Differentiating $\ln(4x)$.
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 $\tfrac{1}{x}$.
$\endgroup$ 44 Answers
$\begingroup$$\ln(4x)=\ln 4 + \ln x$ and $d(\ln 4)/dx=0$ since it is a constant.
$\endgroup$ $\begingroup$By the chain rule, we have
$$\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$$
$\endgroup$ 2 $\begingroup$$$\frac{d(\ln4x)}{dx}=\frac{d(\ln4x)}{d(4x)}\cdot\frac{d(4x)}{dx}=\frac1{4x}\cdot4$$
Or, $$\ln4x=\ln4+\ln x\implies\frac{d(\ln4x)}{dx}=\cdots$$
$\endgroup$ $\begingroup$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.
$\endgroup$ 3
