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 September 23rd, 2012, 07:36 AM #1 Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Find the n'th derivative of a function The question asks: Find the nth derivative of each function by calculating the first few derivatives and observing the pattern that occurs. the function is $f(x)\,=\, x^{(n)}$ So I need to find the function $f^{(n)}(x)$ So if I start taking the derivative I get: $f'(x)\,=\,nx^{n-1}$ $f''(x)\,=\,(n-1)nx^{n-2}$ $f'''(x)\,=\,n(n-1)(n-2)x^{n-3}$ so on and so forth. I see the pattern, but I really don't know the notation to write it out as one function.
September 23rd, 2012, 07:42 AM   #2
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Re: Find the n'th derivative of a function

Quote:
 Originally Posted by jaredbeach The question asks: Find the nth derivative of each function by calculating the first few derivatives and observing the pattern that occurs. the function is $f(x)\,=\, x^{(n)}$ So I need to find the function $f^{(n)}(x)$ So if I start taking the derivative I get: $f'(x)\,=\,nx^{n-1}$ $f''(x)\,=\,(n-1)nx^{n-2}$ $f'''(x)\,=\,n(n-1)(n-2)x^{n-3}$ so on and so forth. I see the pattern, but I really don't know the notation to write it out as one function.
$\text{Use the factorial notation where : } n!=n(n-1)(n-2)...\text{3x2x1.}$

 September 23rd, 2012, 07:53 AM #3 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: Find the n'th derivative of a function Yes, the factorial is one way to go. Another is the permutation function: $n(n-1)(n-2)\cdots(n-k)=\text{nPr}(n,k)$
September 23rd, 2012, 01:07 PM   #4

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Re: Find the n'th derivative of a function

Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by jaredbeach The question asks: Find the nth derivative of each function by calculating the first few derivatives and observing the pattern that occurs. the function is $f(x)\,=\, x^{(n)}$ So I need to find the function $f^{(n)}(x)$ So if I start taking the derivative I get: $f'(x)\,=\,nx^{n-1}$ $f''(x)\,=\,(n-1)nx^{n-2}$ $f'''(x)\,=\,n(n-1)(n-2)x^{n-3}$ so on and so forth. I see the pattern, but I really don't know the notation to write it out as one function.
$\text{Use the factorial notation where : } n!=n(n-1)(n-2)...\text{3x2x1.}$
So is the answer$n!x^{n-1}$?

 September 23rd, 2012, 01:24 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 532 Thanks: 38 Math Focus: Electrical Engineering Applications Re: Find the n'th derivative of a function Where $\rightarrow$ means to take the next derivative. This notation is NOT standard and is used for illustration only: $f(x)\rightarrow 1=1!$ $f(x^2) \rightarrow 2x \rightarrow 2= 2!$ $f(x^3) \rightarrow 3x^2 \rightarrow 6x \rightarrow 6= 3!$
 September 23rd, 2012, 01:47 PM #6 Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Re: Find the n'th derivative of a function Oh! so it's just $n!$?
September 23rd, 2012, 02:00 PM   #7
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Re: Find the n'th derivative of a function

Quote:
Originally Posted by jaredbeach
Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by jaredbeach The question asks: Find the nth derivative of each function by calculating the first few derivatives and observing the pattern that occurs. the function is $f(x)\,=\, x^{(n)}$ So I need to find the function $f^{(n)}(x)$ So if I start taking the derivative I get: $f'(x)\,=\,nx^{n-1}$ $f''(x)\,=\,(n-1)nx^{n-2}$ $f'''(x)\,=\,n(n-1)(n-2)x^{n-3}$ so on and so forth. I see the pattern, but I really don't know the notation to write it out as one function.
$\text{Use the factorial notation where : } n!=n(n-1)(n-2)...\text{3x2x1.}$
So is the answer$n!x^{n-1}$?
I suspect what you meant is:

$f^{\small{(n)}}(x)=n!x^{n-n}=n!$

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