y'?

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

2 answers

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)
...
#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'
...
Similar Questions
    1. answers icon 1 answer
  1. arctan(tan(2pi/3)thanks. arctan(tan(2pi/3) = -pi/3 since arctan and tan are inverse operations, the solution would be 2pi/3 the
    1. answers icon 0 answers
  2. Now we prove Machin's formula using the tangent addition formula:tan(A+B)= tanA+tanB/1-tanAtanB. If A= arctan(120/119) and B=
    1. answers icon 2 answers
  3. Arrange these in order from least to greatest:arctan(-sqrt3), arctan 0, arctan(1/2) So far I got the first two values,
    1. answers icon 2 answers
more similar questions