y'?

Question

1.y=arctan(a+x/1-ax); a is constant
2.y=e^x-e^-x/e^x+e^-x

Answer 1

recall that if
y = arctan(u)
y' = 1/(1+u^2) u'

y=arctan(a+x/1-ax)
y = 1/(1+((a+x)/(1-ax))^2) * (1+a^2)/(1-ax)^2
= ...

amazing!

y=(e^x-e^-x)/(e^x+e^-x) = tanh(x)
...

Answer 2

#1 is not quite so amazing if you recognize that if
a = tan(u)
x = tan(v)
y = arctan(tan(u+v)) = u+v
y' = v'
...