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Math Help - Composite & Quotient Rule -confirm correct answer?

  1. #1
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    Composite & Quotient Rule -confirm correct answer?

    Hi,

    I would be really grateful if someone would help me!

    With regard to the function f(x) = ln(e^x + e^-x)

    By applying the composite rule does this turn into:

    1 / e^x+ e^-x ?

    And by applying the quotient rule, does the first equation turn into:

    e^x + e^ -x / e^x - e^ -x

    Many thanks!!!
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  2. #2
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    You don't need to use the quoient rule in this case, consider

    \frac{d}{dx}(ln(f(x)) = \frac{f'(x)}{f(x)}

    In your case f(x) = e^x + e^{-x} and

    f'(x) = e^x - e^{-x} therefore

    \frac{d}{dx}(ln(e^x + e^{-x})) = \frac{e^x - e^{-x}}{e^x + e^{-x}}
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  3. #3
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    Quote Originally Posted by looking0glass View Post
    Hi,

    I would be really grateful if someone would help me!

    With regard to the function f(x) = ln(e^x + e^-x)

    By applying the composite rule does this turn into:

    1 / e^x+ e^-x ?

    And by applying the quotient rule, does the first equation turn into:

    e^x + e^ -x / e^x - e^ -x

    Many thanks!!!
    Please don't just say "turn into"- I would interpret that as meaning a different form of the same equation or function!

    The composite rule (also called "chain rule") tells us that the derivative of f(x)= [tex]ln(e^x+ e^{-x})[/itex] is \frac{1}{e^x+ e^{-x}} times the derivative of e^x+ e^{-x} which is e^x- e^{-x}.

    So the derivative of f(x)= ln(e^x+ e^{-x}) is f'(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}, the reciprocal of what you gave.

    That derivative is also, by the way, tanh(x).
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