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Why is the derivative of $\ln(4x)$ equal to: $\frac{1}{x}$ Shouldn't it be? $$\frac{1}{4x}$$ This seems so wrong to me because the derivative of $\ln x$ is $\tfrac{1}{x}$.

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You're forgetting to apply the chain rule... – symplectomorphic May 10 '14 at 14:51
This seems so wrong - Well, appearances can be quite deceiving. :-) – Lucian May 10 '14 at 15:21
Actually, it is $\frac{1}{4x}*4$ – Awal Garg May 10 '14 at 17:23
you can take 4 as a constant and take it outside if it helps – Kennan May 11 '14 at 13:23

$\ln(4x)=\ln 4 + \ln x$ and $d(\ln 4)/dx=0$ since it is a constant.

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By the chain rule, we have

$$\frac{d}{dx}\left(\ln(f(x))\right) = \frac{f'(x)}{f(x)}$$

In your case, $f(x) = 4x$. So, $$\frac{d}{dx}\left(\ln(4x)\right) = \dfrac{(4x)'}{4x} = \dfrac 4{4x} = \frac 1x$$

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I added a $d/dx$ where I think you missed one out. – Ben Millwood May 10 '14 at 15:18
Yes, indeed. Thanks @Ben for catching my oversight! – amWhy May 10 '14 at 15:19


Or, $$\ln4x=\ln4+\ln x\implies\frac{d(\ln4x)}{dx}=\cdots$$

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All answers so far (added later: except Marty Cohen's) have (correctly) pointed out that you need to use the chain rule. Here is another reason why you should not be surprised by the outcome. If $c$ is a constant, and $f$ is a differentiable function, the $(c +f(x))^{\prime} = f^{\prime}(x),$ that is $c+f(x)$ and $f(x)$ have the same derivative everywhere. Notice that $\ln(4x) = \ln(4) +\ln(x)$, and $\ln(4)$ is just a constant, so the derivative of $\ln(4x)$ should indeed be the same as the derivative of $\ln(x)$ wherever either derivative exists.

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Oops: I missed Marty Cohen's answer, which gives the same reason – Geoff Robinson May 10 '14 at 15:44
And you missed that @lab was the first to make that point (second half of post, following "or"...) – amWhy May 10 '14 at 17:37
@lab: Sorry for that too. – Geoff Robinson May 10 '14 at 17:56

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