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

Answer 1

Evelyn Agard

The answer is #=(2arcsinx)/(sqrt(1-x^2))#

We need

#(sqrtx)'=1/(2sqrtx)#

#(sinx)'=cosx#

#sin^2x+cos^2x=1#

Let, #y=(arcsinx)^2#

#sqrty=arcsinx#

#sin(sqrty)=x#

#(sin(sqrty))'=x'#

#1/(2sqrty)*cos (sqrty)dy/dx=1#

#dy/dx=2sqrty/cos(sqrty)#

#sin^2(sqrty)+cos^2(sqrty)=1#

#Cos^2(sqrty)=1-x^2#

#cos(sqrty)=sqrt(1-x^2)#

Therefore,

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

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Answer 2

Christopher Allers

The derivative of (arcsin x)^2 is:

2 * (arcsin x) * (1 / sqrt(1 - x^2)) * (1 / sqrt(1 - x^2)) * (1 / sqrt(1 - x^2))

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.