Finding the derivative of arctan

[Bin222]

y=arctan sqrt[(1+x)/(1-x)]


what i did:
let u= sqrt[(1+x)/(1-x)]
du=1/2 [(1-x)/(1+x)]^1/2 [{(1+x)(-1)-(1-x)}/(1+x)^2]dx

I solved this problem many times, but I don't get the right answer....

 

[Steven G]

y=arctan sqrt[(1+x)/(1-x)]


what i did:
let u= sqrt[(1+x)/(1-x)]
du=1/2 [(1-x)/(1+x)]^1/2 [{(1+x)(-1)-(1-x)}/(1+x)^2]dx

I solved this problem many times, but I don't get the right answer....

I really do not know where to start with your result.
1) In computing du, you multiplied by the power, 1/2, which was good but then you did not subtract 1 from the power. Even if you do not know what you should get when you compute 1/2 - 1 you should know that you would not get back 1/2 (since 1/2 - 0 = 1/2).
2) you seem to be using the quotient rule but you did not use the correct formula. What is the correct formula?
3) You tried to take the derivative of sqrt[(1+x)/(1-x)] even though you were not asked to. You were asked to find the derivative of arctan sqrt[(1+x)/(1-x)]. How does the arctan come into play?
4) Why are you finding du which is not the derivative of anything. du/dx, dy/dx, dx/dt, dr/ds are derivatives. You were giving y which in terms of x, so you want to find dy/dx.

Hint: y=arctan sqrt[(1+x)/(1-x)] becomes tan (y) = sqrt[(1+x)/(1-x). Now take the derivative of both sides with respect to x.

 

[stapel]

y=arctan sqrt[(1+x)/(1-x)]

what i did:
let u= sqrt[(1+x)/(1-x)]
du=1/2 [(1-x)/(1+x)]^1/2 [{(1+x)(-1)-(1-x)}/(1+x)^2]dx

I solved this problem many times, but I don't get the right answer....

What happened to the derivative of the arctangent? Aren't you supposed to start (the Chain-Rule process) with that? What is the "right" answer? What did you get instead? Please reply showing ALL of your work. Thank you!