Why is this incorrect (regarding differentiating the natural log)? We must differentiate the following:
$$ [f(x) = \ln (3x^2 +3)]\space '$$
Why is this incorrect? I am just using the product rule:
$ [f(x) = \ln (3x^2 +3)]\space ' = \dfrac{1}{x} \times (3x^2 + 3) + \ln(6x) = \dfrac{3x^2 +3}{x} + \ln(6x)$
My book gives the following answer:
$$\dfrac{6x}{3x^2 +3}$$ 
 A: There is no product here; you should be using the chain rule.
The start of your answer makes it look like you were differentiating $\log(x) \cdot (3x^2 + 3)$ instead of the given function, but the latter part of your attempt clarifies that you are just getting tangled up.
(Also, it's a bit strange that your book didn't reduce its final answer, but it's still correct.)
More precisely:
$[f(g(x))]' = f'(g(x)) \cdot g'(x)$.
In this case,
$$[\log(3x^2 + 3)]' = \frac{1}{3x^2 +3} \cdot (3x^2 + 3)' = \frac{6x}{3x^2 + 3}$$ as your book suggests. Of course, we could divide top and bottom by $3$ to simplify our answer to:
$$\frac{2x}{x^2 + 1}$$
Going back to the original function, note that $\log(3x^2 + 3x) = \log(3) + \log(x^2 + x)$. If you now differentiate the function in this form, the derivative of the constant term $\log(3)$ will be $0$, and you will end up with the same answer as above (already in simplified form).
A: This is not a product, it is a composition. This is a common misconception for students when reading functional notation. The expression in parenthesis in functional notation is not multiplying its function-name, it's being evaluated inside the function.
You are interpreting it as if it were $\ln (x)\times (3x^2+3)$, but it is $3x^2+3$ composed with $\ln (x)$.
$\ln$ can never be written alone, it always has to be read with its input inside the parentheses. Likewise, $\sin(3x)$ is not $(\sin)$ times $3x$, it is $\sin$ evaluated at $3x$.
If original function were $\ln(x) \times (3x^2+3)$,  then you would be correctly applying the product rule, but you would have had to have written this:
$$=\frac{1}{x} \times (3x^2+3)+\ln(x)\times 6x$$
You can see the difference between what you wrote and this is, again, that the $\ln$ had been written without something to be evaluated at.
