How do you find the derivative of $arcsin [x^{\frac{1}{2}}]$ $arcsin[x^(1/2)]$ ?

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Answer 1 · 2016-01-26T19:50:23

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

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:

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

How do you find the derivative of arcsin[x^(1/2)]arcsin[x12]arcsin[x^(1/2)]?