Question

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

Answer 1

induction on m

Answer 2

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

Answer 3

im not sure if i made a mistake , of if the thing im trying to deduce isn't right

Answer 4

for m=1 the statement is correct. Assume it is true for m-1 which means that
\[D^{m-1}x^n=\frac{x^{n+1-m}n!}{n+1-m}\].

Answer 5

Now
\[D^mx^n=D\left(D^{m-1}x^n\right)=\frac{(n+1-m)x^{n-m}n!}{(n+1-m)!}=\frac{x^{n-m}n!}{(n-m)!}\]

Answer 6

There is a typo\[D^{m-1}x^n=\frac{x^{n+1-m}n!}{(n+1-m)!}\]

Answer 7

i dont understand

Answer 8

have you learn mathematical induction?

Answer 9

i dont really like induction,

Answer 10

it always confuses me ,

Answer 11

You'd better learn it again then :D.

Answer 12

do i have to use induction for this question ?

Answer 13

dosent the D^{m-n} disappear because n≥m or something like that?

Answer 14

yours also fine actually. Just remember that D^0(f)=f

Answer 15

hmmm

Answer 16

\(\begin{align*}
&n\in\mathbb N,\qquad m\leq n\\
\\
\mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\
&=\mathrm D^{m-2}n(n-1)x^{n-2}\\
&=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\
&=\qquad\vdots\\
&=\mathrm D^{m-n}n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\
&=\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}\\
&\\
&\\
&\\
\mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\
&=n!
\end{align*}\)

Answer 17

n's are m's

Answer 18

whats the doubt now?

Answer 19

\[\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}=\frac{n!}{(n-m)!}x^{n-m}\]?

Answer 20

\(\begin{align*}
&n\in\mathbb N,\qquad m\leq n\\
\\
\mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\
&=\mathrm D^{m-2}n(n-1)x^{n-2}\\
&=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\
&=\qquad\vdots\\
&=\mathrm D^{m-\color{red}{\large m}} n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\
&=\mathrm D^{m-m}\frac{n!}{(n-m)!}x^{n-m}\\
&=\frac{n!}{(n-m)!}x^{n-m}\\
&\\
&\\
\mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\
&=n!
\end{align*}\)

Answer 21

differentiate 'm' time, not n

Answer 22

*times.

Answer 23

ah that's making some sense now

Answer 24

(i got to go right now , ill come back to this question )

Answer 25

you got the correction in blue also, right ? (n-(m-1)).

Answer 26

\[
\begin{align*}
&m,n\in\mathbb N,\qquad m\leq n\\
\\
\operatorname D^mx^n&=\operatorname D^{m-1}nx^{n-1}\\
&=\operatorname D^{m-2}n(n-1)x^{n-2}\\
&=\operatorname D^{m-3}n(n-1)(n-2)x^{n-3}\\
&=\operatorname D^{m-m}n(n-1)(n-2)\dots(n-m)x^{n-m}\\
&=\frac{n!}{(n-m)!}x^{n-m}\\
&\\
&\\
&\\
\operatorname D^mx^m&=\frac{n!}{(m-m)!}x^{m-m}\\
&=n!\\
\end{align*}
\]

Answer 27

@hartnn , i'm not sure about the stuff in blue ?

Answer 28

@watchmath , deduction is not induction

Answer 29

ah another mistake \[\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(m-m)!}x^{m-m}
=n!\]

Answer 30

Oh, I get it.

Answer 31

good

Answer 32

Isn't that it? You solved your own question.

Answer 33

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

Answer 34

Use the power-rule repeatedly.

Answer 35

\(\operatorname D^n e^x =e^x\)

how does e and ! related

Answer 36

Why?

Answer 37

You can take the factorial of anything.

Answer 38

Gamma Function bro.

Answer 39

yeah

Answer 40

Plus that is not \(x^n\).

Answer 41

\[\operatorname D^n(e^x)^n
=e^xn!\]

Answer 42

Ooooh. The Chain Rule?

Answer 43

\[\color{red}*\]\[\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn})
=n!e^{xn}\]

Answer 44

I don't know what you are trying to say... try the Chain Rule.

Answer 45

In which question, you have what doubt ? mention clearly, plz.....

Answer 46

the blue writing isn't right

Answer 47

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))