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

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Answer 1 · 2016-06-30T17:39:35

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

First I'm going to walk you through a nice way of deriving the inverse trig derivatives.

Start with $y = arcsin(x)$ ,this allows us to construct a triangle shown below. enter image source here

Rearranging we get that $x = sin(y)$

Hence, $(dx)/(dy) = cos(y)$

Note that $cos(y) = sqrt(1-x^2)$

We obtain that:

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

Now that we know the general form, we use the chain rule.

For $y(u(x)), (dy)/(dx) = (dy)/(du)*(du)/(dx)$

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

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

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