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Im reading the Ian Stewart book Concepts of Modern Mathematics, in a page he talks about the function $\ f(x) = log(log(sin(x)))$ and its derivate, $\ f'(x) = \frac{ cot(x)}{log(sin(x))}$. Now, you can´t sketch the graphic of $\ f$ because for all values of $x$, $f(x)$ doesn´t exist. However, $f'$ does make sense for all $x$ such that $sin(x)>0$. If the derivative of a function is the slope of the tangent of every point, but the function $f$ hasn´t points, where is the mistake? if the domain of any function $g$ is the empty set, the function automatically hasn´t derivative?

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This plot may help show how it has an imaginary and real part for all values of $x$: wolframalpha.com/input/… –  apnorton Jan 8 '13 at 3:41
ok, i understand the imaginary part, of course, but, why the plot has a "real part"? for all values of x, i can´t calcule f(x), why Wolfram shows me a graphic of f in blue? –  dwarandae Jan 8 '13 at 3:46
It's still complex for the whole graph. Complex numbers are written in the form $a + bi$. $a$ is the real part, and $b$ is the imaginary part. When the plot says it has a real part, that means $a$ is non-zero. However, $f(x)$ is still complex (that is, $b$ is also non-zero). –  apnorton Jan 8 '13 at 3:59

2 Answers 2

up vote 1 down vote accepted

There's no mistake: the function $\,f'(x)\,$ exists on its own, without any regard of the fantasy called $\,f(x)\,$ which, as you note, doesn't exist...as a real function.

In fact, the complex function $\,f(x)\,\,,\,x\in\Bbb C\,$ exists pretty nicely, but since it seems like you haven't yet studied complex analysis I'm stopping here.

Of course, you can also add to the argument of the first (the outter) logarithm an absolute value...

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$$ f(x) = \log | \log( \sin(x))|$$ works for $0 < x < \frac{\pi}{2}$

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