求隐函数y的导数dy/dx y=x^tanxcos(xy)=x-y所确定的隐函数y=y(x)的导数dy/dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/02 14:44:53
![求隐函数y的导数dy/dx y=x^tanxcos(xy)=x-y所确定的隐函数y=y(x)的导数dy/dx](/uploads/image/z/2625551-71-1.jpg?t=%E6%B1%82%E9%9A%90%E5%87%BD%E6%95%B0y%E7%9A%84%E5%AF%BC%E6%95%B0dy%2Fdx+y%3Dx%5Etanxcos%28xy%29%3Dx%EF%BC%8Dy%E6%89%80%E7%A1%AE%E5%AE%9A%E7%9A%84%E9%9A%90%E5%87%BD%E6%95%B0y%3Dy%28x%29%E7%9A%84%E5%AF%BC%E6%95%B0dy%2Fdx)
求隐函数y的导数dy/dx y=x^tanxcos(xy)=x-y所确定的隐函数y=y(x)的导数dy/dx
求隐函数y的导数dy/dx y=x^tanx
cos(xy)=x-y所确定的隐函数y=y(x)的导数dy/dx
求隐函数y的导数dy/dx y=x^tanxcos(xy)=x-y所确定的隐函数y=y(x)的导数dy/dx
1.y=x^tanx
两边取自然对数得
lny=tanxlnx
两边对x求导得
y'/y=sec^2xlnx+tanx/x
y'=(sec^2xlnx+tanx/x)y=(sec^2xlnx+tanx/x)*x^tanx
2.cos(xy)=x-y,隐函数,两边求导
-sin(xy)*(xy)'=1-y'
-sin(xy)*(y+xy')=1-y'
-ysin(xy)-xcos(xy)*y'=1-y'
y'[1-xsin(xy)]=1+ysin(xy)
y'=[1+ysin(xy)]/[1-xsin(xy)]
也可用设二元函数f(x,y)=cos(xy)-x+y
用隐函数求导法:f'x(x,y)+f,y(x,y)*y'=0
f'x(x,y)=-sin(xy)*(xy)'-1
=-ysin(xy)-1
f'y(x,y)=-sin(xy)*(xy)'+1
=-xsin(xy)+1
∴[-ysin(xy)-1]+[-xsin(xy)+1]*y'=0
y'=-[ysin(xy)-1]/[-xsin(xy)+1]
y'=[1+ysin(xy)]/[1-xsin(xy)]
两边对x求道,得
-sin(xy)(y+xy')=1-y'
得,y'=(1+ysin(xy))/(1-xsin(xy))
y=x^tanx
lny=tanx lnx
y'/y=sec^2x lnx+tanx/x
dy/dx=y'=x^tanx (sec^2x lnx+tanx/x)
cos(xy)=x-y
-sin(xy)* (y+y'x)=1-y'
y'=[1+ysin(xy)] / [1-xsin(xy)]