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

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Answer 1 · 2016-12-27T07:14:25

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))#