设函数y=f(x)由方程x-∫(1,x+y)e^-t^2dt=0确定,求曲线y=y(x)在x=0处的切线方程.
问题描述：
设函数y=f(x)由方程x-∫(1,x+y)e^-t^2dt=0确定,求曲线y=y(x)在x=0处的切线方程.
答
当x = 0,∫(1~y) e^-t² dt = 0 => y = 1
x - ∫(1~x+y) e^-t² dt = 0
e^[-(x+y)²] * (1 + y') = 1
1 + y' = e^(x+y)²
y' = e^(x+y)² - 1
y'|_x=0,y=1) = e^(0+1)² - 1 = e - 1
切线方程为y - 1 = (e - 1)(x - 0)
即(e - 1)x - y + 1 = 0