siny=ln(x y)

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 03:19:05
设siny-e^x+xy^2=0,求dy/dx

siny-e^x+xy^2=0cosy.y'-e^x+2xy.y'+y^2=0(cosy+2xy)y'=e^x-y^2y'=(e^x-y^2)/(cosy+2xy)

已知方程xy-eˆ2x=siny 确定隐函数y=y(x),求dy/dx

xy-eˆ(2x)=siny两边对x求导,得y+x(dy/dx)-2eˆ(2x)=(cosy)*(dy/dx)(x-cosy)*(dy/dx)=2eˆ(2x)-ydy/d

y/x=ln(xy) 求详 dy/dx

方法一(微分法)d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即(xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²

求导 e^x/(e^x +1)dx cosy /siny dy=ln siny

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

y/x=ln(xy) 求dy/dx

两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)

求方程所确定的隐函数的导数dy/dx:siny=ln(x+y)

两边微分cosydy=(dx+dy)/(x+y)[cosy(x+y)-1]dy=dxdy/dx=1/[cosy(x+y)-1]

z= xy ln(xy) 求全微分dz

dz=d(xyln(xy))=xyd(ln(xy))+ln(xy)d(xy)=xyd(xy)/(xy)+ln(xy)d(xy)=d(xy)+ln(xy)d(xy)=(1+ln(xy))d(xy)=(1

求方程xy''=y'ln(y'/x)的通解

设Y=y'降阶:Y'=(Y/x)ln(Y/x)这就是一个一阶齐次方程.设Y/x=u,所以Y=ux,Y'=u+x(du/dx),代回原方程,解得:lnu=C1x+1Y=xe^(C1x+1)所以y=[(C

满足cosx*cosy-sinx*siny=1/2的角xy有可能哪些

前面的那一串式子意思就是cos(x+y),这个是课本上的余弦公式哟.余弦等于1/2的角你根据余弦图表示一下,那么只要x+y与其相等就行啦.

设siny+e的x次方-xy²=0,求dy/dx

dsiny+de^x-dxy²=0cosydy+e^xdx-y²dx-2xydy=0cosydy-2xydy=y²dx-e^xdxdy/dx=(y²-e^x)/

怎么证明 lnx +lny =ln xy

e^(lnx+lny)=e^lnx*e^lny=x*ye^lnxy=xy所以e^(lnx+lny)=e^lnxy所以lnx+lny=lnxy

sinx+ siny=xy求dy/dx

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*

siny+e^x-xy^2=0,求dy/dx

siny+e^x=xy^2,两边求微分,cosydy+e^xdx=d(xy^2)cosydy+e^xdx=y^2dx+2xydy整理,得(e^x-y^2)dx=(2xy-cosy)dydy/dx=(e

求隐函数siny+e的x次方-xy的2次方=0的导数

隐函数求导,就是先左右一起求微分,加个d,然后写出多少dx+多少dy=0,移项变成dy/dx=多少的形式就好了

求隐函数y的二阶导数d^2y/dx^2 siny=ln(x+y)

两边关于x求导,注意y是x的函数y'cosy=[1/(x+y)]*(1+y').①解得y'=1/(x+y)÷[cosy-1/(x+y)].②对①两边关于x求导可得y''cosy-(y')²s

求隐函数的偏导数siny+e^x-xy^2=0,求dy/dx

解两边求导y‘cosy+e^x-y^2-2xyy'=0即y’(cosy-2xy)=y^2-e^xy'=(y^2-e^x)/(cosy-2xy)或者F(x,y)=siny+e^x-xy^2=0Fx=e^

u=ln(xy+z)求du=

u=ln(xy+z)du=d[ln(xy+z)]/dx*dx+d[ln(xy+z)]/dy*dy+d[ln(xy+z)]/dz*dz=y/(xy+z)*dx+x/(xy+z)*dy+1/(xy+z)*

设e^xy-xy^2=Siny,求dy/dx

你好!两边对x求导:e^(xy)*(y+xy')-y^2=y'cosy解得y'=(y^2-ye^(xy))/(xe^(xy)-cosy)

设siny+ex(是e的x次方)-xy=0.求dy/dx

dy/dx=(y^2-e^x)/(cosy-2xy)

e^y+ln(xy)-e^(-x)=0,求y'

两边求导得y'·e^y+(y+xy')/(xy)+e^(-x)=0