Inverse trig function derivative

Question

Find the derivative of the function.

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

Answer 1

In this case u = √(1-x)/(1+x) and consequently requires you to use the chain rule so you must multiply by u':

f '(u) = 1/(1 + u²) • u'

f ' = 1/(1 + [√(1-x)/(1+x)]² • u'

f ' = 1/(1 + [(1-x)/(1+x)] • u'

f ' = 1/[(1+x)/(1+x) + (1-x)/(1+x)] • u'

f ' = 1/[(1 + x + 1 - x)/(1+x)] • u'

f ' = 1/[2/(1+x)] • u'

f ' = (1 + x)/2 • u'

To find u', let's make it easier by using a substitution. let w = (1 - x)/(1 + x) so

u = w^1/2

Using the power rule and chain rule:

u' = 1/2w^-1/2 • w'

u' = 1/(2w^1/2) • w' then back-substituting:

u' = 1/(2[(1 - x)/(1 + x)]^1/2) • w'

u' = 1/[2(1 - x)^1/2/(1 + x)^1/2] • w'

u' = (1 + x)^1/2/(2(1 - x)^1/2) • w'

To find w', use the quotient rule:

(a/b)' = (a'b - ab')/b²

a = 1 - x meaning a' = -1

b = 1 + x meaning b' = 1

b² = (1 + x)²

So w' = [(-1)(1 + x) - (1 - x)(1)]/(1 + x)²

w' = (-1 - x - 1 + x)/(1 + x)²

w' = -2/(1 + x)²

Then plugging that into our u' = (1 + x)^1/2/(2(1 - x)^1/2) • w' we get:

u' = (1 + x)^1/2/(2(1 - x)^1/2) • -2/(1 + x)²

u' = -(1 + x)^1/2/(1 - x)^1/2 • 1/(1 + x)²

u' = -1/[(1 - x)^1/2 • (1 + x)^3/2]

And plugging this into our equation f ' = (1 + x)/2 • u' we get:

f '(x) = (1 + x)/2 • -1/[(1 - x)^1/2 • (1 + x)^3/2]

f'(x) = -1/[2(1 - x)^1/2 • (1 + x)^1/2]