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From "some proof that uses the definition" of a derivative, I understand that:

$$\ (\log_a x)' = \frac{1}{x \ln a} $$

Since this is the case, I should also be able to prove that:

$$\ \left(\frac{\ln a}{\ln a}\right)' = \frac{1}{x \ln a} $$

So I did some stuff with $\ \left(\frac{\ln a}{\ln a}\right)'$ until I managed to get to:

$$\ \frac{1 - \frac{x (\ln a)' \ln x}{\ln a}}{x \ln a} $$

Which would mean, if I'm not mistaken, that:

$$\ \frac{x (\ln a)' \ln x}{\ln a} = 0 $$

Can someone explain how this is possible to me, or if I made a mistake of some kind?

jayphur
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1 Answers1

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I overlooked that $\ (\ln a)′ = 0 $, that would, of course make $\ \frac{x (\ln a)' \ln x}{\ln a} = 0 $.

jayphur
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