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Math Help - derviatives

  1. #1
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    Post derviatives

    y(x)=(1-t)/(t+1)^3

    y'(x)=(t+1)^3(-1)(1-t) - (1-t)(3)(t+1)^2/(t+1)^6

    y'(x)=(t+1)^3(t-1) - (1-t)(3)(t+1)^2/(t+1)^6
    y'(x)=(t+1)^2[3t^2(t+1)]/(t+1)^6
    y'(x)=3t^2(t+1)/(t+1)^4
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  2. #2
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    Quote Originally Posted by becky
    y(x)=(1-t)/(t+1)^3
    It is not correct.
    When you take the derivative of,
    (1-t) you have,
    -1
    Not,
    (-1)(1-t)
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  3. #3
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    Red face trying again

    y(x)=(1-t)/(t+1)^3

    y'(x) = (t+1)^3 (-1) - (1-t)(3)(t+1)^2/(t+1)^6

    Y'(x) = -3(t+1)^2[(t+1)(1-t)]/(t+1)^6
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  4. #4
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    Quote Originally Posted by becky
    y(x)=(1-t)/(t+1)^3

    y'(x) = (t+1)^3 (-1) - (1-t)(3)(t+1)^2/(t+1)^6
    Correct

    Y'(x) = -3(t+1)^2[(t+1)(1-t)]/(t+1)^6
    Wrong. (Factorization was wrong).
    ---Let Die Meister attemptum,

    \frac{-(t+1)^3-3(t+1)^2(1-t)}{(t+1)^6}
    Thus,
    \frac{-(t+1)^2[(t+1)+3(1-t)]}{(t+1)^6}
    Thus,
    \frac{-[(t+1)+3(1-t)]}{(t+1)^4}
    Thus,
    \frac{-(t+1+3-3t)}{(t+1)^4}
    Thus,
    \frac{2t-4}{(t+1)^4}
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