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This question looks so simple, yet it confused me.

If $f(x) = e^{\ln(1/x)}$, then $f'(x) =$ ?

I got $e^{\ln(1/x)} \cdot \ln(1/x) \cdot (-1/x^2)$.

And the correct answer is just the plain $-1/x^2$. But I don't know how I can cancel out the other two function.

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but how do I cancel out the original e and ln ? edit: nvm –  user64698 Mar 2 '13 at 20:45
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They way you did, use chain rule, $\displaystyle \frac{d}{dx} \left( \log \left( \frac 1 x \right)\right) = ??$ –  Santosh Linkha Mar 2 '13 at 20:49
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The answers below are the way to go. As to your solution, your second term of $\ln(\frac1x)$ is incorrect. –  Mike Mar 2 '13 at 20:52
    
Yes. I see the solution now. I just can't select my best answer yet in 2 minutes. Thank to all who helped me. –  user64698 Mar 2 '13 at 20:53
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7 Answers

up vote 6 down vote accepted

Hint: $$\quad e^{\ln(a)}=a\quad $$ Solving the more complicated way also gives the right answer: $$\frac{d}{dx}(e^{\ln(1/x)})=e^{\ln(1/x)}\cdot\frac{d}{dx}(\ln(1/x))=e^{\ln(1/x)}\cdot \left[\frac{1}{1/x}\cdot \frac{d}{dx}(1/x)\right]=\frac{1}{x}\cdot \left[x\cdot \left(-\frac{1}{x^2}\right)\right]$$ where we have used that $$\frac{d}{dt}(e^{f(t)})=e^{f(t)}\left(\frac{d}{dt} f(t)\right)$$ and that $$\frac{d}{dt}(\ln(g(t)))=\frac{1}{g(t)}\left(\frac{d}{dt} g(t)\right)$$

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Hint: the exponential and logarithm are inverse functions: $$e^{\ln u}=u$$ for any $u>0$.

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For $a>0$, $e^{\ln(a)} = a.\;\;$ Similarly, $\;\;\ln(e^a) = a\ln(e) = a\;\;$ because for $$\;f(x) = e^x,\;\;f^{-1}(x) = \ln(x)\quad\text{and}\quad g(x) = \ln x,\;\;g^{-1}(x) = e^x$$

That is, $\ln(x)$ and $e^x$ are each others' inverse function on the set of positive real numbers.

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The other hints provided will help you find the correct solution the easy way.

However, I'd like to point out that your approach has a flaw: $$\frac{d}{dx}\left(e^{f(x)}\right)\ne e^{f(x)}\cdot f(x)$$

Rather, it is: $$\frac{d}{dx}\left(e^{f(x)}\right)= e^{f(x)}\cdot f'(x)$$

Thus, we'd have (this is the hard way, applying the line of thought you tried to use):

$$\begin{align} \frac{d}{dx}\left(e^{\ln 1/x}\right) & = e^{\ln 1/x}\cdot\left(\frac{d}{dx}\left(\ln 1/x\right)\right)\\ & = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\frac{d}{dx}\left(1/x\right)\right)\\ & = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\ & = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\ \end{align}$$ This is the correct answer, we don't simplify anything. Simplifying (using the same property as other have hinted that you should use:

$$\begin{align} e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right) & = \frac{1}{x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\ & = \frac{1}{x}\cdot\left(x\cdot\left(\frac{-1}{x^2}\right)\right)\\ & = \frac{-1}{x^2} \end{align}$$

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$f(x) = e^{\ln(1/x)}=1/x$, then $f'(x) =(1/x)'=\frac{1'\cdot x-x'\cdot 1}{x^2}=\frac{- 1}{x^2}$

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Nice my friend. +1 –  B. S. Mar 6 '13 at 11:44
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As mentioned by others you use that $$ e^{\ln(x)} = x $$ so that $$ f(x) = \frac{1}{x}. $$ Note here though that the domain of $f$ is all positive numbers. The original expression for $f$ is not defined for negative $f$. So also the domain for the derivative will (only) be all positive numbers.

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oh well.. Yup, it's that simple, but I don't see it T_T. Thanks all –  user64698 Mar 2 '13 at 20:46
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Applying the chain rule, one gets $$ \frac{d}{dx} e^{\ln(1/x)} = e^{\ln(1/x)}\cdot \ln'(1/x)\cdot \frac{d}{dx} \frac1x =\cdots\cdots $$ So: $$ \frac{d}{dx} e^{\ln(1/x)} = e^{\ln(1/x)}\cdot \frac{1}{1/x}\cdot \frac{d}{dx} \frac1x =\cdots\cdots $$

You were simply failing to apply the chain rule correctly.

All of the above is of course the hard way. Here's the easy way: $$ \frac{d}{dx} e^{\ln(\text{whatever})} = \frac{d}{dx}(\text{whatever}) = \cdots\cdots $$

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