y=sin^2(2x),求dy-搜答案-搜狗做题网

y=2^cosx+sin√x,复合函数求导数y'=(2^cosx)ln2(﹣sinx)+cos√x*1/(2√x)dy={(2^cosx)ln2(﹣sinx)+cos√x*1/(2√x)}dx

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&

y=sinx/(1+x^2)先求导：y'=cosx*(1+x^2)-sinx*(2x)/(1+x^2)^2那么,dy=y'dx=[cosx*(1+x^2)-sinx*(2x)]dx/(1+x^2)^2

sin(x^2+y^2)+e^x-xy^2=0左右微分得到cos(x^2+y^2)*(2xdx+2ydy)+(e^x)dx-(y^2)dx-2xydy=0余下的求出dy就可以了

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^

dy/dx相当于对x进行求导：dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于：sinx=cosx,sin(x^2)=2x*cos(x^2)

你只要知道arcsinx和sinx的原函数不就能求解了吗再问：还是不懂能写写过程么再答：你都求到这一步∫dy/(2sin(y/2))=-∫sin(x/2)dx了，答案不就在下一步么再问：等式左边积分不

y=sin(x^2+1) 求dy

加油啊,不然大学考不起啊!

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(

两边同时对x微分得dcos(x^2+y)=dx,即-sin(x^2+y)(2dx+dy)=dx,将dy移过去,变形得到-（1+2sin(x^2+y)）dx/sin(x^2+y)=dy

y=sin^2 {x/(x+1)} 求dy

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

等式两边对x求导：cos(xy)*(y+x*y')-(2x*2y+x^2*2*y'=0解出y'即为所求

dy=(sin(^2)x+2x)'dx=(2sinxcosx+2)dx=(sin2x+2)dx再问：步骤的话，就是一步么？再答：嗯..就是求个导因为dy=f'(x)dx

y=sin(x^2),dy/dx=?

公式

dy=cos(2x+e^2x)*(2+2e^2x)dx

y=x^2x,求dy

y=x^(2x)lny=2xlnx(1/y)dy=(2+2lnx)dxdy=x^(2x).(2+2lnx)dx

y=sin(x^2),求dy/d(x^3)

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

令k=x+yy=k-xdy/dx=d(k-x)/dx=dk/dx-1原方程变为dk/dx-1=-sin^2kdk/dx=1-sin^2k=cos^2k(两边同乘sec^2kdx)sec^2kdk=dx