求隐函数y的导数dy/dx y=x^tanx

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)]