# Derivation of the derivative formula for logarithms

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.

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

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

• Jeez how did I miss this. Thanks a lot man. – Arkilo Jun 20 '19 at 18:13

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

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.

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