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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1this is what i have so far \[\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{mn}n(n1)(n2)\dots(nm)x^{nm}\\ &=\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots(mm)x^{mm}\\ &=n! \end{align*}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1im not sure if i made a mistake , of if the thing im trying to deduce isn't right

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0for m=1 the statement is correct. Assume it is true for m1 which means that \[D^{m1}x^n=\frac{x^{n+1m}n!}{n+1m}\].

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0Now \[D^mx^n=D\left(D^{m1}x^n\right)=\frac{(n+1m)x^{nm}n!}{(n+1m)!}=\frac{x^{nm}n!}{(nm)!}\]

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0There is a typo\[D^{m1}x^n=\frac{x^{n+1m}n!}{(n+1m)!}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1i dont understand

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0have you learn mathematical induction?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1i dont really like induction,

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1it always confuses me ,

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0You'd better learn it again then :D.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1do i have to use induction for this question ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1dosent the D^{mn} disappear because n≥m or something like that?

watchmath
 one year ago
Best ResponseYou've already chosen the best response.0yours also fine actually. Just remember that D^0(f)=f

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{mn}n(n1)(n2)\dots\color{blue}{(n(m1))}x^{nm}\\ &=\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots\color{blue}{(n(n1))}x^{mm}\\ &=n! \end{align*}\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}=\frac{n!}{(nm)!}x^{nm}\]?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{m\color{red}{\large m}} n(n1)(n2)\dots\color{blue}{(n(m1))}x^{nm}\\ &=\mathrm D^{mm}\frac{n!}{(nm)!}x^{nm}\\ &=\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots\color{blue}{(n(n1))}x^{mm}\\ &=n! \end{align*}\)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0differentiate 'm' time, not n

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1ah that's making some sense now

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1(i got to go right now , ill come back to this question )

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0you got the correction in blue also, right ? (n(m1)).

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align*} &m,n\in\mathbb N,\qquad m\leq n\\ \\ \operatorname D^mx^n&=\operatorname D^{m1}nx^{n1}\\ &=\operatorname D^{m2}n(n1)x^{n2}\\ &=\operatorname D^{m3}n(n1)(n2)x^{n3}\\ &=\operatorname D^{mm}n(n1)(n2)\dots(nm)x^{nm}\\ &=\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \operatorname D^mx^m&=\frac{n!}{(mm)!}x^{mm}\\ &=n!\\ \end{align*} \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1@hartnn , i'm not sure about the stuff in blue ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1@watchmath , deduction is not induction

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1ah another mistake \[\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(mm)!}x^{mm} =n!\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Isn't that it? You solved your own question.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1i kinda want to understand this result a little better, \[\operatorname D^nx^n =n!\], the nth derivative of x to the power n is equal to n factorial

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Use the powerrule repeatedly.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\(\operatorname D^n e^x =e^x\) how does e and ! related

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0You can take the factorial of anything.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Gamma Function bro.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Plus that is not \(x^n\).

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\operatorname D^n(e^x)^n =e^xn!\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Ooooh. The Chain Rule?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\color{red}*\]\[\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn}) =n!e^{xn}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what you are trying to say... try the Chain Rule.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0In which question, you have what doubt ? mention clearly, plz.....

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1the blue writing isn't right

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0for m2>n(n(21)) for m3>n(n1)(n(31)) for m4>n(n1)(n2)(n(41)) . . for mm>n(n1)....(n(m1))
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