Differentiate: 2arcsin(y/2)+√(4-y^2)
Set f(y) = 2sin−1(y2)+4−y22 sin^{-1}(\frac {y}{2}) + \sqrt{4-y^2}2sin−1(2y)+4−y2
Since sin−1(y)=11−y2sin^{-1}(y) = \frac{1}{\sqrt{1-y^2}}sin−1(y)=1−y21
We have that f′(y)=2(12)(11−(y2)2)+12(−2y)(14−y2)=24−y2−y4−y2f'(y) = 2(\frac {1}{2})(\frac {1}{\sqrt {1- (\frac {y}{2})^2}}) + \frac {1}{2}(-2y)(\frac {1}{\sqrt {4- y^2}}) = \frac {2}{\sqrt {4- y^2}} - \frac{y}{\sqrt {4- y^2}}f′(y)=2(21)(1−(2y)21)+21(−2y)(4−y21)=4−y22−4−y2y
f′(y)=2−y4−y2=2−y2+yf'(y) = \frac {2-y}{\sqrt {4- y^2}} = \frac {\sqrt {2-y}}{\sqrt {2+y}}f′(y)=4−y22−y=2+y2−y
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