calculus

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Please help, my live tutor and I could spent 2 hours trying to solve these two:
find an equation for tangent line y=-1-8^2@ (-4,-129)

find the derivative of the function:----we both kept getting the same answer, but Mathlab kept saying that it was wrong: -4lnl3xl/2+3x
we did the quotient rule, and the chain rule...the derivative and our sign would be wrong...

  • calculus -

    I cannot make sense out of your equation, check your typing.

  • calculus -

    -4 ln l3xl
    2+3x
    -4ln absolute value of 3x over 2+3x

    y= -1-8x^2
    (-4,-129)
    the first set is in fraction form.
    the 3x is absolute.
    the 8x^2 is 8x squared.

  • calculus -

    for the derivative using the quotient rule I got (-8 - 12x + 12x(ln(3x))) / (x(2+3x)^2)


    for the tangent question
    dy/dx = -16x

    at the point the slope is -16(-4) = 64

    so using y = mx + b for the tangent equation
    -129 = 64(-4) + b
    b = 127

    so the tangent equation is y = 64x + 127

  • calculus -

    Thanks. One small error? I can see what we did wrong..we did not do the
    -129= part.....guess we were tired, or tried to do it in our heads.

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