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The formula under discussion is this:

$$ \frac{d}{dx} [ \log_b x] = \frac {1}{x \ln b}$$

The author does not derive the formula completely, instead he states that:

$$\frac {d}{dx} [\log_b x ] = \frac {d}{dx} \left[\frac{\ln x}{\ln b}\right] \qquad\qquad (1)$$

I understand it up to this. You let $\log_b x$ equal to $y$ and then $y$ can be rewritten as $b^y$ and it equals $x.$ After that simply taking the natural log on both sides will arrange it into the form expressed in (1).

However after this the author states:

$$ \frac {d}{dx} \left[\frac{\ln x}{\ln b}\right] = \frac{1}{\ln b} \frac{d}{dx} [\ln x]$$

This is the step that I don't understand. I would suppose that the quotient rule would be applied but that just makes it all very weird and I can't get the L.H.S.

I would appreciate if anyone can explain what is exactly happening here.

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In these steps, $b$ is a constant. It is the base of the logarithm. You can take a constant out when you differentiating, and so you get the last step, where

$$\frac{d}{dx}\frac{\ln{x}}{\ln{b}}=\frac{1}{\ln{b}}\frac{d}{dx}\ln{x}$$

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  • $\begingroup$ Jeez how did I miss this. Thanks a lot man. $\endgroup$ – Arkilo Jun 20 '19 at 18:13
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He uses the rule that $$\log_{b}{x}=\frac{\ln(x)}{\ln(b)}$$ for $$b>0$$ and $$b\ne 1$$ and $b$ is constant, so we can write $$\left(\frac{\ln(x)}{\ln(b)}\right)'=\frac{1}{\ln(b)}(\ln(x))'=\frac{1}{x\ln(b)}$$'

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This is simply an application of $\frac{d}{dx}(kf(x)) = k\frac{d}{dx}(f(x))$, where $k$ is a constant. Here $\frac{1}{\ln b}$ is a constant.

This is a basic "rule" in differentiation. You can, of course, show it with either product or quotient rule, but there's no need to.

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  • $\begingroup$ Yeah I weirdly know all the rules but I wasn't paying attention to the fact that b is a constant. $\endgroup$ – Arkilo Jun 20 '19 at 18:27

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