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Being: $\ln f(x)=\log_ef(x)$

I started derivating $$\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}$$ but I get to a point that I don't know how to follow.

I try to get it by derivating the logarithm directly and by using the logarithmic properties such us: $$\ln f(x)^n=n\ln f(x)$$ and $$\ln \frac{f(x)}{Q(x)}=\ln f(x)-\ln Q(x)$$ but I don't get it.

By the time I have done this: \begin{align} f(x)&=\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}\\ &=\frac{\ln e^{x^2}-\ln (e^x+2)}{2}\\ &=\frac{x^2-\ln (e^x+2)}{2}\\ \text{And derivating:}\\ f'(x)&=\frac{1}{2}\left(\frac{d}{dx}x^2-\frac{d}{dx}\ln (e^x+2)\right)\\ &=x-\frac{1}{2}\left(\frac{e^x}{e^x+2}\right) \end{align}

So I finally get this:


But WolframAlpha says that $$f'(x)=\frac{x(x^2-5)}{x^2-4}$$

If I'm wrong, what do I do wrong? And if I'm right, how can I get from $x-\dfrac{e^x}{2e^x+4}$ to $\dfrac{x(x^2-5)}{x^2-4}$

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Just a very small non-mathematical point: the verb is to differentiate, not to derivate: by differentiating the logarithm, etc. – Brian M. Scott May 9 '12 at 17:51
Incidentally, what was your input to WA? I can't reproduce the error... – David Mitra May 9 '12 at 17:52
Instead of writting $e^x+2$ I wrote $x^2-4$. – Garmen1778 May 9 '12 at 17:53
up vote 5 down vote accepted

Your calculation is perfectly correct, and efficient. The answer that you say Alpha gives is incorrect, and in particular not equivalent to yours. It is possible that Alpha interpreted what you typed as something different from what you intended. If you look carefully at what Alpha thinks it is differentiating, you might find where missing parentheses led Alpha astray.

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Do you think the error might come from putting $e^{x^2}$ in WolframAlpha? – Garmen1778 May 9 '12 at 17:44
@Garmen1778 WA returns the same answer as yours (after simplifying it somewhat). You made a typo somewhere in the input to WA. – David Mitra May 9 '12 at 17:45
@DavidMitra I've seen it. It was my typing fault. – Garmen1778 May 9 '12 at 17:47
@Garmen: next time you ask anything related to Wolfram Alpha, please don't forget to include the exact thing you typed into it... – J. M. May 9 '12 at 17:53

A check, if you trusting your integration skills more than your derivating ones:

Integrate your $f'(x)$, and if you compute your derivative correctly, you will then get your original function, $f(x)$ with some constant. (For more reading: Fundamental theorem of calculus)

$$\int\frac{x(x^2-5)}{x^2-4}\ dx=\int (x-\frac{1}{2(x-2)}+\frac{1}{2(x+2)})\ dx = $$



Now, the question, is there $C\in\mathbb{R}$, such that:


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