Dr. M.G.R. Educational and Research Institute - 1st Year B ...



UNIT-IVDIFFERENTIATIONBASIC CONCEPTS OF DIFFERTIATIONConsider a function y=f(x) of a variable x. Suppose x changes from an initial value x0 to a final value x1. Then the increment in x defined to be the amount of change in x. It is denoted by ?x. The increment in y namely ?y depends on the values of x0 and ?x.If the increment ?y is divided by ?x the quotient ?y?x is called the average rate of change of y with respect to x, as x changes from x0 to x0+?x. The quotient is given by ?y?x = fx0+?x- fx0?xThis fraction is also called a difference quotient.Differentiation using standard formulaeDefine differentiation. Solution: The rate of change of one variable quantity with respect to another variable quantity is called Differentiation.(i.e)If y is a function of x, then the rate of change of y with respect to x is called the differential co-efficient of y. It is denoted by dy/dx (or) d(f(x))/dx (or) f’x (or) Df(x)Find dydx if y=3sinx+4cosx-ex. Solution: dydx=3cosx -4sinx-exFind dydx if y=ex+3tanx+logx6. Solution: Given y=ex+3tanx+6logx dydx= ex+ 3sec2x+6x If y=x3-6x2+7x+6 +cosx+ 1xx, find dydx. Solution:Given y= x3-6x2+7x+6+cosx+x-32 dydx=3x2-12x+7-sinx+-32x-32-1=2x-sinx+-32x52 If y= 3x+3x4 -13x, find dydx . Solution: Given y=3x+3x4 -13x dydx=3-12x-12+12x3--13x-43 =-32x32+12x3+13x43 Find the derivative of y= e7x+sin3x+e5x+3. Solution: dydx=ddxe7x+sin3x+e5x+3. =7e7x+3cos3x+5e5x+3 Find dydx if y= log7x . Solution: we know that ,if y= logax then dydx = 1x logae dydx=1xlog7eChain Rule (or) Differential coefficient of a function of function If u=f(x), y=f(u) then dydx=dydu. dudxFind dydx , if y= x+1 x+x+1x3. (L1) Solution: dydx=1+-1x-2+3x+1x2ddx x+1x =1-1 x2+3x+1x21-1x2 Find dydx ,if y= loge(2x+3). Solution: dydx=12x+3 ddx(2x+3) =12x+32=2(2x+3)Differentiate y= cos(x+y) . Solution:Given, y= cos(x+y) Differentiating both sides w.r.to x dydx= -sinx+y(1+dydx)=-sinx+y-dydx(sinx+y) dydx1+sinx+y= -sinx+y ?dydx=-sinx+y1+sinx+yDifferentiate y=log?( sin2x). Solution: Given y=log?( sin2x) Let u=sin2x ? dudx=2 sinx cosx ∴y=logu ? dydu=1u dydx=dydu.dudx=1u. 2 sinx cosx =1sin2x. 2 sinx cosx =2cosxsinx=2cotxFind dydx if Y=sec(ax+b). Solution: Let y= sec(ax+b) dydx = a sec(ax+b) tan(ax+b)Differentiate y=2x+3 +etanx. Solution: y=etanx1/2 dydx=122x+3-1/2 2+etanx1/2ddxtanx1/2 =12x+3 +etanx 12tanx-1/2ddx(tanx) =12x+3 +etanx2√tanx sec2xFind dydx,given y=3x3+x+1 . Solution: Given y=3x3+x+1 dydx=13x3+x+113-1.dx3+x+1dx dydx=13x3+x+113-1.3x2+1 =3x2+13x3+x+123Differentiate y=logcos53x4 with respect to `x'. Solution: Given y=logcos53x4 Differentiate both sides with respect to x, dydx = ddxlogcos53x4 = 1cos53x4.ddxcos53x4 = 1 cos53x4ddxcos3x45 = 1cos53x45(cos3x44ddxcos?(3x4) = 5 cos4(3x4)cos53x4-sin?(3x4) ddx(3x4) =5cos3x4-sin?(3x4)12 x3 =-60 x3 sin?(3x4)cos?(3x4) =-60 x3 tan3x4Differentiate y=4x+x-513,with respect to `x'.Hence show that dydx=4x6-53x83(4X6+1)23 . Solution: Given y=4x+x-513 ∴ dydx=134x+x-513-1ddx4x+x-5 =134x+x-5-23(4-5x-6) =134x+1x5-234-5x6 =134x6+1x5-234x6-5x6 =13x54x6+1234x6-5x6 =13x1034x6+1234x6-5x6 =x103-634x6+1234x6-5 =x-834x6-534x6+123 =4x6-53x83(4X6+1)23 Thus proved.Find dydx if y= log1+sinx1-sinxSolution: Given y=log1+sinx1-sinx dydx=11+sinx1-sinxddx1+sinx1-sinx =1-sinx1+sinx1-sinxcosx-1+sinx(-cosx)(1-sinx)2 =cosx-sinxcosx-(-cosx-sinx cosx)1+sinx(1-sinx) =2cosx1-sin2x =2cosxcos2x=2secx Find dydx if y=1-sin2x . Solution: Given y= 1-sin2x y=(1-sin2x)1/2 dydx=12(1-sin2x)12-1ddx(1-sin2x) =12(1-sin2x)-12(0-2sinx cosx) =-sin2x21-sin2xFind dydx if y= log1+x1-x. Solution: y= log(1+x) –log(1-x) dydx=11+xddx(1+x)- 11-xddx(1-x) = 11+x 12x - 11-x (-12x) = 12√x 11+x+11-x =22x((1-(x)2) = 1x(1-x)Differentiation using product ruleLet u & v be differentiable functions of x. Then the product of a functionY = u(x).v(x) is differentiable.d(uv)=udv+vduIf y=x2sinx,find dydx. Solution: By product rule, dydx=x2ddxsinx+sinxddx(x2) dydx=x2cosx+2xsinx =x(xcosx+2sinx). Find dydx if y=extanx . Solution: Let y=extanx By product rule, dydx= exddxtanx+tanxddx(ex) dydx=exsec2x+extanx =ex(sec2x+tanx)If y=3x4 ex ,find dydx. Solution: By product rule, dydx=3 exddxx4 +x4 ddx(ex) dydx=3(x4 ex+4x3 ex) =3x3 ex(x+4)If y=cosx ex , find dydx. Solution: By product rule, dydx= exddxcosx+cosxddx(ex dydx= ex(-sinx)+ excosx =ex(cosx-sinx) If y= xlogex , find dydx . Solution:By product rule, dydx= xddxlogex+logexddx(x) dydx=x1x+logex =1+ logexDifferentiate Y=(x2-2) (3x+1). (L4) Solution: Let y= (x2-2) (3x+1). dydx=x2-23+3x+1(2x ) =3 (x2-2) +2x 3x+1 =3x2-6 + 6x2+2x =9x2+2x-6 If y=cosecx cotx, find dydx . Solution:By product rule, dydx= cosecxddxcotx+cotxddx(cosecx) dydx=cosecx(-cosec2x)+cotx (-cosecx cotx) =- cosecx ( cosec2x+ cot2x) If y= (x2+7x+2) (ex-logx), finddydx. Solution:By product rule, dydx=(x2+7x+2) ddx(ex-logx)+(ex-logx)ddx(x2+7x+2 = (x2+7x+2)( ex-1x) +ex-logx(2x+7), If y=(6sinx log10x ) ,find dydx . Solution:By product rule, dydx=6 sinxddxlog10x +log10x ddx(sinx) = 6 sinx1xlog10e+log10x(cosx)If y=(ex logxcotx) ,find dydx . Solution:By product rule, dydx=exlogxddxcotx+ex cotxddxlogx+logxcotxddxex = ex logx(-cosec2x)+ ex cotx1xlogx cotx ex.Differentiatesin2x cos3x. Solution: Let y=sin2x cos3x dydx=sin2x ddxcos3x+cos3x ddxsin2x =sin2x -sin3x.3+cos3x[2sinx. ddxsinx] =-3sin2x sin3x+cos3x [2sinx cosx] =sinx[-3sinx sin3x+2cosx cos3x]Differentiate e4xsin4x. Solution: Let y= e4xsin4x dydx=e4x d dxsin4x+sin4xd dxe4x =e4xcos4xd dx4x+sin4x[e4x.d dx4x] =e4xcos4x(4)+sin4x[e4x.4] =4e4x[cos4x+sin4x]Quotient Rule for DifferentiationLet u & v be differentiable functions of x, then uv is also differentiable d(uv) = vu,-uv,v2Find y’ , if y= x21+x2 . Solution:By quotient rule, dydx=1+x2ddxx2-x2ddxx2+11+x22 dydx = y’= 1+x22x- x2(2x)(1+x2)2=1(1+x2)2Find y’ , if y= x2-11+x2. Solution:By quotient rule, dydx=1+x2ddxx2-1-(x2-1)ddxx2+11+x22 dydx = y’ =1+x22x- (x2-1)(2x)(1+x2)2 =4x(1+x2)2Find y’ , if y= 2x-34x+5. Solution:By quotient rule ,dydx=4x+5ddx2x-3-(2x-3)ddx4x+54x+52 dydx = y’ =4x+52-2x-34(4x+5)2 = 8x+10-8x+12(4x+5)2 =22(4x+5)2Find y’ , if y= logxsinx. Solution:By quotient rule dydx=sinxddxlogx-(logx)ddxsinxsinx2 dydx = y’ = sinx1x-logx(cosx)(sinx)2 = sinx-x cosx logxx sin2xFind y’ , if y= logx2ex . Solution: y= 2logxex By quotient rule, dydx=exddx2logx-(2logx)ddxexex2 dydx = y’= ex21x-2logx(ex)(ex)2=2ex-x2logx(ex)x(ex)2 =2ex(1-xlogx)xe2xFind y’ , if y= x2+ex(cosx+logx) . Solution:By quotient rule, dydx=(cosx+logx)ddxx2+ex-(x2+ex)ddx(cosx+logx)(cosx+logx)2 dydx = y’= cosx+logx2x+ex+(x2+ex)-sinx+1x(cosx+logx)2Find y’ , if y= sinx+cosxsinx-cosx . Solution:By quotient rule, dydx=sinx-cosxddxsinx+cosx-(sinx+cosx)ddxsinx-cosxsinx-cosx2dydx =y’= sinx-cosxcosx-sinx-sinx+cosx(cosx+sinx)(sinx-cosx)2 =-(sinx-cosx)2-(sinx+cosx)2(sinx-cosx)2Differentiate y= (x7-47)(x-4) . Solution:By quotient rule, dydx=x-4ddxx7-47-(x7-47)ddxx-4x-42 dydx = y’= x-47x6-x7-47(1)(x-4)2 =7x7-28x6-x7+47)(x-4)2=(6x7-28x6+47)(x-4)2Differentiate Y= (x+1)(x2+1) . Solution: y=(x+1)(x2+1) dydx=x2+1-x+12x(x2+1)2 =x2+1-2x2-2x(x2+1)2 =-x2-2x+1(x2+1)2Differentiate x2-x+1x2+x+1 Solution: Let y=x2-x+1x2+x+1 dydx=x2+x+12x-1-x2-x+1(2x+1)x2+x+12 =2x3-x2+2x2-x+2x-1-2x3+x2-2x2-x+2x+1x2+x+12 =2x3+x2+x-1-2x3+x2-x-1x2+x+12 =2x2-2x2+x+12 =2(x2-1)x2+x+12Differentiate secxlogx. Solution: Let y=secxlogx dydx=logxsecx tanx-secx.1xlogx2 =x logx secx tanx-secxxlogx2 =secx[xtanx logx-1]xlogx2Find the derivative of ex+e-xex-e-x . Solution: Let y= ex+e-xex-e-xdydx=ex-e-xex-e-x-(ex+e-x)(ex+e-x)ex-e-x2 =ex-e-x2-ex+e-x2ex-e-x2 =(e2x+e-2x-2)-(e2x+e-2x+2)ex-e-x2 =(e2x+e-2x-2-e2x-e-2x-2)ex-e-x2 =-4ex-e-x2Differentiate y=te2t2 cost with respect to `t'.Solution: Given y=te2t2 cost Let u=te2t, v=2 cost dudt=t2e2t+e2t1=2te2t+e2t, dvdt=-2sint ∴ dydt=vu'-uv'v2 =2cost2te2t+e2t-te2t(-2sint)2 cost2 =4t e2tcost+2e2tcost+2t e2tsint4cos2t =2e2t2t cost+cost+t sint4cos2t i.e dydt=e2t2cos2t2t cost+cost+t sintDifferentiation of Parametric functions If x and y are expressed in terms of a third variable t , then the third variable is called the parameter, equation containing a parameter is known as parametric equation. (ie) If x = f(t), y = g(t) then dydx = dydtdxdt If x=t,y=t+1t ,then find dydx. Solution: dydx=dydtdxdt x=t ? dxdt= 12t y=t+1t ? dydt=1-1t2 dydx=(1-1t2)12t =2t2-1√tt2=2t2-1t32If x=a(1+cosθ) ; y= a(θ+sinθ) then, find dydx . Solution: dxdθ=a(0-Sinθ)= -asinθ dydθ =a(1+cosθ) dydx=dydθdxdθ=a(1+cosθ) -asinθ =-cos2θ2sinθ2cosθ2 =-cotθ2Find dydx ,if x=at-sint,y=a(1-cost). Solution: Given x=at-sint, y=a(1-cost) dxdt=a1-cost , dydt=a(0+sint) ∴dydx=dydtdxdt=a(sint)a(1-cost)=sint1-cost Find dydx,if x=ct ,y=ct . Solution: Given x=ct , y=ct dxdt=c , dydt=c(-1)t2 ∴dydx=dydtdxdt=-ct2×1c=-1t2 Find dydx,if x=acost,y=bsint . Solution: Given x=acost, y=bsint dxdt=-a sint , dydt=b cost ∴dydx=dydtdxdt=b cost-a sint=-bacott Find dydx,if x=acos2t,y=bsin2t. Solution: Given x=acos2t, y=bsin2t dxdt=a 2cost- sint , dydt=2bsint cost ∴dydx=dydtdxdt=2 b sint cost-2 a cost sint=-baLogarithmic Differentiation Take the logarithm of the given function, then differentiate. This method is useful for those functions in which the base and index both are variables.Differentiate Y= sinxx. Solution: y= sinxxTaking log on both sides ,we get, log y= logsinxx log y= xlogsinxDifferentiate y=xx . Solution: Let y= xxTaking log on both sides ,we get, log y= logxx log y= x logxDifferentiating both sides w.r.to x, 1ydydx=x1x+ logx 12x =1x1+logx2=xx1x1+logx2Differentiate x-2(x-1)x+1(x-3). Solution: Let y=x-2(x-1)x+1(x-3) Taking log on both sides logy= logx-2(x-1)x+1(x-3) =logx-2x-1-logx+1(x-3) =log?(x-2)+log?(x-1)-log?(x+1)-log?(x-3) Differentiate both sides with respect to `x', 1y dydx=1x-2+1x-1-1x+1-1x-3 dydx=y1x-2+1x-1-1x+1-1x-3=x-2(x-1)x+1(x-3)1x-2+1x-1-1x+1-1x-3Find dydx ,given y=x-1x-2x-3(x-4) . Solution: Given y=x-1x-2x-3(x-4) By taking log on both sides logy=12 logx-1x-2x-3(x-4) =12logx-1+logx-2+logx-3+log?(x-4) Differentiate both sides with respect to `x'1y dydx=121x-1+1x-2+1x-3+1x-4∴ dydx=y. 121x-1+1x-2+1x-3+1x-4 =x-1x-2x-3(x-4). 121x-1+1x-2+1x-3+1x-4 If xmyn=x+ym+n,then show that y,=yx . Solution: Given xmyn=x+ym+n Taking log on both sides logxmyn=logx+ym+n log xm+ log yn=m+nlog?(x+y) m logx+n logy=m+n log?(x+y) Differentiate both sides with respect to x, m.1x+n.1y.y,=m+n.1x+y(1+y,) ? ny. y,-m+nx+y. y,=m+nx+y-mx ? y,ny-m+nx+y=xm+n-m(x+y)x(x+y) ? y,nx+y-y(m+n)y(x+y)=xm+n-m(x+y)x(x+y) ? y, nx+ny-my-nyy(x+y)=mx+nx-mx-myx(x+y) ? y,nx-myy(x+y)=nx-myx(x+y) ? y,=yxDifferentiate (tanx)secx. Solution: Let y=(tanx)secx Taking log on both sides logy=log(tanx)secx logy=secx log?(tanx) Differentiate both sides with respect to x,1ydydx=secx1tanxsec2x+secxtanx.log?(tanx)? dydx=y1cosxcosxsinxsec2x+secxtanx.log?(tanx) =(tanx)secxcosecxsec2x+secxtanx.log?(tanx)If xy=yx ,then prove that dydx=y(y-xlogy)x(x-ylogx) . Solution: Given xy=yx Taking log on both sides y logx=x logy Differentiate both sides with respect to `x' y1x+logx.dydx=x.1y dydx+logy.1 ? yx+dydx.logx=xy.dydx+logy dydxlogx-xy=logy-yx dydxy log x-xy=x logy-yx ∴dydx=y(xlogy-y)x(ylogx-x)=y(y-xlogy)x(x-ylogx) Find dydx,if cosxy=sinyx. Solution: Given cosxy=sinyx By taking log on both sides log cosxy=logsinyx ?ylogcosx=x log?(siny) Differentiate both sides with respect to `x'y1cosx.-sinx+logcosx.dydx=x.1sinycosydydx+logsiny.1 dydxlogcosx-x coty=logsiny+y tanx ?dydx=logsiny+y tanxlogcosx-x cotyIf xy=ex-y Prove that y,=logx(1+logx)2 . Solution: Given xy=ex-y By taking log on both sides logxy=logex-y y logx=x-y.loge ?y logx=x-y --------------------------(1) Differentiate both sides with respect to x,by product rule y.ddxlogx+logx.dydx=1-dydx ?y.1x+logxdydx=1-dydx ?logx.dydx+dydx=1-yx ? dydx1+logx=1-yx ----------(2) From (1) , we get, y logx+y=x ? y1+logx=x ? y=x1+logx∴ Equation 2 ?dydx1+logx=1-x1+logxx =1-11+logx =1+logx-11+logx =logx1+logx ∴ dydx=logx1+logx2 Differentiate x3x+2 with respect to x. Solution: Let y=x3x+2 By taking log on both sides logy=logx3x+2 i.e logy=3x+2logx Differentiate both sides with respect to x, ∴1y dydx=3x+2 1x+logx 3 Hence , dydx=y3x+2x+3 logx =x3x+23x+2x+3 logxDifferentiation of Implicit functions If the relation between x and y is given by an equation of the form f(x,y) = 0, then the function is called implicit functionFind dydx if x2a2 +y2b2 = 1. Solution:Given, x2a2 +y2b2 = 1, Differentiating both sides w.r.to x 1a22x+1b2(2y) dydx =0 2yb2dydx=-2xa2 dydx=-2xa2 xb22y=-ba2xyDifferentiate xy2 =k . (L4) Solution: Given , xy2 =k , Differentiating both sides w.r.to x x2ydydx+y2.1=0 2xy dydx= -y2 dydx=-y2x Find dydx,given x2+y2+x+y+λ=0. (L4) Solution: Given x2+y2+x+y+λ=0 Differentiate both sides with respect to x, 2x+2ydydx+1+dydx+0=0 ? dydx1+2y=-1+2x ∴dydx=-1+2x1+2yMaxima (or) MinimaStationary point (or) Turning point The point at which the function changes its nature is called the turning point. Stationary points on a graph where the gradient is zero.Maxima (or) Minima: At a point where the function changes from an increasing function to a decreasing function , the function attains its maximum value (ie) the value of the function that point is greater than all other values in the neighbourhood on either side of the turning point.Working rule to find maxima or minima of a given function;Find dydx and equate it to zeroFind the roots of dydx = 0. Let it be a1,a2…..an. These points x=a1,x=a2…. x=anare called turning pointsFind d2ydx2Find( d2ydx2)at x=a1, ( d2ydx2)at x=a2,…… ( d2ydx2)at x=anIf ( d2ydx2)at x=a1= _ ve, then we have a max value at x=a1 = +ve, then we have a min value at x=a1If ( d2ydx2)at x=a1= 0 then find ( d3ydx3)at x=a1= _ ve, then we have a max value at x=a1 = +ve, then we have a min value at x=a1 Define stationary points. Solution: Stationary points are points on a graph where the gradient is zero.Find the stationary points on the graph of y=2x2+4x3. (L1)Solution:Given y= 2x2+4x3 dydx=4x+12x2 At stationary points, dydx=0? 4x+12x2=0 ? 4x(1+3x) =0 ? 4x=0 (or)( 1+3x)=0? x=0 (or) x =-13 d2ydx2=4+24x When x=0, d2ydx2=4>0 is positive. ?x=0 is a point of minimum value. When x=-13, d2ydx2=-4<0 is negative. ?x=-13 is a point of maximum value. At x=0 ? y=0. At x=-13?y=219+4-127=227 ∴ Maximum point is-13,227 Minimum point is (0,0) Determine the stationary points on the graph of y=x3-3x+1. State their nature. (L6)Solution: Given y= x3-3x+1 dydx=3x2-3 At stationary points , dydx=0?3x2-3=0?3x2-1=0?x2=1?x=±1 d2ydx2=6x. When x=1, d2ydx2=6>0 is positive. ?x=1 is point of minimum value. When x=-1, d2ydx2=-6<0 is negative.?x=-1 is a point of maximum value.At x=1; y=-1 At x=-1 , y=3. ∴ Maximum point is (-1,3). Minimum point is (1,-1).Leibnitz theorem If u and v be any two functions of x,then the nth derivative of the function of y=uv is Dn(uv)= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+-- -------+UDn(v) It is useful for finding the nth differential coefficient of a productState Leibnitz theorem. (L1) Solution:If u and v be any two functions of x,then the nth derivative of the function of y=uv is Dn(uv)= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+-- -------+UDn(v)Find the nth differential coefficient of y=xm. (L1) Solution: Let y=xm,Then y1=mxm-1, y2=mm-1xm-2,y3=mm-1(m-2)xm-3 In general, yn=mm-1m-2…(m-n+1)xm-n Determine the nth differential coefficient of x2eax using Leibnitz's rule. (L6)Solution: Leibnitz's Formula is Dn(uv)= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+-- -------+uDn(v) Let y=x2eax , u= eax, v=x2 D(u)= a eax, D(v)=2x D2(u)=a2eax, D2v=2 D3u=a3eax D3v=0 . . . . . . . . . . . . . . . . . . Dnu=aneax, Dn(uv)= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+-- -------+uDn(v) =aneaxx2+ nc1 an-1eax2x+ nc2 an-2eax2+0 =anx2eax+ nc1 2xan-1eax+ nc2 2 an-2eax.Determine the nth differential coefficient of x3logex ,using Leibnitz’s rule. (L6)Solution: Let y=x3logex u=logx v= x3 Du=1x D(v)=3x2 D2u= -1x2 D2v=6x D3u=2x3 D3v=6 D4u=-6x4 D4v=0 D5u=24x5 D6u=-120x6 Dnu=(-1)n-1n-1!xn, Dn-1u=(-1)n-2n-2!xn-1, Dn-2u=(-1)n-3n-3!xn-2….. Leibnitz’s rule is Dnuv= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+-- -------+uDnv Dn(uv)= Dnuv+ nc1 Dn-1uDv+ nc2 Dn-2uD2v+ nc3Dn-3uD3v+nc4Dn-4uD4v =(-1)n-1n-1!xnx3+nc1 (-1)n-2n-2!xn-13x2+nc2 (-1)n-3n-3!xn-26x+nc3(-1)n-4n-4!xn-36. ................
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