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Okay, I have find the derivative of $$\frac{x}{\ln x}$$ so I used the quotient rule and got: $$\frac{\ln x(1)-x(\frac{1}{x})}{(\ln x)^2}$$

I then simplified this to: $$\frac{-x}{\ln x}.$$

Now here come the problem, when I put $e$ into $f'(x)$ I am supposed to get $-1.5271$ and I do not and I cannot figure out why. Could someone point out what I am doing wrong?

Thanks

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    $\begingroup$ Be careful. You cannot simplify as you have. The log terms do not cancel. It is not true that $\frac{a-b}{a^2}=\frac{-b}{a}$. Try it with a few numbers to see why. $\endgroup$ – Jared May 21 '13 at 18:46
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    $\begingroup$ Isn't your numerator $ \ \ln x - 1 \ $ , making your derivative $ \frac{(\ln x) - 1}{(\ln x)^2} \ $? $\endgroup$ – colormegone May 21 '13 at 18:46
  • $\begingroup$ Be careful: is $ln x^2$ the same as $[\ln(x)]^2$ or $ln(x^2)$? $\endgroup$ – leonbloy May 21 '13 at 18:48
  • $\begingroup$ I see my mistake straight away and I feel pretty stupid, thanks everyone for helping. $\endgroup$ – user May 21 '13 at 18:51
  • $\begingroup$ Just a note, "derive" is not the same as "take the derivative," at least in most English usage of which I am aware. $\endgroup$ – Thomas Andrews May 21 '13 at 18:58
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It's $$ {\ln(x) - 1\over (\ln(x))^2}.$$ The $\ln(x)\cdot 1$ parses to $\ln(x)$.

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  • $\begingroup$ But if I put $e$ into that I get zero right? $ln(e) -1 =0$ and zero divided by anything equals zero. $\endgroup$ – user May 21 '13 at 18:54
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You incorrectly cancelled during your last step. It is not true that $$\frac{ab+c}{a^2}=\frac{b+c}{a}$$

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The answer you were given is wrong.

$f'(e) = 0$.

Plot it and see.

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