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The product rule is defined as $$(f \cdot g)' = f' \cdot g + g' \cdot f.$$

I have the following function $u(x) = x\cdot \ln(3)$. I understand that you can derive it by implicit differentiation and have $\ln(3)$ as the result.

I, however, do not understand why I get the wrong result by applying the product rule:

$$ f(x) = x\\ g(x) = \ln(3)\\ f'(x) = 1\\ g'(x) = 1/3\\ D(f(x) * g(x))=\\ = 1 * \ln(3) + 1/3 * x \\ = \ln(3) + 1/3x \neq ln(3)$$

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The derivative of $\ln(3)$ is 0 – LeeNeverGup Feb 21 '14 at 14:17
You can, of course...but is overkill. – DonAntonio Feb 21 '14 at 14:17
You made a mistake, the derivative od $g$ in your case is $0$. – 5xum Feb 21 '14 at 14:28
If $g(x)=\ln(3)$ then $g'(x)=0$. You make it wrongly $g'(x)=1/3$. Probably inspired by the next thing: If you would have $g(x)=\ln(x)$ then $g'(3)=1/3$ since in that case $g'(x)=1/x$. – drhab Feb 21 '14 at 14:32
up vote 4 down vote accepted

$\ln(3)$ is a constant.. There's no need for the product rule.

$(x\cdot c)' = c$ when $c$ is a constant.

You can use the product rule, but there's no need for it, since the derivative of any constant $c$ is given by $(c)' = 0$. That gives us $(\ln 3)' = 0$.

If $x = f(x)$ and $\ln 3 = g(x)$, then the derivative, using the product rule, is given by: $$f'(x)g(x) + f(x)g'(x) = (1)\cdot\ln(3) + x\cdot (0) = \ln(3)$$

Note that for $g(x) = \ln 3, \;g'(x) \neq \frac 13$. It is true that if $h(x) = \ln x,\;$ then $h'(x) = \frac 1x$, but $x$ is the variable with respect to which we are differentiating. In contrast, the argument of $\ln(3)$, $3$, is a constant, as is $\ln 3$, and like any constant or constant function, in this case, its derivative with respect to $x$ is $0$

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I wasn't the downvote, but if $\ln(3)$ is not a function of $x$, then it doesn't make sense to take the derivative of it with respect to $x$. – Cam McLeman Feb 21 '14 at 14:20
@CamMcLeman. Why doesn't it make sense to say that the derivative of a constant is $0$ ? – Tom-Tom Feb 21 '14 at 14:22
Writing $(5)'$ is not completely rigorous in mathematic terms, but it is just as rigurious as writing $(\sin x)'=\cos x$, because you are not deriving $\sin x$ but the function $\sin$. What i want to say is that such nitpicking is utterly stupid. – 5xum Feb 21 '14 at 14:22
@V.Rossetto: That comment was in reference to an earlier version of the answer, in which the author claimed that constants were not function of $x$. If they are not functions of $x$, then it is does not make sense to take their derivative. – Cam McLeman Feb 21 '14 at 15:36
@CamMcLeman. Okay, that was not easy to catch... – Tom-Tom Feb 21 '14 at 15:39

It is not necessary to use the productrule (see the answer of amWhy). You can however consider $\ln(3)$ to be a function of $x$. It is a constant function. The derivative of any constant function is $0$ and applying the productrule with $f(x)=x$ and $g(x)=\ln(3)$ gives: $$f'(x)g(x)+f(x)g'(x)=1\times \ln(3)+x\times 0=\ln3$$

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