y=y(x)是由sin(xy)=ln[(x+e)/y]+1确定的隐函数,则y'(0)=?

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y=y(x)是由sin(xy)=ln[(x+e)/y]+1确定的隐函数,则y'(0)=?
y=y(x)是由sin(xy)=ln[(x+e)/y]+1确定的隐函数,则y'(0)=?

y=y(x)是由sin(xy)=ln[(x+e)/y]+1确定的隐函数,则y'(0)=?
sin(xy)=ln[(x+e)/y]+1
两边对x求导,cos(xy)*(y+x*y’)=y/(x+e)*[y-(x+e)*y’]/y^2,
cos(xy)*y+ cos(xy)*x*y’=1/(x+e)*[y-(x+e)*y’]/y,
cos(xy)*y+ cos(xy)*x*y’=1/(x+e)-y’/y,
cos(xy)*x*y’+y’/y =1/(x+e) -cos(xy)*y,
(cos(xy)*x*y+1)y’ =y/(x+e) -cos(xy)*y^2
y’ =[y/(x+e) -cos(xy)*y^2]/ (cos(xy)*x*y+1)
把x=0代入原方程得：ln[(0+e)/y]+1=1-lny+1=2-lny=0,lny=2,y=e^2
把x=0,y=e^2代入导数式中得：
y’(0)= [e^2/(0+e) -cos(0)* e^4]/ (cos(0)*0* e^2+1)
=[e-e^4]/1
=e-e^4

sin(xy)=ln[(x+e)/y]+1, y(0）=1
sin(xy)=ln(x+e)-lny+1 两边求导:
cos(xy)(y+xy')=1/(x+e)-y'/y
y(0）=1代入得：
1=1/e-y'
y'(0)=1/e-1

1、按照二元函数求导法对等式两边分别求导，注意y是x的函数，2、然后移项将含有y'的项移到等号右边其他的移到左边3、将0代入

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