UNIT-IV
DIFFERENTIATION
BASIC CONCEPTS OF DIFFERTIATION
Consider 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∆x
This fraction is also called a difference quotient.
Differentiation using standard formulae
· Define 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-ex
· Find 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=1xlog7e
Chain Rule (or) Differential coefficient of a function of function
If u=f(x), y=f(u) then dydx=dydu. dudx
· Find 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+y
· Differentiate 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=2cotx
· Find 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 sec2x
· Find 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+123
· Differentiate 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 tan3x4
· Differentiate 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-sinx
Solution:
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-sin2x
· Find 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 rule
Let u v be differentiable functions of x. Then the product of a function
Y = u(x).v(x) is differentiable.
d(uv)=udv+vdu
· If 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+ logex
· Differentiate 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 Differentiation
Let u & v be differentiable functions of x, then uv is also differentiable
d(uv) = vu,-uv,v2
· Find 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)2
· Find 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)2
· Find 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)2
· Find y’ , if y= logxsinx.
Solution:
By quotient rule
dydx=sinxddxlogx-(logx)ddxsinxsinx2
dydx = y’ = sinx1x-logx(cosx)(sinx)2
= sinx-x cosx logxx sin2x
· Find 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)xe2x
· Find 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)2
· Find y’ , if y= sinx+cosxsinx-cosx .
Solution:
By quotient rule,
dydx=sinx-cosxddxsinx+cosx-(sinx+cosx)ddxsinx-cosxsinx-cosx2
dydx =y’= sinx-cosxcosx-sinx-sinx+cosx(cosx+sinx)(sinx-cosx)2 =-(sinx-cosx)2-(sinx+cosx)2(sinx-cosx)2
· Differentiate 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)2
· Differentiate 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)2
· Differentiate 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+12
· Differentiate secxlogx.
Solution:
Let y=secxlogx
dydx=logxsecx tanx-secx.1xlogx2 =x logx secx tanx-secxxlogx2 =secx[xtanx logx-1]xlogx2
· Find the derivative of ex+e-xex-e-x .
Solution:
Let y= ex+e-xex-e-x
dydx=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-x2
· Differentiate 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 sint
Differentiation 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=2t2-1t32
· If 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 =-cotθ2
· Find 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=-ba
Logarithmic 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= sinxx
Taking log on both sides ,we get,
log y= logsinxx
log y= xlogsinx
· Differentiate y=xx .
Solution:
Let y= xx
Taking log on both sides ,we get,
log y= logxx
log y= x logx
Differentiating both sides w.r.to x,
1ydydx=x1x+ logx 12x
=1x1+logx2=xx1x1+logx2
· Differentiate 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-3
· Find 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,=yx
· Differentiate (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 coty
· If 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 logx
Differentiation 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 function
· Find 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=-ba2xy
· Differentiate 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+2y
Maxima (or) Minima
Stationary 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 zero
· Find the roots of dydx = 0. Let it be a1,a2…..an. These points x=a1,x=a2…. x=anare called turning points
· Find d2ydx2
· Find( d2ydx2)at x=a1, ( d2ydx2)at x=a2,…… ( d2ydx2)at x=an
· If ( d2ydx2)at x=a1= _ ve, then we have a max value at x=a1
= +ve, then we have a min value at x=a1
· If ( 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 product
· State 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
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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.
183