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Math Help - Differenntiate equation, help

  1. #1
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    Differenntiate equation, help

    Hello, Can anyone please help me to differentiate  <br />
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  <br />
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.

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

    Leave the constants for now

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

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

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

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

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

     \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  p = .96^t

     lnp = t ln .96

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

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

    Subbing back in

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

    Finally,

     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


     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:

    lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e)<br />

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

    \frac{1}{y} y^{ \prime} = ln(.96) - <br />
(6.984)(.96^{3.3})[.96^{t}]^{\prime}<br />

    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:

    lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e) <br />

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

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

     y = lnt

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

    In this case, we have

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

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

    By the same token,

     tln(.96)

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

     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  y^{ \prime }
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