How do you find the derivative of $(arcsin x)^2$ ?

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Answer 1 · 2016-11-14T21:46:47

The derivative is $f'(x) = (2arcsinx)/sqrt(1 - x^2)$ .

Let $y = u^2$ and $u = arcsinx$ . By the chain rule, you have:

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

Let's start by finding the derivative of $u = arcsinx$ .

$u = arcsinx -> sinu = x$

Differentiate implicitly:

$cosu((du)/dx) = 1$

$(du)/dx = 1/cosu$

Rearrange the identity $sin^2x + cos^2x = 1$ for $cosx$ to get $cosx = sqrt(1 - sin^2x)$ .

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

Since $sinu = x$ , we have:

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

Now to $dy/(du)$ . By the product rule, we have, $dy/(du) = 2u$ .

Call the function $y = (arcsinx)^2$ $f(x)$ .

$f'(x) = dy/(du) xx (du)/dx$

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

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

Hopefully this helps!

How do you find the derivative of (arcsin x)^2(arcsin x)^2?