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cal 2

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find the derivative


ln [x^3 +((x+3)^3)((x^2)+4)^7

  • cal 2 - ,

    You left out a ] somewhere

  • cal 2 - ,

    Hmm the logical fix would be
    ln [x^3 +((x+3)^3)]((x^2)+4)^7

    a^3 + b^3 = (a+b)(a^2-ab+b^2), so

    x^3 + (x+3)^3 = (x+(x+3))(x^2 - x(x+3) + (x+3)^2)
    = (2x+3)(x^2 + 3x + 9)

    and the log of the product then becomes

    y = ln(2x+3) + ln(x^2+3x+9) + 7ln(x^2+4)

    y' = 2/(2x+3) + (2x+3)/(x^2+3x+9) + 14x/(x^2+4)

    If my placement of [] is wrong, please feel free to clarify

  • cal 2 - ,

    Find the derivative of the function
    y=ln(2x/x+3)

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