设y=arctan(a/x)+1/2[ln(x-a)-ln(x+a)],求dy|x=0

y=arctan(a/x)+1/2[ln(x-a)-ln(x+a)],
利用复合函数求导的链锁规则,有
y'=1/(1+(a/x)^2)*(-a/x^2)+1/2[1/(x-a)]-1/(x+a)]
=-a/(x^2+a^2)+a/(x^2-a^2)
=2a^3/(x^4-a^4)
y'|x=0 =-2/a
dy|x=0 =-2/a dx