We have d/dxarcsin^2x
According to the chain rule, (df)/dx=(df)/(du)*(du)/dx, where u is a function within f.
Here, f=u^2 where u=arcsinx
So we have d/(du)u^2*d/dxarcsinx
2u*d/dxarcsin x
To find the derivative of arcsinx, take arcsinx=sin^-1x. We have:
y=sin^-1x, and so:
siny=x
Taking the derivative of both sides, we have:
(cosy)*y'=1
y'=1/cosy
We need to write cosy in terms of siny, in order to change the variable to x.
Remember that sin^2y+cos^2y=1. So:
cos^2y=1-sin^2y
cosy=sqrt(1-sin^2y)
We can take the positive square root, as sin^-1x is normally given on the interval -pi/2,pi/2 and as the cosine function is positive in that interval.
Inputting that into our earlier function:
y'=1/sqrt(1-sin^2y)
y'=1/sqrt(1-x^2)
So as d/dxarcsinx=1/sqrt(1-x^2), we can write the derivative of arcsin^2x as:
2u*1/sqrt(1-x^2)
But remember that u=arcsinx, so we have:
(2arcsinx)/sqrt(1-x^2)
Done.
