Question: It is known that f'(x) = 2 + (square root of 4 - x^2) for -2 < x < 2. Show that f'(x) is equivalent

It is known that f'(x) = 2 + (square root of 4 - x^2) for -2 < x < 2. Show that f'(x) is equivalent to r = 4sin(theta) on [-2,2]

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