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Thread: trouble taking a derivative

  1. #1
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    trouble taking a derivative

    I need help taking the following derivatives:


    d/dt of (e^-2)*(e^(1+e^x))*(e^t)

    and


    d/dt of (e^-2)*(e^(1+e^t))*(e^x)


    thanks

    Fred1956


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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Fred1956 View Post
    I need help taking the following derivatives:


    d/dt of (e^-2)*(e^(1+e^x))*(e^t)
    If $\displaystyle x \neq f(t)$ let $\displaystyle u = e^{-2}(e^{1+e^x}) $ which is a constant which gives:

    $\displaystyle \frac{d}{dt}(ue^t)$ where u is a constant, should be easy enough.




    d/dt of (e^-2)*(e^(1+e^t))*(e^x)
    $\displaystyle \frac{d}{dt}(e^{-2}e^x)(e^{1+e^t})$

    Let $\displaystyle y = (e^{-2}e^x)(e^{1+e^t})$

    take logs:

    Spoiler:
    $\displaystyle ln(y) = ln[(e^{-2}e^x)(e^{1+e^t})] $

    (Because $\displaystyle ln(abc) = ln(a)+ln(b)+ln(c)$)
    $\displaystyle = ln(e^{-2}) + ln(e^x) + ln(e^{1+e^t})$

    (Because $\displaystyle ln(a^k) = k\,ln(a)$ and $\displaystyle ln(e^k) = k$)
    $\displaystyle = -2 + x + 1+e^t = e^t+x-1$


    $\displaystyle ln(y) = e^t+x-1$

    Differentiate implicitly:

    $\displaystyle \frac{1}{y} \frac{dy}{dt} = e^t$

    $\displaystyle
    \frac{dy}{dt} = ye^t = e^t(e^{-2}e^x)(e^{1+e^t})$
    Last edited by e^(i*pi); Oct 29th 2009 at 12:00 PM. Reason: Changed a minus sign to a plus. Doesn't make a difference when differentiating because the x term is treated as constant.
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