# logarithmic differentiation trouble

• Jan 23rd 2009, 04:47 PM
littlejodo
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!
• Jan 23rd 2009, 04:56 PM
Mush
Quote:

Originally Posted by littlejodo
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:

$\frac{d}{dx} e^{Cx} = Ce^{Cx}$
• Jan 23rd 2009, 05:55 PM
littlejodo
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!
• Jan 23rd 2009, 05:58 PM
Mush
Quote:

Originally Posted by littlejodo
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.
• Jan 23rd 2009, 06:04 PM
littlejodo
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.
• Jan 23rd 2009, 06:16 PM
Mush
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 $\frac{dy}{dx} = g'(x) + h'(x) = 2^{e^x}e^x\ln(2)+ e\ln(2) (2^e)^x$