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Thread: Differenntiate equation, help

  1. #1
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    Differenntiate equation, help

    Hello, Can anyone please help me to differentiate $\displaystyle
    e^{(10.234 - 6.984*0.96^{(t+3.3)})}* 1000 * 0.96^t$

    Any help would be appreciated.

    Thanks
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dgamma1 View Post
    Hello, Can anyone please help me to differentiate $\displaystyle
    e^{(10.234 - 6.984*0.96^{(t+3.3)})}* 1000 * 0.96^t$

    Any help would be appreciated.

    Thanks
    Let's make it look a little nicer by bringing all the constants out front.

    $\displaystyle 1000 e^{10.234} [.96^t][e^{(-6.984).96^{t+3.3}} ] $

    Leave the constants for now

    $\displaystyle [.96^t][e^{(-6.984).96^{t+3.3}} ] $

    $\displaystyle [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] $

    Let $\displaystyle y = [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] $

    $\displaystyle lny = ln [.96^t] + ln[e^{(-6.984)(.96^{3.3}).96^{t}} ] $

    $\displaystyle lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e) $

    $\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) - (6.984)(.96^{3.3})[.96^{t}]^{\prime} $

    To find the unknown derivative on the right hand side

    Let $\displaystyle p = .96^t $

    $\displaystyle lnp = t ln .96 $

    $\displaystyle \frac{1}{p} p^{ \prime } = ln .96 $

    $\displaystyle p^{ \prime } = .96^t ln .96 $

    Subbing back in

    $\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] $

    Finally,

    $\displaystyle y^{ \prime} = [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] $

    But remember we have that constant out front, so in total we get


    $\displaystyle y^{ \prime} = 1000 e^{10.234} [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] $
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  3. #3
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    wow thanks for that. although there is a step I don't understand:


    1. going from here:

    $\displaystyle lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e)
    $

    to here: what did you do with the t in front of the Ln, and where did the Ln go from the left side.

    $\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) -
    (6.984)(.96^{3.3})[.96^{t}]^{\prime}
    $

    Thanks for all your help so far!
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dgamma1 View Post
    wow thanks for that. although there is a step I understand:


    . going from here:

    $\displaystyle lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e)$$\displaystyle
    $

    to here: what did you do with the t in front of the Ln, and where did the Ln go from the left side.

    $\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) - $$\displaystyle
    (6.984)(.96^{3.3})[.96^{t}]^{\prime}
    $

    Thanks again for all your help!


    No problem mate.

    This is known as implicit differentiation.

    Let's say we have

    $\displaystyle y = lnt $

    The derivative of $\displaystyle lnt $ with respect to t is $\displaystyle \frac{1}{t} t` = \frac{1}{t} (1) $

    In this case, we have

    $\displaystyle lny $ and the derivative of which (with respect to t) is $\displaystyle \frac{1}{y} y^{ \prime } $

    In this case, however, y depends on t so $\displaystyle y^{ \prime } $ doesn't equate to 1 like in the above example. So we must keep it.

    By the same token,

    $\displaystyle tln(.96) $

    We are taking the derivative of the above with respect to t, which is equal to

    $\displaystyle ln(.96) $

    So basically we're taking the derivative of the left side with respect to t, and of the right side with respect to t. But since the left side is depdent on t and we dont have an equivilent expression for it, we leave it as $\displaystyle y^{ \prime } $
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