`f(x) = arctan(sqrt(x))` Find the derivative of the function

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)` then
`u^ 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))`