# Thread: How To Get The Derivative of Y...? (Tougher Ones..)

1. ## How To Get The Derivative of Y...? (Tougher Ones..)

I checked the answers in the back of the book, and we got different ones.

Can anyone help?

#1) y = s*(sqrt(1-s^2) + arccos s
y' = sqrt(1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2)
y' = sqrt(1-s^2)* (2*(sqrt1-s^2)/(2*(sqrt1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) * (2/2)
y' = (2(1-s^2) + s - 2)/(2*(sqrt(1-s^2)
y' = (-2s^2+s)/(2*(sqrt(1-s^2)

#2) y = arccotangent(1/x) - arctangent(x)
y' = -1/(1+(1/x)^2) * (1/x)' - 1/(1+x^2)
y' = -1/(1+(1/x))*(-1/x) - 1/(1+x^2)
y' = 1/(x^2+1) - 1/(1+x^2)
y' = 0

These two are wrong. But I tried.

2. Originally Posted by AlphaRock
I checked the answers in the back of the book, and we got different ones.

Can anyone help?

#1) y = s*(sqrt(1-s^2) + arccos s
y' = sqrt(1-s^2) - s^2/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) <-- problem here
y' = sqrt(1-s^2)* (2*(sqrt1-s^2)/(2*(sqrt1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) * (2/2)
y' = (2(1-s^2) + s - 2)/(2*(sqrt(1-s^2)
y' = (-2s^2+s)/(2*(sqrt(1-s^2)