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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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    @MichaelHardy my bad, cleaned it2011-12-10

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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 you2011-12-10