How do you find the derivative of #arctan sqrt [ (1-x)/(1+x)]#?

Question

6743 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-09T10:06:33

# \ y' = - 1/(2 sqrt [(1-x^2)] #

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

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

#tan^2 y = (1-x)/(1+x)#

#tan^2 y = 2/(1+x) - 1#

differentiating each side

#2 tan y sec^2 y\ y' = -2/(1+x)^2#

# tan y (tan^2 y + 1)\ y' = -1/(1+x)^2#

# sqrt [ (1-x)/(1+x)] ( (1-x)/(1+x) + 1)\ y' = -1/(1+x)^2#

# sqrt [ (1-x)/(1+x)] ( (2)/(1+x))\ y' = -1/(1+x)^2#

# \ y' = - 1/(1+x)^2 * sqrt [ (1+x)/(1-x)] * (1+x)/2#

# \ y' = - 1/(2 sqrt(1+x) * sqrt [(1-x)] #

# \ y' = - 1/(2 sqrt (1-x^2) #