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$\log_ax$

I know there is a formula for this, but I am trying to understand the logic behind the formula.

I know that $$\log_ax = \frac{\ln x}{\ln a} = \ln x \cdot \ln a ^{-1}$$ Because this is in the form $f(x)g(x)$, I have to use $f'(x)g(x)+f(x)g'(x)$ to solve this, and so I have:

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

So I did:

$$\frac{1}{x} \cdot \ln a ^{-1} + \ln x \cdot -\frac{1}{a\ln^2 a} = \frac{1}{x \ln a} - \frac{\ln x}{a\ln^2 a} = \frac{a \ln^2(a)-\ln(x)x\ln(a)}{ax \ln^3(a)}= \frac{a \ln(a) - \ln (x)x}{ax\ln^2(a)} = \text{???}$$

I think this is wrong. I put it on Wolfram Alpha http://www.wolframalpha.com/input/?i=(aln(a)-ln(x)x)%2F(axln(a)%5E2) and it doesn't seem to simplify to $\frac{1}{\ln(a)x}$, which is the correct answer.

What did I do wrong?

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    $(\frac{1}{\ln(a)} )' = 0$ – 2017-01-14
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    It is wrong because $a$ is a constant. – 2017-01-14
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    g(x) is a constant function – 2017-01-14
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    @Arnaldo Oh, I see. I should have used $(kg(x))' = kg'(x)$ instead, because $\ln a ^{-1}$ Is a constant. I'll try, thanks – 2017-01-14
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    Have you learned implicit differentation? – 2017-01-14
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    You should treat 'a' as a constant. – 2017-01-14
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    In standard usage, $\ln a^{-1}$ means $\ln(a^{-1}),$ but you're using that notation to mean $(\ln a)^{-1},$ without even mentioning that that's what you meant instead. – 2017-01-14

2 Answers 2

4

You simply treat $(\ln a)^{-1}$ as a constant. That is $\frac{d}{dx} ((\ln a)^{-1} \ln x )= (\ln a)^{-1} \frac{d}{dx} \ln x = (\ln a)^{-1} \cdot \frac{1}{x}$

3

Since $\log_a{x}=\frac{\ln{x}}{\ln{a}}$, and $a$ is just a constant, not a variable, you do not need to use the product rule.

Therefore:

$$\frac{d}{dx}(\log_a{x})=\frac{d}{dx}\left(\frac{1}{\ln{a}}\cdot \ln{x}\right)=\frac{1}{\ln{a}}\cdot \frac{d}{dx}(\ln{x})=\frac{1}{\ln{a}}\cdot \frac{1}{x}=\frac{1}{x\ln{a}}$$