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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

UnkleRhaukus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Isn't that it? You solved your own question.

UnkleRhaukus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Use the powerrule repeatedly.

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

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

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

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

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

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

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

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

hartnn
 2 years 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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