Nth derivative formula for high degree power rule
$begingroup$
I was wondering if there is a general formula for finding some $n$-th derivative. I came up with this (the Google doc has the math with proper formatting).
Can someone tell me if this is correct and if it has been done before (it probably has)? If it has been done before, can someone please link the source?
Also, is there any way to make it more general?
Thanks.
derivatives
edited Jan 18 at 22:21
Dog_69
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
$endgroup$
add a comment |
$begingroup$
I was wondering if there is a general formula for finding some $n$-th derivative. I came up with this (the Google doc has the math with proper formatting).
Can someone tell me if this is correct and if it has been done before (it probably has)? If it has been done before, can someone please link the source?
Also, is there any way to make it more general?
Thanks.
derivatives
edited Jan 18 at 22:21
Dog_69
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
$endgroup$
add a comment |
$begingroup$
I was wondering if there is a general formula for finding some $n$-th derivative. I came up with this (the Google doc has the math with proper formatting).
Can someone tell me if this is correct and if it has been done before (it probably has)? If it has been done before, can someone please link the source?
Also, is there any way to make it more general?
Thanks.
derivatives
edited Jan 18 at 22:21
Dog_69
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
$endgroup$
I was wondering if there is a general formula for finding some $n$-th derivative. I came up with this (the Google doc has the math with proper formatting).
Can someone tell me if this is correct and if it has been done before (it probably has)? If it has been done before, can someone please link the source?
Also, is there any way to make it more general?
Thanks.
derivatives
derivatives
edited Jan 18 at 22:21
Dog_69
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
edited Jan 18 at 22:21
Dog_69
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
edited Jan 18 at 22:21
Dog_69
6361523
edited Jan 18 at 22:21
Dog_69
6361523
edited Jan 18 at 22:21
Dog_69
6361523
6361523
asked Jan 18 at 21:56
FIREREEDFIREREED
83
asked Jan 18 at 21:56
FIREREEDFIREREED
83
asked Jan 18 at 21:56
FIREREEDFIREREED
83
83
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=xcdot e^{-x}$$
$$f^{n)}(x)=((-1)^n)(xcdot e^{-x})+((-1)^{n+1}cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=frac{1}{1-x}$$
$$g^{n)}(x)=frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.
answered Jan 18 at 22:23
idriskameniidriskameni
749321
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add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=xcdot e^{-x}$$
$$f^{n)}(x)=((-1)^n)(xcdot e^{-x})+((-1)^{n+1}cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=frac{1}{1-x}$$
$$g^{n)}(x)=frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.
answered Jan 18 at 22:23
idriskameniidriskameni
749321
$endgroup$
add a comment |
$begingroup$
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=xcdot e^{-x}$$
$$f^{n)}(x)=((-1)^n)(xcdot e^{-x})+((-1)^{n+1}cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=frac{1}{1-x}$$
$$g^{n)}(x)=frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.
answered Jan 18 at 22:23
idriskameniidriskameni
749321
$endgroup$
add a comment |
$begingroup$
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=xcdot e^{-x}$$
$$f^{n)}(x)=((-1)^n)(xcdot e^{-x})+((-1)^{n+1}cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=frac{1}{1-x}$$
$$g^{n)}(x)=frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.
answered Jan 18 at 22:23
idriskameniidriskameni
749321
$endgroup$
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=xcdot e^{-x}$$
$$f^{n)}(x)=((-1)^n)(xcdot e^{-x})+((-1)^{n+1}cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=frac{1}{1-x}$$
$$g^{n)}(x)=frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.
answered Jan 18 at 22:23
idriskameniidriskameni
749321
answered Jan 18 at 22:23
idriskameniidriskameni
749321
answered Jan 18 at 22:23
idriskameniidriskameni
749321
answered Jan 18 at 22:23
idriskameniidriskameni
749321
749321
add a comment |
add a comment |
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