y=sin(5x π 4)求dy-搜答案-拍题作业网

sin(x^2+y^2)=x,求dy

sin(x^2+y^2)=x两边同时求导,得(x^2+y^2)'cos(x^2+y^2)=dx(2xdx+2ydy)cos(x^2+y^2)=dx2xdx+2ydy=dx/cos(x^2+y^2)2y

y = x^2 sin 4x,求dy,

y'=2xsin4x-x²cos4x·4所以dy=(2xsin4x-4x²cos4x)dxy=ln√4＋t²=1/2ln(4+t²)y'=1/2·1/(4+t&

求通解dy/dx=sin(x-y)

∵dy/dx=sin(x-y)==>dy=sin(x-y)dx==>dx-dy=dx-sin(x-y)dx==>d(x-y)=(1-sin(x-y))dx==>d(x-y)/(1-sin(x-y))=

dx/dt=4(cost)^3*(cost)'dy/dt=4(sint)^3*(sint)'而(cost)'=-sint(sint)'=cost于是dy/dx=(dy/dt)/(dx/dt)=4(co

cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]

x=sin(y/x)+e^2 求dy/dx

x=sin(y/x)+e^2求dy/dxd(x)=d(sin(y/x)+e^2)dx=dsin(y/x)+de^2dx=cos(y/x)d(y/x)dx=cos(y/x)(xdy-ydx)/x^2x^

y=sin(lnx) +1/x,求dy

dy=dsinlnx+d(1/x)=cos(lnx)dlnx+(-1/x²)dx=cos(lnx)*1/xdx-1/x²dx=[xcos(lnx)-1]/x²dx

y=sin(e∧x+1),求dy

y=sin(e∧x+1)dy=cos(e^x+1)d(e^x+1)=e^xcos(e^x+1)dx

y=sin(x^2+1) 求dy

加油啊,不然大学考不起啊!

dy=4sinxdsinx-3xdx=4sinx·cosxdx-3xdx=(4sinxcosx-3x)dx再问：计算不定积分∫乘(1/x1nx)dx再问：帮帮忙呀再问：计算极限lim下面是x→4乘x-

y=x*sin(lnx) 求dy

y=x*sin(lnx)y'=sin(lnx)+x*cos(lnx)*(lnx)'=sin(lnx)+x*cos(lnx)*1/x=sin(lnx)+cos(lnx)dy=[sin(lnx)+cos(

y=sin^2 {x/(x+1)} 求dy

dy=2sin[x(x+1)]cos[x(x+1)](2x+1)

d[ln(y)]=d[e^(6y)sin(4x)]dy/y=6[e^(6y)][sin(4x)]dy+4[e^(6y)][cos4x]dxdy/dx=[e^(6y)][4cos4x]/{1/y-6[e

等式两边求导6x+12x^2y+4x^3dy/dx-10ydy/dx+cosydy/dx=0(4x^3-10y+cosy)dy/dx=-6x-12x^2ydy/dx=(6x+12x^2y)/(10y-

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²

y=sin(x+y) 求dy/dx

y=sin(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dydy/dx=cos(x+y)/(1-cos(x+y))

设y=sin[x/(1=x)]求dy

1=x?

y=(x/lnx)+sin e 求dy

e是自然对数底,约等于2.71828,是sine是常数,其导数是0,故dy=(x/Lnx)'*dx=dx*(lnx-1)/ln^2x注意:求导数和微分是有区别的.

第二步确实应该是d（tanx²）=sec²x²d（x²）

y=sin(x^2),求dy/d(x^3)

dy/d(x^3)=(dy/dx)/(d(x^3)/dx)=cosx/3(x^2)