Question

不会啊函数题。

题1，已知 （1+tanhx)/(1-tanhx)=e的2x次方，推论 tanhy的反函数= 1/2 log [ (1+y)/(1-y)].
题二，已知 tanhx= sinhx/coshx 证明 tanh(x+y)= (tanhx+tanhy)/(1+tanhxtanhy)

Answer 1

（1+tanhx)/(1-tanhx)=e^(2x)
取对数得log[（1+tanhx)/(1-tanhx)]=2x,
所以tanhx=y的反函数为x= (1/2) log [ (1+y)/(1-y)].
(2)tanhx= sinhx/coshx
所以tanh(x+y)
=sinh(x+y)/cosh(x+y)
=[e^(x+y)-e^(-x-y)]/[e^(x+y)+e^(-x-y)]
(tanhx+tanhy)/(1+tanhxtanhy)
={[(e^x-e^(-x)]/[e^x+e^(-x)]+[e^y-e^(-y)]/[e^y+e^(-y)]}/{1+[(e^x-e^(-x)]/[e^x+e^(-x)][e^y-e^(-y)]/[e^y+e^(-y)]}
={[(e^x-e^(-x)][e^y+e^(-y)]+[e^y-e^(-y)][e^x+e^(-x)]}/{[(e^x+e^(-x)][e^y+e^(-y)]+[e^y-e^(-y)][e^x-e^(-x)]}
=2[e^(x+y)-e^(-x-y)]/{2[e^(x+y)+e^(-x-y)]}
=[e^(x+y)-e^(-x-y)]/[e^(x+y)+e^(-x-y)],
所以等式成立。