What is the derivative of $arcsin[x^(1/2)]$ ?

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Answer 1 · 2015-06-06T04:04:13

To find the derivative we will need to use the Chain Rule

$dy/dx=dy/(du)*(du)/(dx)$

We want to find

$d/(dx)(arcsin(x^(1/2)))$

Following the chain rule we let $u=x^(1/2)$

Deriving u we get

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

Now we substitute u in place of x in the original equation and derive to find $dy/(du)$

$y=arcsin(u)$

$(dy)/(du)=1/(sqrt(1-u^2)$

Now we substitute these derived values into the chain rule to
find $dy/(dx)$

$dy/dx=dy/(du)*(du)/(dx)$

$dy/dx=1/(sqrt(1-u^2))*1/(2sqrt(x))$

Substitute x back into the equation to get the derivative in terms of x only and simplify

$u=x^(1/2)$

$dy/dx=1/(sqrt(1-(x^(1/2))^2))*1/(2sqrt(x))$

$dy/(dx)=1/(sqrt(1-x))*1/(2sqrt(x))$

$dy/(dx)=1/(2sqrt(x)*sqrt(1-x))$

$dy/(dx)=1/(2sqrt(x-x^2))$

What is the derivative of arcsin[x^(1/2)]arcsin[x^(1/2)]?