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