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

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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- Jan 23rd 2009, 03:47 PMlittlejodologarithmic differentiation trouble
find y' if $\displaystyle y = 2^{e^{x}} + (2e)^{x}$

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! - Jan 23rd 2009, 03:56 PMMush

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

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

$\displaystyle \frac{d}{dx} e^{Cx} = Ce^{Cx} $ - Jan 23rd 2009, 04:55 PMlittlejodo
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.

$\displaystyle (2^{e})^x$

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

Thank you! - Jan 23rd 2009, 04:58 PMMush
- Jan 23rd 2009, 05:04 PMlittlejodo
Okay, so the final answer I came to is:

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

Is this correct? Again, I really appreciate your help. - Jan 23rd 2009, 05:16 PMMush
Actually I just realised the first term is a bit more difficult than that:

It might be better to treat each term separately!

$\displaystyle y = 2^{e^{x}} + (2^e)^x $

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

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

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

Take the first term.

$\displaystyle g(x) = 2^{e^x} $

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

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

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

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

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

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

$\displaystyle h(x) = (2^e)^x $

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

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

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

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

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

Hence $\displaystyle \frac{dy}{dx} = g'(x) + h'(x) = 2^{e^x}e^x\ln(2)+ e\ln(2) (2^e)^x$