering the container is20=4dxdt5cm/min Boardworks Ltd 200650 of 53Examination-style question 2a)find f(x),b)find the coordinates of any stationary points and determine their nature,c)sketch the curve y=f(x).Given that ,f22()=+4xxxa)Using the quotient rule:f2()=dudvdxdxvuxvf2222(+4)2(2)()=(+4)xxxxxxxx22222+84=(+4)2222(4)=(+4)xx Boardworks Ltd 200651 of 53Examination-style question 2b)When f(x)=0,xx2222(4)=0(+4)x22(4)=0 x24=0 x=2When x=2,y4=81=2When x=2,y4=81=2Therefore,the graph of the function has turning points at(2,)and(2,).xxx22f()=+41212 Boardworks Ltd 200652 of 53Examination-style question 2Looking at the gradient just before and just after x=2:12So(2,)is a maximum point.x1.922.1Value of2222(4)=(+4)dyxdxxSlopeLooking at the gradient just before and just after x=2:12So(2,)is a minimum point.x2.121.9Value of2222(4)=(+4)dyxdxxSlopeive0+ive0.0100.010.0100.01ive0+ive Boardworks Ltd 200653 of 53Examination-style question 2When x=0,c)The curve crosses the axes when x=0 and when y=0.22=+4xyxy=0.(Also,when y=0,x=0).Therefore the curve has one crossing point at the origin,a minimum at(2,)and a maximum at(2,):1212Also,as,xy 00and,as,xy 0+.02 2=+4xyx0221212xy