If $ y= x^{n-1}\log(x)$ , then prove that $D^n y $ is $ \frac{(N-1)!}{x}$ $ y= x^{n-1} \log(x)$ , then prove that $D^n y $ is $  \frac{(N-1)!}{x}$">
I don't understand this notation of "$D^n$".I searched it on internet and didn't got any useful information (What is the operator "capital D" and how can the chain rule be used in this way) Please tell me what this notation stands for so that I can solve this question.
 A: Usually $D^n y $ means $\frac{d^ny}{dx^n}$. 
$D$ is refers to the differential operator. Applying $n$ times means differentiate $n$ times.
A: The exponent you are familiar with expresses "repeated multiplication by", like in
$$a^3\times b=a\times (a\times (a\times b))$$ ($b$ can be $1$).
By similarity, it is used to express "repeated application" of some operation, such as differentiation:
$$D^3f=D(D(Df)))=f'''.$$
In some contexts, it can be the repeated application of a function, like
$$\sin^3(x)=\sin(\sin(\sin(x))).$$
(Caution: can be confused with $(\sin(x))^3)$).

Now by induction,
$$D^{n+1}(x^n\log x)=D^n(D(x^n\log x))=D^n\left(nx^{n-1}\log x+\frac{x^n}x\right)=nD^n\left(x^{n-1}\log x\right).$$
The base case
$$D(\log x)=\frac1x$$ completes the proof.
A: In addition to understand $D^ny=\frac{d^ny}{dx^n}$, you'll want to use Leibniz's General Product Rule
$$\frac{d^n}{dx^n}(fg)=\sum_{k=0}^nC(n,k)(D^{n-k}f)(D^kg)$$
where $C(n,k)$ is a binomial coefficient, $f=x^{n-1}$, $g=\ln x$
Keeping in mind $D^p(x^n)=\frac{n!}{(n-p)!}x^{n-p}$ or $0$ if $p>n$.
Also note $D^p(\ln x)=\frac{(p-1)!(-1)^{p-1}}{x^p}$
