i understand the 2nd part, but why is the derivative 1/ x log e (e^ log x)?

by chain rule, i get 1/x (e^ log x)...becuase u keep the e^u and multiply by the derivative of u, which in this case is 1/x

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- Nov 17th 2009, 10:42 PM #1

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- Nov 17th 2009, 10:43 PM #2

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- Nov 17th 2009, 10:44 PM #3

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- Nov 17th 2009, 10:48 PM #4
Recall that $\displaystyle \left(\log_a x\right)^{\prime}=\frac{1}{x\ln a}$.

In the case of $\displaystyle \log x$ (which I assume is in base $\displaystyle e$), we see the derivative is $\displaystyle \frac{1}{x\log e}=\frac{1}{x}$. But for the sake of this problem, we're interested in the derivative of the form $\displaystyle \frac{1}{x\log e}=\frac{1}{\log e^x}$.

Does this clarify things?

- Nov 17th 2009, 10:58 PM #5

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oh ok that makes sense..thanks...the ln you threw in there initially confused me... in HS, we used ln alot, but now my college professor claims we were misguided by ln, and that log should be used in its place..

i dont know if that makes sense...anyways i digress,,, thanks for all the help- i got it now

- Nov 17th 2009, 11:22 PM #6

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## why is e^ log x = x

i know the derivative e^ logx = 1

but i need to show that in my work somehow...by chain rule, i got 1/x e^logx

thats the exact same problem i started with, because if i can do something to show that e^logx =x then i get x/x which is 1

also, i got another similar log question that i asked a few mins. ago, that i accidently replied to myself- and thereford im thinking people are disregarding it as answered, so id appreciate it if someone could give that a look if its still unanswered

any help is very appreciated!

- Nov 17th 2009, 11:32 PM #7
Its also good to know that $\displaystyle \log e^x = e^{\log x}$

It turns out that if you differentiate $\displaystyle e^{\log x}$, you get $\displaystyle \frac{1}{x\log e}e^{\log x}$, but that is the same as $\displaystyle \frac{1}{\log e^x}e^{\log x}$. If we can assume the equality I mentioned, then we see that we end up with $\displaystyle \left[e^{\log x}\right]^{\prime}=1$.

Does this make sense?