设函数y=y(x)由方程y^2 f(x)+xf(x)=x^2确定,其中f(x)为可微函数,求dy.若改成y^2f(x)+xf(y)=x^2,仍求dy?
问题描述:
设函数y=y(x)由方程y^2 f(x)+xf(x)=x^2确定,其中f(x)为可微函数,求dy.
若改成y^2f(x)+xf(y)=x^2,仍求dy?
答
两边对x求导得:
2yy'*f(x)+y^2f'(x)+f(x)+xf'(x)=2x
得:y'=[2x-xf'(x)-y^2f'(x)]/(2yf(x)]
dy=[2x-xf'(x)-y^2f'(x)]/(2yf(x)] * dx
若改成y^2f(x)+xf(y)=x^2,
两边对x求导得:2yy'*f(x)+y^2 f'(x)+f(y)+xf'(y)y'=2x
得:y'=[2x-f(y)-y^2f'(x)]/[2yf(x)+xf'(y)]
dy=[2x-f(y)-y^2f'(x)]/[2yf(x)+xf'(y)]*dx