Question

求由方程∫(y到0)e^tdt+∫(x^2到x)1/tdt=0所确定的隐函数的二阶导数,

Answer 1

两边同时求导即可得

(e^y)y'+(1/x^2)(x^2)'-(1/x)(x)'=0

y'e^y+1/x=0

y'=-e^(-y)/x

y''=e^(-y) y'/x+e^(-y)/x²=e^(-y)[-e^(-y)/x]/x+e^(-y)/x²=e^(-y)[1-e^(-y)]/x²