# Derivative of $g(x) = \frac{\log x}{\log a}$

I am currently reading Spivak's Calculus. I have an older version. In chapter 17 the author presents a definition of the logarithm function. There he presents the derivative of $$g(x) = \frac{\log x}{\log a}$$ as

$$g'(x) = \frac{1}{x \log a}$$

But by the quotient rule, if I am correct, the derivative should be

$$g'(x) = \frac{\frac{1}{x} \cdot \log a - \log x \cdot \frac{1}{a}}{(\log a)^2}$$

So if both equations are correct, we need $$\log x = 0$$. But I don't know, why this is so. Can anyone explain it to me or point me to my mistake? Thanks in advance.

• Presumably, $a$ is constant May 25, 2022 at 18:16
• Quotient rule is corrected. Sorry for the misnomer. May 25, 2022 at 18:37

## 3 Answers

The obtained result is wrong because $$(\log(a))'$$ is $$0$$, since $$a$$ is a constant.

Having said that, you can still apply the proposed method, which yields the desired result: \begin{align*} g(x) = \frac{\log(x)}{\log(a)} & \Rightarrow g'(x) = \frac{\frac{1}{x}\times\log(a) - \log(x)\times 0}{\log^{2}(a)} = \frac{1}{x\log(a)} \end{align*}

and we are done

Hopefully this helps!

In this context, $$a$$ is a constant, and so is $$\log a$$. Therefore$$g(x)=\frac{\log x}{\log a}=\frac1{\log a}\times\log x\implies g'(x)=\frac1{\log a}\times\frac1x=\frac1{x\log a}.$$

No, the derivative is with respect to $$x$$, thus $$a$$ and also $$\log a$$ are constants and hence $$(\log_a)' = 0$$. After that, shorten out $$\log a$$.

Or use that $$\log a$$ is a constant from the very start and that $$(cx)' = cx'$$ when $$c$$ is constant. resp not a function of $$x$$.