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What would be the further steps for the case like this: I am finding the first derivative of a function: $f(x) = \ln(1+x^2)$ So the procedure would then be:

  • $f'(x) = \frac{2x}{1+x^2}$.
  • $f'(x) = 0 \implies \frac{2x}{1+x^2}$ , where $1+x^2 \neq 0$.
  • Next step would be to show points at which $f'(x) \neq 0$, so I would have $1+x^2 = 0$ which results in $x^2 = -1$, which does not have any solutions, since the root is always positive.

Am I doing something wrong?

Graph shows clearly that at $y = 0$ at $x = 0$, but how to show it in calculations?

enter image description here

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I do not understand what you are aiming at. Are you trying to locate the extrem value(s) of your original function $f$? Then just search for $x$ such that the derivative vanishes. Only for those you need to check whether the denominator is zero. If not please tell us what the question is. – user20266 Dec 10 '11 at 9:30
$\frac{2x}{1+x^2}=0$ $\Leftrightarrow$ $2x=0$ $\Leftrightarrow$ $x=0$; Is this what you mean in your question. (Note that for a fraction $\frac ab$ to be zero you want $a=0$, not $b=0$, so there is no reason for solving $1+x^2=0$ in relation to this.) – Martin Sleziak Dec 10 '11 at 9:38
I wanted to see when denominator equals zero so that I could avoid(ignore) these "zero" points. My mistake – Mocialov Boris Dec 10 '11 at 9:47
You shouldn't write $2x/1+x^2$ if you mean $2x/(1+x^2)$. Those are two different things. – Michael Hardy Dec 10 '11 at 16:59
@MichaelHardy my bad, cleaned it – Mocialov Boris Dec 10 '11 at 17:02
up vote 2 down vote accepted

Are you trying to solve $f'(x)=0$? A fraction is zero if and only if the numerator is 0 and the denominator is non-zero. Here, since $1+x^2$ is never zero, the equation $${2x\over 1+x^2}=0$$is equivalent to $$2x=0.$$ So, $x=0$ is the (only) solution to $f'(x)=0$. For any other point $x$, $f'(x)\ne 0$.

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I wanted to see when denominator equals zero so that I could avoid(ignore) these points, where functions denominator equals zero, but now I see, that I should do it for numerator only, thank you – Mocialov Boris Dec 10 '11 at 9:46

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