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...
 A: 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.
A: Here is another way, using Taylor expansion (with, maybe, the assistance of a Computer Algebra System). I will take the example of $n=3$, i.e. $f(x)=x^2 \ln(x)$ in order to simplify computations. Let us formally expand $f(x+y)$ in two ways:
$$f(x+y)=\begin{cases}f(x)+\tfrac{f'(x)}{1!}y+\tfrac{f''(x)}{2!}y^2+\cdots \\ (x+y)^2 \ln(x+y)=(x+y)^2 (\ln(x)+\ln(1+\tfrac{y}{x}))\end{cases}$$
the last expression being $$(x^2+2xy+y^2)(\ln(x)+\tfrac{y}{x}-\tfrac{y^2}{2x^2}+\tfrac{y^3}{3x^3}-\cdots)=$$
$$=\underbrace{x^2 \ln(x)}_{f(x)}+y\underbrace{(x+2x\ln(x))}_{f'(x)}+y^2\underbrace{(\ln(x)+\tfrac{3}{2})}_{f''(x)/2}+\cdots$$
by identification.
Taing a general value of $n$ instead of a particular value like $n=3$ here,  the general pattern given by @Amitai Yuval is clearly provable.
A: After checking for a few $n$, you can easily prove by induction that the answer is A. Is obviously true for $n = 1$. If is true for $n$, then:
$$f_{n+1}^{(n+1)}(x) = \frac{d^n}{dx^n}f_{n+1}'(x) = \frac{d^n}{dx^n}(nx^{n-1}\log x +x^n\frac1x) = nf_n^{(n)}(x) + 0 = \cdots$$
