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))))#