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    I can't do this question and I understand how I would go about doing this question, but am I missing something that I didn't learn when saying I don't know how I would differentiate 4^x in terms of x?
    Thanks for any help
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    (Original post by mnaj123)
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    I can't do this question and I understand how I would go about doing this question, but am I missing something that I didn't learn when saying I don't know how I would differentiate 4^x in terms of x?
    Thanks for any help
    \displaystyle 4^x = e^{x \ln 4}.

    Now use the fact that \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left(e^{ax}\right) = ae^{ax}.
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    (Original post by Zacken)
    \displaystyle 4^x = e^{x \ln 4}.

    Now use the fact that \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left(e^{ax}\right) = ae^{ax}.
    So 4^x differentiates to (x/4)e^(xln4)?
    Sorry, I'm really confused
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    (Original post by mnaj123)
    So 4^x differentiates to (x/4)e^(xln4)?
    Sorry, I'm really confused
    No. 4^x = e^{x \ln 4}, so 4^x differentiates to \ln 4 e^{x \ln 4} = \ln 4 \cdot 4^x

    Do you know how to differentiate e^{2x}?
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    (Original post by Zacken)
    No. 4^x = e^{x \ln 4}, so 4^x differentiates to \ln 4 e^{x \ln 4} = \ln 4 \cdot 4^x

    Do you know how to differentiate e^{2x}?
    Yeah, just 2e^2x, right?
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    (Original post by mnaj123)
    Yeah, just 2e^2x, right?
    Yeah, exactly, so how do you differentiate e^{x \ln 4}? Remember that \ln 4 is just another number, like 2 is?
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    (Original post by Zacken)
    Yeah, exactly, so how do you differentiate e^{x \ln 4}? Remember that \ln 4 is just another number, like 2 is?
    Rightttt, yeah, I always misinterpret ln values as not being constants.
    Thanks.
    So it would be:
    ln4.e^xln4
    therefore, using your first response,
    = ln4.4^x
    Is that right?

    Last question, sorry, would I leave it as ln4.e^xln4 in the case of this question to make it easier or in the form ln4.4^x?
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    (Original post by mnaj123)
    Rightttt, yeah, I always misinterpret ln values as not being constants.
    Thanks.
    So it would be:
    ln4.e^xln4
    therefore, using your first response,
    = ln4.4^x
    Is that right?
    First class work. Correct.

    In general, a^x = e^{x \ln a} for any a.

    Last question, sorry, would I leave it as ln4.e^xln4 in the case of this question to make it easier or in the form ln4.4^x?
    I would use \ln 4 \cdot 4^x. :yes:
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    (Original post by Zacken)
    First class work. Correct.

    In general, a^x = e^{x \ln a} for any a.



    I would use \ln 4 \cdot 4^x. :yes:
    Thanks so much for your help haha
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    (Original post by mnaj123)
    Thanks so much for your help haha
    My pleasure. :yep:
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     \displaystyle \frac{d}{dx} \left ( a^x \right ) =(\ln a) a^x

     \displaystyle \int a^x dx = \frac{a^x}{\ln a} + c .
 
 
 
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