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:
$$ f(x) = x\\ g(x) = \ln(3)\\ f'(x) = 1\\ g'(x) = 1/3\\ D(f(x) * g(x))=\\ = 1 * \ln(3) + 1/3 * x \\ = \ln(3) + 1/3x \neq ln(3)$$
$\endgroup$ 42 Answers
$\begingroup$$\ln(3)$ is a constant.. There's no need for the product rule.
$(x\cdot c)' = c$ when $c$ is a constant.
You can use the product rule, but there's no need for it, since the derivative of any constant $c$ is given by $(c)' = 0$. That gives us $(\ln 3)' = 0$.
If $x = f(x)$ and $\ln 3 = g(x)$, then the derivative, using the product rule, is given by: $$f'(x)g(x) + f(x)g'(x) = (1)\cdot\ln(3) + x\cdot (0) = \ln(3)$$
Note that for $g(x) = \ln 3, \;g'(x) \neq \frac 13$. It is true that if $h(x) = \ln x,\;$ then $h'(x) = \frac 1x$, but $x$ is the variable with respect to which we are differentiating. In contrast, the argument of $\ln(3)$, $3$, is a constant, as is $\ln 3$, and like any constant or constant function, in this case, its derivative with respect to $x$ is $0$
$\endgroup$ 5 $\begingroup$It is not necessary to use the productrule (see the answer of amWhy). You can however consider $\ln(3)$ to be a function of $x$. It is a constant function. The derivative of any constant function is $0$ and applying the productrule with $f(x)=x$ and $g(x)=\ln(3)$ gives: $$f'(x)g(x)+f(x)g'(x)=1\times \ln(3)+x\times 0=\ln3$$
$\endgroup$
