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若y=1+Cos x则Dy=?

y=1+cosx y'=(1+cosx)' =1'+(cosx)' =0+(-sinx) =-sinx dy=-sinxdx

y=1+cosx y'=(1+cosx)' =1'+(cosx)' =0+(-sinx) =-sinx dy=-sinxdx

dy=d(cos(x+y)) dy=-sin(x+y)d(x+y) dy+sin(x+y)(dx+dy)=0 (1+sin(x+y))dy=-sin(x+y)dx dy=-sin(x+y)dx/(1+sin(x+y))

y=cos√x+2^x y'=-sin√x/(2√x)+ln2*2^x dy=[-sin√x/(2√x)+ln2*2^x]dx

解:dy=y' =

y'=-2cos(3x-1)sin(3x-1)*3 dy=(-6cos(3x-1)sin(3x-1))dx

你好 y=x^3/2+cos3x dy=(3/2x^1/2-3sin3x)dx dy=(3/2√x-3sin3x)dx 很高兴为您解答,祝你学习进步!有不明白的可以追问! 如果有其他问题请另发或点击向我求助,答题不易,请谅解. 如果您认可我的回答,请点击下面的【采纳为满意回答】按钮...

若y=f(x)由方程ysinx-cos(x y)=0所确定,求dy y'sinx+ycosx+sin(x-y) (1-y') = 0 y'[sinx -sin(x-y)] = -ycosx - sin(x-y) 所以dy=ycosx + sin(x-y)

换元u=y-x,则dy/dx=du/dx+1,原方程化为du/dx=-cosu,分离变量du/cosu=-dx,两边积分ln(secu+tanu)=-lnx+lnC,所以x(secu+tanu)=C,代入u=y-x得通解x(sec(y-x)+tan(y-x))=C

y=u² u=cos2x 所以dy=2udu =2cos2xdcos2x 选D

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