siny=lnx y
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z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3.此时z=3
dy/dx=-x/siny-sinydy=xdx两边取积分cosy=ln|x|+c再问:详细些再答:囧算错了-sinydy=xdxS-sinydy=Sxdxcosy=x^2/2+c再问:要一步一步来再
公式:sin²x+cos²x=1-1=
两边求Ln,得到cosy*ln(sinx)=siny*ln(cosx),化简得y=acrtan(lnsinx/lncosx)公式:y=arctanxy'=1/1+x^2带入上面y'=(1/1+(lns
两边求导:cos(x+y)*(1+y')=cosx+cosy*y'y'=(cosx-cos(x+y))/(cos(x+y)-cosy)e^x+1=e^y*y'+y'y'=(e^x+1)/(e^y+1)
sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2(sinx)^2-1+2(siny)^2)=(si
(cosxsiny)dx+(sinxcosy)dy=0sinydsinx+sinxdsiny=0dsinx/sinx+dsiny/siny=0d(lnsinx)+d(lnsiny)=0d(ln(sin
求导?是求积分吧∫e^x/(e^x+1)dx=∫1/(e^x+1)d(e^x+1)=ln|e^x+1|+C,C为常数∫cosy/sinydy=∫1/sinyd(siny)=ln|siny|+C,C为常
因为SinX+SinY=2/3,两边平方,得:1+SinX*SinY=4/9所以SinX*SinY=-5/9而(SinX-SinY)的平方等于1-SinX*SinY等于14/9所以SinX-SinY=
由sinx+siny=2/3得siny=2/3-sinx代入,2/3+siny-cos²x=1/3-sinx-cos²x=1/3-sinx+sin²x=(sinx-1/2
由已知条件有siny=13−sinx且siny=13−sinx∈[−1,1](结合sinx∈[-1,1])得−23≤sinx≤1,而siny-cos2x=13−sinx-cos2x═sin2x−sin
隐函数的导数求法~
∫e^ysinydy=-∫e^yd(cosy)=-[e^y*cosy-∫cosyd(e^y)]=∫cosy*e^ydy-e^ycosy=∫e^yd(siny)-e^ycosy=e^ysiny-∫sin
1+y'=cosy*y'y'=1/(cosy-1)dy/dx=1/(cosy-1)
两边对x求导得cosx+y'cosy=y+xy'解出来y'就可以了再问:z=f(xy^2,x^2y)求δz/δx,δz/δy这个呢再答:令u=xy^2,v=x^2yδz/δx=f'u*u'x+f'v*
y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx
y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx
siny=1/3-sinx则-1
z=xy+lnxy=xy+lnx+lny所以zy=x+1/y对的.
定义域需满足:xy>0√(1-x^2-y^2)>0即xy>0,且x^2+y^2