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Math Help - simple question - derivative - quotient rule

  1. #1
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    simple question - derivative - quotient rule

    I'm having problems using the quotient rule and am hoping that someone can help.

    g(t) = A / 1+ BE^-t

    Using the quotient rule the numerator simplifies to

    0 - A(-BE^-t) = ABE^-t

    I don't see how (1 + BE^-t)(A') simplifies to 0,

    and how -A(1 + BE^-t)' simplifies to -A(-BE^-t).

    In the latter if e is raised to the -t shouldn't that come down using the power rule and then it's e ^t-1?

    Will someone please help?

    Ted
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  2. #2
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    Quote Originally Posted by tedmetro View Post
    I'm having problems using the quotient rule and am hoping that someone can help.

    g(t) = A / 1+ BE^-t

    Using the quotient rule the numerator simplifies to

    0 - A(-BE^-t) = ABE^-t

    I don't see how (1 + BE^-t)(A') simplifies to 0,

    and how -A(1 + BE^-t)' simplifies to -A(-BE^-t).

    In the latter if e is raised to the -t shouldn't that come down using the power rule and then it's e ^t-1?

    Will someone please help?

    Ted
    review your rules for derivatives ...

    the derivative of a constant is 0 ... A is a constant.

    the derivative of e^{-t} is -e^{-t} ... the power rule does not apply to exponential functions of this type.

    g(t) = \frac{A}{1+Be^{-t}}

    where A and B are constants.

    you do not need the quotient rule to find g'(t) ... the chain rule will suffice.

    rewrite as

    g(t) = A(1+Be^{-t})^{-1}

    g'(t) = -A(1+Be^{-t})^{-2} \cdot (-Be^{-t})

    g'(t) = \frac{ABe^{-t}}{(1+Be^{-t})^2}

    g'(t) = \frac{A}{1+Be^{-t}} \cdot \frac{Be^{-t}}{1+Be^{-t}}

    g'(t) = g(t) \cdot \frac{1}{1 + Ce^t} where C = \frac{1}{B}<br />
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    Thank you for the quick, and very helpful, reply!!
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