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    I did this in the exam in January, but can't seem to do it now :confused:

    I = 16 - 16 (0.5^t)

    find dI/dt when t=3.

    I've totally forgotten how to differentiate this. Can anyone help?
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    For a > 0 you can write a^x = (e^{\ln a})^x = e^{x\ln a}, which you can differentiate using the chain rule.
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    (Original post by chaz1992)
    I did this in the exam in January, but can't seem to do it now :confused:

    I = 16 - 16 (0.5^t)

    find dI/dt when t=3.

    I've totally forgotten how to differentiate this. Can anyone help?
    Note that a^t = e^{\ln a^t} = e^{t\ln a}. Then differentiate like you would any exponential.
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    (Original post by nuodai)
    For a > 0 you can write a^x = (e^{\ln a})^x = e^{x\ln a}, which you can differentiate using the chain rule.
    That's a really nice way of doing it, never of thought of that.
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    Thanks for the help guys. Thumbs up given.
 
 
 
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