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Find the derivative of the following function f(x)=(x-3/2x+3)^2

  • Calculus -

    Assuming the usual carelessness with parentheses, I gather you mean

    f(x) = ((x-3)/(2x+3))^2

    If f = u^2, f' = 2u u'

    So, f' = 2(x-3)/(2x+3) * ((x-3)/(2x+3))'
    = 2(x-3)/(2x+3) * [(2x+3)-2(x-3)/(2x+3)^2]
    = 2(x-3)/(2x+3)*(9/(2x+3)^2)
    = 18(x-3)/(2x+3)^3

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