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!
