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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.

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    $\begingroup$ Presumably, $a$ is constant $\endgroup$
    – Doobius
    May 25, 2022 at 18:16
  • $\begingroup$ Quotient rule is corrected. Sorry for the misnomer. $\endgroup$
    – Incompl33t
    May 25, 2022 at 18:37

3 Answers 3

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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!

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

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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 $x$.

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