(1)y=(10^x-1)/(10^x+1)
y10^x+y=10^x-1
(y-1)10^x=-1-y
x=log10[(1+y)/(1-y)]
=1/ln10[ln(1+y)-ln(1-y)]
∴反函数的解析式:f(x)=1/ln10[ln(1+x)-ln(1-x)]
(2) h(x)=1/x-1/ln10[ln(1+x)-ln(1-x)]
h'(x)=-1/x^2-1/ln10[1/(1+x)+1/ln(1-x)]
(1)y=(10^x-1)/(10^x+1)
y10^x+y=10^x-1
(y-1)10^x=-1-y
x=log10[(1+y)/(1-y)]
=1/ln10[ln(1+y)-ln(1-y)]
∴反函数的解析式:f(x)=1/ln10[ln(1+x)-ln(1-x)]
(2) h(x)=1/x-1/ln10[ln(1+x)-ln(1-x)]
h'(x)=-1/x^2-1/ln10[1/(1+x)+1/ln(1-x)]