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    Is it just 6^x?
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    (Original post by mil88)
    Is it just 6^x?
    Think of it in exponential form.
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    (Original post by zetamcfc)
    Think of it in exponential form.
    Do you mean e^x?
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    (Original post by mil88)
    Do you mean e^x?
    Yes so how would you write 6^x in the form e^ax ?
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    (Original post by zetamcfc)
    Yes so how would you write 6^x in the form e^ax ?
    I'm slightly confused.

    Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
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    (Original post by mil88)
    I'm slightly confused.

    Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
    Ok, differentiating 6^x =/= 6^x, only e has this property. Which is why you need to get it into that form, e^ax.
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    (Original post by zetamcfc)
    Ok, differentiating 6^x =/= 6^x, only e has this property. Which is why you need to get it into that form, e^ax.
    Only e has that property? Oh, I never knew that!

    Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

    So would it be e^(ln6)x?
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    (Original post by mil88)
    I'm slightly confused.

    Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
    No. 6^x and e^x are different functions, so they will have different slopes. They are closely related functions though. Note that:

    1) since logs and exponentials are inverse functions, we have e^{\log_e u} = u

    2) if we put u=6^x we have:

    6^x = e^{ \log_e 6^x} = e^{x \log_e 6} = e^{ax} with a=\log_e 6
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    (Original post by mil88)
    Only e has that property? Oh, I never knew that!

    Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

    So would it be e^(ln6)x?

    :borat:
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    Alternatively you could let y=a^x and take natural logs of both sides and differentiate impicitly to show that if y=a^x, dy/dx=(a^x)lna.
    So d/dx(6^x) =(6^x)ln6.
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    (Original post by mil88)
    Only e has that property? Oh, I never knew that!

    Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

    So would it be e^(ln6)x?
    If what you mean is  e^{x\ln{6}} then yes! Now just simplify using the chain rule.
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    (Original post by Louisb19)
    If what you mean is  e^{x\ln{6}} then yes! Now just simplify using the chain rule.
    Yeah thanks I know the differentiating part, I just wasn't sure about the conversion to e form but the other users helped me!
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    (Original post by B_9710)
    Alternatively you could let y=a^x and take natural logs of both sides and differentiate impicitly to show that if y=a^x, dy/dx=(a^x)lna.
    So d/dx(6^x) =(6^x)ln6.
    Thanks
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    (Original post by mil88)
    Yeah thanks I know the differentiating part, I just wasn't sure about the conversion to e form but the other users helped me!
    The idea just incase you were wondering was that

     e^{\ln{f(x)}} = f(x) so for f(x) = 6^x gives  e^{\ln{6^x}} = e^{x\ln{6}}
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    Isn't the answer just 6^xln(6)?

    6^x
    d/dx = 6^xln(6)
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    (Original post by atsruser)
    No. 6^x and e^x are different functions, so they will have different slopes. They are closely related functions though. Note that:

    1) since logs and exponentials are inverse functions, we have e^{\log_e u} = u

    2) if we put u=6^x we have:

    6^x = e^{ \log_e 6^x} = e^{x \log_e 6} = e^{ax} with a=\log_e 6
    Thanks but still dont understand this. What can I do with the info that a = log_e 6? where do I go from there in terms of differentiating
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    (Original post by sunburner)
    Thanks but still dont understand this. What can I do with the info that a = log_e 6? where do I go from there in terms of differentiating
    From there, it's just the chain rule. If y=e^{ax} and u=ax then \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \cdots
 
 
 
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