Question

How do you find the derivative of `arcsin[x^(1/2)]`?

Answer 1

`f'(x) = (1)/(2sqrt(x(1-x))`

Explanation:

Using the chain rule, we have that the derivative of `f(u)` when `u` is a function of x `-u(x)-` is:

`f(u) = arcsin(u)`

`f'(x) = (df)/(du)(du)/(dx) = (1)/(sqrt(1-u^2))(du)/(dx)`

In this case, we have:

`u(x) = x^(1/2)`

So, its derivative is:

`(du)/(dx) = (1)/(2)x^((-1)/(2)) = (1)/(2x^(1/2)) = (1)/(2sqrt(x))`

And the final expression:

`f'(x) = (1)/(sqrt(1-x))(1)/(2sqrt(x)) = (1)/(2sqrt(x(1-x))`