$f(x) = x \ln(\cos3x - x^3)$, then $f'(x)$ = ?
I got
$$\begin{align} &X * \left(\frac{1}{\cos3x - x^3}\right) * \left({-3}({\sin3x -3x^2})\right)\\ &= \frac{-(3x \sin3x) - 3x^3}{(\cos3x) -x^3}\\ \end{align}$$
and I don't know what to to next. The correct answer on my homework should be : $\ln(\cos3x) - (3x\tan3x) -3x^2$
But I have no idea how to arrive it. Please help.
Edit: The first answer below is to the problem as originally stated. The remark has to do with what the actual problem probably is.
Your function is a product, $x$ times a messy function. To differentiate, use the Product Rule.
Remark: On the assumption that the reported answer is correct, the function you are asked to differentiate is actually $x\ln(\cos 3x)-x^3$.
Use the Product Rule on $x\ln(\cos 3x)$.
$f(x) = x \ln(\cos3x - x^3)$, then $$f'(x) =\ln(\cos3x - x^3)+x\frac{1}{\cos3x-x^3}(-3\sin3x-3x^2)$$ $$=\ln(\cos(3x) - x^3)-\frac{3x(\sin(3x)+x^2)}{\cos(3x)-x^3}$$
Note that you first have to use the product rule: $$\begin{align} f'(x) &= \frac{d}{dx}[x]\ln(\cos(3x) - x^3) + x \frac{d}{dx}\ln(\cos(3x) - x^3)\\ &= \ln(\cos(3x) - x^3) + x\frac{1}{\cos(3x) - x^3}(-3\sin(3x) - 3x^2). \end{align} $$
