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Calculate \(\mathrm D^mx^n\)

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induction on m
this is what i have so far \[\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-n}n(n-1)(n-2)\dots(n-m)x^{n-m}\\ &=\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots(m-m)x^{m-m}\\ &=n! \end{align*}\]

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Other answers:

im not sure if i made a mistake , of if the thing im trying to deduce isn't right
for m=1 the statement is correct. Assume it is true for m-1 which means that \[D^{m-1}x^n=\frac{x^{n+1-m}n!}{n+1-m}\].
Now \[D^mx^n=D\left(D^{m-1}x^n\right)=\frac{(n+1-m)x^{n-m}n!}{(n+1-m)!}=\frac{x^{n-m}n!}{(n-m)!}\]
There is a typo\[D^{m-1}x^n=\frac{x^{n+1-m}n!}{(n+1-m)!}\]
i dont understand
have you learn mathematical induction?
i dont really like induction,
it always confuses me ,
You'd better learn it again then :D.
do i have to use induction for this question ?
dosent the D^{m-n} disappear because n≥m or something like that?
yours also fine actually. Just remember that D^0(f)=f
\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-n}n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\ &=\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\ &=n! \end{align*}\)
n's are m's
whats the doubt now?
\[\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}=\frac{n!}{(n-m)!}x^{n-m}\]?
\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-\color{red}{\large m}} n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\ &=\mathrm D^{m-m}\frac{n!}{(n-m)!}x^{n-m}\\ &=\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\ &=n! \end{align*}\)
differentiate 'm' time, not n
ah that's making some sense now
(i got to go right now , ill come back to this question )
you got the correction in blue also, right ? (n-(m-1)).
\[ \begin{align*} &m,n\in\mathbb N,\qquad m\leq n\\ \\ \operatorname D^mx^n&=\operatorname D^{m-1}nx^{n-1}\\ &=\operatorname D^{m-2}n(n-1)x^{n-2}\\ &=\operatorname D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\operatorname D^{m-m}n(n-1)(n-2)\dots(n-m)x^{n-m}\\ &=\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \operatorname D^mx^m&=\frac{n!}{(m-m)!}x^{m-m}\\ &=n!\\ \end{align*} \]
@hartnn , i'm not sure about the stuff in blue ?
@watchmath , deduction is not induction
ah another mistake \[\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(m-m)!}x^{m-m} =n!\]
Oh, I get it.
Isn't that it? You solved your own question.
i kinda want to understand this result a little better, \[\operatorname D^nx^n =n!\], the n-th derivative of x to the power n is equal to n factorial
Use the power-rule repeatedly.
\(\operatorname D^n e^x =e^x\) how does e and ! related
You can take the factorial of anything.
Gamma Function bro.
Plus that is not \(x^n\).
\[\operatorname D^n(e^x)^n =e^xn!\]
Ooooh. The Chain Rule?
\[\color{red}*\]\[\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn}) =n!e^{xn}\]
I don't know what you are trying to say... try the Chain Rule.
In which question, you have what doubt ? mention clearly, plz.....
the blue writing isn't right
for m-2--->n(n-(2-1)) for m-3--->n(n-1)(n-(3-1)) for m-4--->n(n-1)(n-2)(n-(4-1)) . . for m-m--->n(n-1)....(n-(m-1))

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