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Question Number 106449 by mohammad17 last updated on 05/Aug/20

Answered by Dwaipayan Shikari last updated on 05/Aug/20

y(√x)−x(√y)=9  y(1/(2(√x)))+(√x).(dy/dx)−(x/(2(√y))).(dy/dx)−(√y)=0  (dy/dx)((√x)−(x/(2(√y))))=(√y)−(y/(2(√x)))  (dy/dx)=((2(√(xy))−y)/(2(√x))).((2(√y))/(2(√(xy))−x))  (dy/dx)=(√(y/x)) (((2(√(xy))−y)/(2(√(xy))−x)))

yxxy=9y12x+x.dydxx2y.dydxy=0dydx(xx2y)=yy2xdydx=2xyy2x.2y2xyxdydx=yx(2xyy2xyx)

Commented by mohammad17 last updated on 05/Aug/20

sir the question want the secend dervitive

sirthequestionwantthesecenddervitive

Answered by Dwaipayan Shikari last updated on 05/Aug/20

x=v^2 ,1=2v.(dv/dx)  y=p^2 ,(dy/dx)=2p.(dp/dx)  y(√x)−x(√y)=9  (d^2 /dx^2 )(p^2 v−v^2 p)=0  (d/dx)(v.2p.(dp/dx)+p^2 .(dv/dx)−v^2 .(dp/dx)−p.2v.(dv/dx))=0  (d/dx)(v.(dy/dx)+(p^2 −v^2 )(dp/dx)−p)=0  v.(d^2 y/dx^2 )+(dy/dx).(dv/dx)+(2p(dp/dx)−2v(dv/dx)).(dp/dx)+(d^2 p/dx^2 )−(dp/dx)=0  (√x).(d^2 y/dx^2 )+(dy/dx)((dv/dx)−2v((dv/dx))^2 ).(1/(2p)).(dy/dx)+(d^2 p/dx^2 )−(dy/dx).(1/(2p))=0  (√x).(d^2 y/dx^2 )+m^2 ((1/(2v))−(1/(2v))).(1/(2p))+(d/dx)((1/(2p)).(dy/dx))−(m/(2p))=0  ((dy/dx)=m)    (√x)(d^2 y/dx^2 )+(d^2 y/dx^2 ).(1/(2p))+(dy/dx).((1/(−2p^2 ))).(dp/dx)−(m/(2p))=0  (d^2 y/dx^2 )((√x)+(1/(2p)))−(1/(2p^2 .2p))((dy/dx))^2 −(m/(2p))=0  (d^2 y/dx^2 )((√x)+(1/(2(√y))))=m((m/(4(y)^(3/2) ))−(1/(2(√y))))  (d^2 y/dx^2 )=m((m/(4(y)^(3/2) ))−(1/(2(√y)))).(((2(√y))/(2(√(xy))+1)))     (m=(√(y/x)) (((2(√(xy))−y)/(2(√(xy))−x)))

x=v2,1=2v.dvdxy=p2,dydx=2p.dpdxyxxy=9d2dx2(p2vv2p)=0ddx(v.2p.dpdx+p2.dvdxv2.dpdxp.2v.dvdx)=0ddx(v.dydx+(p2v2)dpdxp)=0v.d2ydx2+dydx.dvdx+(2pdpdx2vdvdx).dpdx+d2pdx2dpdx=0x.d2ydx2+dydx(dvdx2v(dvdx)2).12p.dydx+d2pdx2dydx.12p=0x.d2ydx2+m2(12v12v).12p+ddx(12p.dydx)m2p=0(dydx=m)xd2ydx2+d2ydx2.12p+dydx.(12p2).dpdxm2p=0d2ydx2(x+12p)12p2.2p(dydx)2m2p=0d2ydx2(x+12y)=m(m4(y)3212y)d2ydx2=m(m4(y)3212y).(2y2xy+1)(m=yx(2xyy2xyx)

Answered by mathmax by abdo last updated on 05/Aug/20

in this case is better to use bospital rule let  u(x) =1−cos(4x)sin^2 (x)−cos^2 (x) and v(x)=x^4  ⇒  u(x) =1−cos(4x)(((1−cos(2x))/2))−((1+cos(2x))/2)  =(1/2)−(1/2)cos(4x)+(1/2)cos(4x)cos(2x)−(1/2)cos(2x)  =(1/2)−(1/2)cos(2x)−(1/2)cos(4x)+(1/4)(cos(6x)+cos(2x))  =(1/2)−(1/4)cos(2x)−(1/2)cos(4x)+(1/4)cos(6x) ⇒  u^′ (x) =(1/2)sin(2x)+2sin(4x)−(3/2)sin(6x)  u^((2)) (x) =cos(2x)+8cos(4x)−9cos(6x)  u^((3)) (x) =−2sin(2x)−32sin(4x) +54 sin(6x)  u^((4)) (x) =−4cos(2x)−128 cos(4x)+6×54cos(6x)  ⇒lim_(x→0) u^((4)) (x) =−4−128 +6×54 =54×6−132  =324−132 =192  v^′ (x) =4x^3  ⇒v^((2)) (x) =12x^2  ⇒v^((3)) (x) =24 x ⇒v^((4)) (x) =24  ⇒lim_(x→0)   ((u(x))/(v(x))) =((192)/(24))

inthiscaseisbettertousebospitalruleletu(x)=1cos(4x)sin2(x)cos2(x)andv(x)=x4u(x)=1cos(4x)(1cos(2x)2)1+cos(2x)2=1212cos(4x)+12cos(4x)cos(2x)12cos(2x)=1212cos(2x)12cos(4x)+14(cos(6x)+cos(2x))=1214cos(2x)12cos(4x)+14cos(6x)u(x)=12sin(2x)+2sin(4x)32sin(6x)u(2)(x)=cos(2x)+8cos(4x)9cos(6x)u(3)(x)=2sin(2x)32sin(4x)+54sin(6x)u(4)(x)=4cos(2x)128cos(4x)+6×54cos(6x)limx0u(4)(x)=4128+6×54=54×6132=324132=192v(x)=4x3v(2)(x)=12x2v(3)(x)=24xv(4)(x)=24limx0u(x)v(x)=19224

Commented by mathmax by abdo last updated on 05/Aug/20

⇒lim_(x→0)   ((1−cos(4x)sin^2 x−cos^2 x)/x^4 )=8

limx01cos(4x)sin2xcos2xx4=8

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