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second derivative of -3/x-3
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-3/x - 3? f'(x) = 3/x^2 - 0 = 3/x^2 f''(x) = -6/x^3
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how did you get x^2?
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-3/x = -3(x)^-1 So, to differentiate that: (-3)(-1) * (x)^(-1-1) = 3x^(-2) = 3/x^2
my bad the question was -3/(x-3)
Write those in! -3 / (x-3) -3(x-3)^-1 f'(x) = 3(x-3)^-2 f''(x) = -6(x-3)^-3
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