If $f(x)=x^{n-1}\log x$, then the $n$-th derivative of $f$ is equals? The options are:
A) $\dfrac{(n-1)!}{x}$;
B) $\dfrac{n}{x}$;
C) $(-1)^{n-1}\dfrac{(n-1)!}{x}$;
D) $\dfrac{1}{x}$
My attempt:
$$f'(x)= (n-1)x^{n-2}\log x+ x^{n-2}$$
$$f''(x)=(n-2)(n-1)x^{n-3}\log x+ (n-1)x^{n-3}+(n-2)x^{n-3}$$
But I fail to see any pattern...

Here is the pattern:
The second summand in $f'(x)$, is $x^{n-2}$. This summand will not survive $n-1$ more derivatives, and so, you may ignore it. This leaves you with $$f'(x)=(n-1)x^{n-2}\log x+(\mathrm{irrelevant\; stuff}).$$
Likewise, for the second derivative you have$$f''(x)=(n-2)(n-1)x^{n-3}\log x +(\mathrm{irrelevant\;stuff}).$$ It is not too hard to continue this until you reach the $n$-th derivative.
