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This is the question: What is the derivative of this function? $$\frac{2^{x}}{e^{x}}$$

I have seen two answers for this question and I would like to know which one is correct and which one is not.

This is one:

$$\frac{d}{dx} \left( \left( \frac{2}{e} \right)^x \right) = \left( \frac{2}{e} \right)^x \log \left( \frac{2}{e} \right)$$

since the derivative of $(2/e)^x$ is $$\left( \frac{2}{e} \right)^x \log \left( \frac{2}{e} \right).$$

And the other one is where you re-write the function to be $2^{x}e^{-x}$ and use the product rule. Which one is it, I'm confused on how to approach this question.

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    $\begingroup$ Both should arrive at the same answer, without worrying about the form that it is in... $\endgroup$ – abiessu Dec 23 '13 at 19:24
  • $\begingroup$ So the first one is correct? $\endgroup$ – forzamilan Dec 23 '13 at 19:25
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    $\begingroup$ Yes, as long as "log" is the natural logarithm. $\endgroup$ – David Mitra Dec 23 '13 at 19:26
  • $\begingroup$ but the natural logarithm is ,e. $\endgroup$ – forzamilan Dec 23 '13 at 19:29
  • $\begingroup$ The derivative of a function is unique. The technique that you use to arrive at such a derivative may put the answer in a different form than another technique, but they should be equivalent. $\endgroup$ – abiessu Dec 23 '13 at 19:30
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They're both correct. \begin{align*} \frac{d}{dx}\left(\left(\frac{2}{e}\right)^x\right) &= \left(\frac{2}{e}\right)^x\log\frac{2}{e}\\ &= \left(\frac{2}{e}\right)^x(\log 2 - \log e)\\ &= \left(\frac{2}{e}\right)^x\log 2 - \left(\frac{2}{e}\right)^x\\ &= 2^x e^{-x}\log 2 - 2^x e^{-x}\\ &= \color{red}{2^x\log 2}\cdot e^{-x} + 2^x\color{blue}{(-e^{-x})}\\ &= e^{-x}\color{red}{\frac{d}{dx}\left(2^x\right)} + 2^x\color{blue}{\frac{d}{dx}\left(e^{-x}\right)}\\ &= \frac{d}{dx}\left(2^x\cdot e^{-x}\right) \end{align*}

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  • $\begingroup$ $+1=+1$, yup everything checks out. $\endgroup$ – Dylan Yott Dec 31 '13 at 3:03
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$$\frac {d\left(\frac 2e\right)^x}{dx}=\left(\frac 2e\right)^x\ln \frac 2e=\left(\frac 2e\right)^x(\ln 2-1)$$

$$\frac {d\frac {2^x}{e^x}}{dx}=\frac {2^xe^x\ln 2-2^xe^x}{(e^x)^2}=\frac {2^x}{e^x}(\ln 2-1)$$

$$\frac {d(2^xe^{-x})}{dx}=2^xe^{-x}\ln 2+2^xe^{-x}(-1)=\frac {2^x}{e^x}(\ln 2-1)$$

It should be apparent that these are equivalent.

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Write $2^x = e^{x \ln(2)}$. The function becomes $$ e^{(-1+\ln(2))x}$$ The derivative is $$ (-1+\ln(2))e^{(-1+\ln(2))x}$$ You can simplify it by writing it as $$ (-1+\ln(2)) \left(e^{-x} e^{x \ln(2)}\right) = (\ln(2)-1) \left(e^{-x} 2^x\right) =(\ln(2)-1) \left(\frac{2}{e}\right)^x = \ln\left(\frac{2}{e}\right)\left(\frac{2}{e}\right)^x $$

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    $\begingroup$ Minor pedantic comment: $$\ln\left(\frac 2e\right)\left(\frac 2e\right)^x$$ is not necessarily clear as being the same as $$\left(\frac 2e\right)^x\ln\left(\frac 2e\right)$$ $\endgroup$ – abiessu Dec 23 '13 at 20:55
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$$\frac{d}{dx}\left(\left(\frac2e\right)^x\right)=\frac d{dx}\left(e^{x\log\frac2e}\right)=\log\frac2ee^{x\log\frac 2e}=\left(\frac2e\right)^x\log\frac2e$$

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