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Slope of Tangent Line

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A curve is given by the equation: y^3+1004=(e^x+1)^2
    Find the slope of the tangent line at the point (0,-10).

    2. Relevant equations

    3. The attempt at a solution
    I took the derivative of ((e^x+1)^2-1004)^(1/3) and that is (2e^x(1+e^x))/(3((1+e^x)^2-1004)^(2/3)) but plugging in 0 for x does not give me the right answer.
  2. jcsd
  3. May 26, 2010 #2


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    Homework Helper

    I've done it both by implicit differentiation and by your method and the answers both have given me 1/75. You probably just plugged in wrong!
  4. May 26, 2010 #3
    Thanks that is the right answer, just didn't get into the calculator right.
  5. May 26, 2010 #4


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    Homework Helper

    There's no need for a calculator. Since you're dealing with exponentials and x=0, it is clear that the answer will be simple to solve by hand.


    x=0, [tex]\frac{dy}{dx}=\frac{2e^0(e^0+1)}{3\left((e^0+1)^2-1004\right)^{2/3}}[/tex]

    e^0=1 so this simplifies to [tex]\frac{dy}{dx}=\frac{4}{3(4-1004)^{2/3}}[/tex]

    Now [itex](-1000)^{1/3}=-10[/itex] and [itex](-10)^2=100[/itex] so [itex](4-1004)^{2/3}=100[/itex].
    That gets you the answer 1/75 as required, and no need for throwing a messy long expression into the calculator which, as you've seen, can easily lead to errors.
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