skynugurq7

2022-07-15

Differentiating logarithms
I am trying to prove that
$f\left(x\right){=}^{a}log\left(x\right)=>{f}^{\prime }\left(x\right)=\frac{1}{ln\left(a\right)\ast x}$
So I start at
$f\left(x\right){=}^{a}log\left(x\right)$
Then I move to:
$f\left(x\right)=\frac{ln\left(x\right)}{ln\left(a\right)}$
And there I get stuck: I want to use the quotient rule, but the entire internet tells me to use the chain rule. And indeed, with the quotient rule I get stuck on an island far away. But still: to me this looks a lot more like:
$f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$
Then
$f\left(x\right)=g\left(h\left(x\right)\right)$
So why do I need to use the chain rule from here? How can I use it in this situation?

jugf5

Expert

No need for the quotient rule: $\frac{1}{\mathrm{ln}\left(a\right)}$ is a constant.
$f\left(x\right)=\frac{\mathrm{ln}\left(x\right)}{\mathrm{ln}\left(a\right)}=\frac{1}{\mathrm{ln}\left(a\right)}\cdot \mathrm{ln}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{f}^{\prime }\left(x\right)=\frac{1}{\mathrm{ln}\left(a\right)}\cdot \frac{1}{x}=\frac{1}{\mathrm{ln}\left(a\right)x}$

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