设可导函数y=y(x)由方程∫(下标0,上标x+y)e^(-x^2)dx=∫(下标0,上标x)xsint^2dt确定,当x=0时,求y'(x).
问题描述:
设可导函数y=y(x)由方程∫(下标0,上标x+y)e^(-x^2)dx=∫(下标0,上标x)xsint^2dt确定,当x=0时,求y'(x).
答
由∫(下标0,上标x+y)e^(-x^2)dx=∫(下标0,上标x)xsint^2dt两边对x求导
[e^-(x+y)²](1+y'(x))=xsinx²+∫(下标0,上标x)sint^2dt将x=0代入其中,[e^-y(0)²](1+y'(0))=0.因为[e^-y(0)²]>0,所以1+y'(0)=0,故y'(0)=-1