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y=Cos(x+y)的导数怎么求?

本题考查复合函数的求导法则 给题主重新算一遍 y = cos ( x+ y) y' = [ cos ( x + y )]' * ( x + y)' 链式法则,先求外面的函数的导数,再求里面函数的导数.y' = -sin ( x + y ) * ( 1 + y') 函数求导法则,cos ( x+y)的导数是-sin(x+y),后面括号里 面x的导数是1

对x求偏导,得-sin(x+y)对y求得,-sin(x+y)

cosxy=x-sinxy*(y+xy')=1 y+xy'=-cscxy y'=-(cscxy+y)/x.y=cos(x+y) y'=-sin(x+y)*(1+y') y'[1+sin(x+y)]=-sin(x+y) y'=-sin(x+y)/[1+sin(x+y)].

y=cos(x+y)两边同时对x求导得y'=-sin(x+y)(1+y') (*)得y'=-sin(x+y) /[1+sin(x+y)] (#)(*)式两边同时对x求导得y''=-sin(x+y)y'' -cos(x+y)(1+y')(1+y')即y''=-cos(x+y)(1+y') / [1+sin(x+y)]将(#)代入上式得y''=-cos(x+y) / [1+sin(x+y)]^3

y=cosx/xxy=cosx,两边求导数得到:y+xy'=-sinx所以:y'=(-sinx-y)/x=(-sinx-cosx/x)/x=-(xsinx+cosx)/x^2.

y'=[cos(x+y)]'=-sin(x+y)*(1+y')=-sin(x+y)+-sin(x+y)*y'把含y'的部分移到等式的右边,所以:y'=-sin(x+y)/1+sin(x+y)

y'=-sin(x+y)*(x+y)'y'=-sin(x+y)*(1+y')y'/(1+y')=-sin(x+y)y'=-sin(x+y)/(1+sin(x+y))

先对cos求导=-sin x^2再对x^2求导=2x所以y'=-2x*cos x^2

两边对x求导,有cosy-xsiny*y'=cos(x+y)(1+y')cosy-cos(x+y)=y'*[cos(x+y)+xsiny]y'=[cosy-cos(x+y)]/[cos(x+y)+xsiny]希望对你有所帮助如有问题,可以追问.谢谢采纳

y'=-(1+y')sin(x+y)

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