We use the difference rule to differentiate the entire relation. However, let's differentiate arcsinx and sqrt(1 - x^2) individually.
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y = arcsinx -> siny = x
cosy(dy/dx) = x
dy/dx = 1/cosy
Using the identity cos^2y = sqrt(1 - sin^2y):
dy/dx = 1/sqrt(1 - sin^2y)
Since siny = x, sin^2y = x^2.
dy/dx = 1/sqrt(1 - x^2)
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y = sqrt(1 - x^2)
We let y = sqrtu = u^(1/2) and u = 1 - x^2. Differentiating each, we get y' = 1/(2u^(1/2)) and u = -2x.
By the chain rule:
dy/dx =dy/(du) xx (du)/dx
dy/dx = 1/(2u^(1/2)) xx -2x
dy/dx= -(2x)/(2(1- x^2)^(1/2))
dy/dx= -(2x)/(2sqrt(1 - x^2))
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We can now combine the two derivatives.
Call the initial function f'(x):
f'(x) = 1/sqrt(1 - x^2) - (-(2x)/(2sqrt(1 - x^2)))
f'(x) = 1/sqrt(1 - x^2) + (2x)/(2sqrt(1 - x^2))
f'(x) = (2)/(2sqrt(1 - x^2)) + (2x)/(2sqrt(1 -x^2))
f'(x) = (2 + 2x)/(2sqrt(1 - x^2))
f'(x) = (2(1 + x))/(2sqrt(1 - x^2)
f'(x) = (1 + x)/sqrt(1 - x^2)
You may want to rationalize the denominator, depending on your teacher's wishes.
Hopefully this helps!
