设Y=Y(x)由参数方程则dy
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d(y^2)/dx=d(y^2)/dy*dy/dx=2y*dy/dx这个复合函数求导法则正如ovtr0001仁兄所说那样,你可以翻翻课本这个……还要详细点呀?你有书么?你看书那里不懂可以提出来,我可能
左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx
lny+x/y=0等式两边求导:y'*1/y+1/y+x*y'(-1/y²)=0(1/y-x/y²)y'=-1/y∴y'=(-1/y)/(1/y-x/y²)=-y/(y-
dx=-2tdtdy=(1-3t²)dt所以dy/dx=(3t²-1)/2t
dy/dx=(dy/dt)/(dx/dt)=[2t/(1+t^2)]/[1-1/(1+t^2)]=2/t
两边对x求导2x+2y*dy/dx=0dy/dx=-x/y有不明白的追问再问:刚学不太明白,2x+2y*dy/dx=0里的dy/dx哪来的,是y'吗?再答:是的复合函数求导注意这里y是x的函数不妨换个
由隐函数微分法可得:-sin(x+y)(1+y′)+y′=0-sin(x+y)+[1-sin(x+y)]y′=0∴y′=sin(x+y)/[1-sin(x+y)].
两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)
对两边求导:[-sin(x+y)](1+dy/dx)+dy/dx=0-sin(x+y)-[sin(x+y)]dy/dx+dy/dx=0dy/dx=[sin(x+y)]/[1-sin(x+y)]
为你提供精确解答e^y+xy=e两边对x求导知:(e^y)(dy/dx)+y+x(dy/dx)=0解出:dy/dx=-y/(e^y+x)
dy/dt=cost-cost+tsint=tsintdx/dt=-sintdy/dx=(dy/dt)/(dx/dt)=-t再问:为什么-tcost会分解成-cost+tsint~~~+_+知道了==
dy/dt=e^t+te^t=(1+t)e^tdx/dt=e^tdy/dx=(dy/dt)/(dx/dt)=1+t=2
xy+y^2-2x=0y+xy'+2yy'-2=0(x+2y)y'=2-yy'=(2-y)/(x+2y)dy/dx=(2-y)/(x+2y)
分别对y求导,求左边为1+【e^(x+y)×(dx/dy+1)】右边为2×dx/dy推的dx/dy:自己算下,没得草稿纸.
两端对x求导数(把y看作x的函数),则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]
本题将方程的两边对x求导数左右为dy/dx右边为0+e^y+x*e^y*dy/dx提取dy/dx得:dy/dx=e^y/(1-xe^y)整理得:dy/dx=e^y/(2-y)由此,可以确定x和y的函数
dy=lnt+1dx=1-sintdy/dx=(lnt+1)/(1-sint)
ln(x+y)=x·lny(1+y‘)/(x+y)=lny+x/y·y‘y+y·y‘=y(x+y)lny+x(x+y)·y‘y‘=【y(x+x)lny-y】/【y-x(x+y)】再问:лл����
dx/dy=(dx/dt)*(dt/dy)dx/dt=2tdt/dy=1所以dx/dy=2tdy/dt=1/2t
化为:e^(ylnx)-e^y=sin(xy)两边对x求导:e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[