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Math Help - logarithmic differentiation trouble

  1. #1
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    logarithmic differentiation trouble

    find y' if y = 2^{e^{x}} + (2e)^{x}

    I'm not sure at all what to do here. Can anyone tell me what definition/property is key to use? I would really appreciate it!
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  2. #2
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    Quote Originally Posted by littlejodo View Post
    find y' if y = 2^{e^{x}} + (2e)^{x}

    I'm not sure at all what to do here. Can anyone tell me what definition/property is key to use? I would really appreciate it!

     (2e)^x = e^{\ln(2e)^x} = e^{x\ln(2e)}

    Now both the term has been converted into the form  e^{Cx} , where C is just some constant. These can be easily differentiated using the rule that:

     \frac{d}{dx} e^{Cx} = Ce^{Cx}
    Last edited by Mush; January 23rd 2009 at 06:00 PM.
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  3. #3
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    I just realized I made a mistake in the function. The second term was supposed to read 2raised to the e, all raised to x.

    (2^{e})^x

    How would that change things? I was thinking that it might end up looking like the other term?

    Thank you!
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  4. #4
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    Quote Originally Posted by littlejodo View Post
    I just realized I made a mistake in the function. The second term was supposed to read 2raised to the e, all raised to x.

    (2^{e})^x

    How would that change things? I was thinking that it might end up looking like the other term?

    Thank you!
    It doesn't really change things:

    (2^{e})^x = e^{\ln((2^{e})^x)}

    (2^{e})^x = e^{x\ln((2^{e})}

    Same rules apply.
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  5. #5
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    Okay, so the final answer I came to is:

    y' = 2(\ln{2^{e}}e^{x\ln{2^{e}}})

    Is this correct? Again, I really appreciate your help.
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  6. #6
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    Actually I just realised the first term is a bit more difficult than that:

    It might be better to treat each term separately!

     y = 2^{e^{x}} + (2^e)^x

    Let  2^{e^x} = g(x) and  (2^e)^x = h(x)

     \frac{dy}{dx} = \frac{d}{dx}2^{e^{x}} + \frac{d}{dx}(2^e)^x

     \frac{dy}{dx} = g'(x)+ h'(x)

    Take the first term.

     g(x) = 2^{e^x}

     \ln(g(x)) = \ln(2^{e^x})

     \ln(g(x)) = e^x\ln(2)

     \frac{d}{dx} \ln(g(x)) = \frac{d}{dx}e^x\ln(2)

     \frac{1}{g(x)} \times g'(x) = e^x\ln(2)

     g'(x) = g(x)e^x\ln(2)

     g'(x) = 2^{e^x} e^x \ln(2)


     h(x) = (2^e)^x

     h(x) = e^{\ln((2^e)^x)}

     h(x) = e^{x\ln((2^e))}

     h'(x) = \ln(2^e)e^{x\ln((2^e))}

     h'(x) = e\ln(2)e^{\ln((2^e))^x}

     h'(x) = e\ln(2) (2^e)^x

    Hence \frac{dy}{dx} = g'(x) + h'(x) = 2^{e^x}e^x\ln(2)+  e\ln(2) (2^e)^x
    Last edited by Mush; January 23rd 2009 at 06:29 PM.
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