Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 44

The given function f(x) = arctan(sqrt(x)) is in a inverse trigonometric form.
The basic derivative formula for inverse tangent is:
d/(dx) arctan(u) = ((du)/(dx))/sqrt(1-u^2) .
Using u-substitution, let u = sqrt(x) thenu^ 2 = (sqrt(x))^2 = x
For the derivative of u or (du)/(dx) , we apply the Power Rule:
d/dx(x^n)=n*x^(n-1) * d(x)
This is possible since sqrt(x) = x^(1/2)
Then d/(dx) x^(1/2) = 1/2 * x^(1/2-1)* 1 Note: d/(dx) (x) = 1
=1/2 x^(-1/2)
Applying the Law of Exponents: x^(-n)=1/x^n .
(du)/(dx) =1/2 * x^(-1/2)
= 1/2 * 1/ x^(1/2)
=1/(2 x^(1/2) ) or1/(2 sqrt(x))

We now have: u=sqrt(x) and (du)/(dx) =1/(2sqrt(x)) .
Applying the formula d/(dx) arctan(u)= ((du)/(dx))/(1+u^2) :
f'(x) = d/(dx)arctan(sqrt(x))=(1/(2sqrt(x)))/(1+(sqrt(x))^2)
f'(x) =(1/(2sqrt(x)))/(1+x)
f'(x) =1/(2sqrt(x))* 1/(1+x)
f'(x) =1/(2sqrt(x)(1+x))