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 January 25th, 2019, 03:22 PM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 829 Thanks: 113 Math Focus: Elementary Math n-th derivative Find n-th derivative of function $\displaystyle f(x)=\sin^{n}(x)$ . n-natural number January 25th, 2019, 03:54 PM   #2
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Quote:
 Originally Posted by idontknow Find n-th derivative of function $\displaystyle f(x)=\sin^{n}(x)$ . n-natural number
So let n = 1. What is f'(x)? Let n = 2, what is f''(x)?

Rinse and repeat. Work out several of them and see if you can spot a pattern.

-Dan January 25th, 2019, 04:19 PM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 829 Thanks: 113 Math Focus: Elementary Math Worked four and got it. $\displaystyle y^{(1)}=n\sin^{n-1}(x)\cos(x)$ . Last edited by idontknow; January 25th, 2019 at 04:36 PM. Reason: Wrong January 25th, 2019, 04:29 PM   #4
Math Team

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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by idontknow Worked four and got it. $\displaystyle y^{(n)}=n\sin^{n-1}(x)\cos(x)$ .
Good start, but that's only $\displaystyle y^{(1)}$, not $\displaystyle y^{(n)}$.

$\displaystyle y^{(0)}(x) = \sin^n(x)$

$\displaystyle y^{(1)}(x) = n ~ \sin^{n - 1}(x) \cdot \cos(x)$ by the chain rule

$\displaystyle y^{(2)}(x) = n(n - 1)~\sin^{(n - 2)}(x) \cdot \cos^2(x) + n ~ \sin^{n - 1}(x) \cdot - \sin(x)$

etc.

-Dan

Last edited by skipjack; January 26th, 2019 at 03:54 AM. January 26th, 2019, 07:33 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 I'll assume $n$ is a non-negative integer. If $n$ is odd, $\displaystyle \sin^n(x) = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} (-1)^{\left(\frac{n-1}{2}-k\right)} \binom{n}{k} \sin{\left((n-2k)x\right)}$ If $n$ is even, $\displaystyle \sin^n(x) = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{\left(\frac{n}{2}-k\right)} \binom{n}{k} \cos{\left((n-2k)x\right)}$ The individual terms in the above sums are easy to differentiate $n$ times. Thanks from topsquark and idontknow Last edited by skipjack; January 26th, 2019 at 04:22 PM. January 26th, 2019, 12:10 PM   #6
Math Team

Joined: May 2013
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Posts: 2,340
Thanks: 983

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by skipjack I'll assume $n$ is a non-negative integer. If $n$ is odd, $\displaystyle \sin^n(x) = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} (-1)^{\left(\frac{n-1}{2}-k\right)} \binom{n}{k} \sin{\left((n-2k)x\right)}$ If $n$ is even, $\displaystyle \sin^n(x) = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{\left(\frac{n}{2}-k\right)} \binom{n}{k} \cos{\left((n-2k)x\right)}$ The individual terms in the above sums are easy to differentiate $n$ times.
That's a much better way. Thank you.

-Dan

Last edited by skipjack; January 26th, 2019 at 04:22 PM. Tags derivative, nth Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post daltyboy11 Calculus 2 July 10th, 2014 07:57 PM lackofimagination Calculus 1 July 6th, 2014 09:05 PM Hustler333 Calculus 4 November 7th, 2013 01:17 PM supaman5 Linear Algebra 0 November 26th, 2012 11:14 AM rgarcia128 Calculus 4 September 24th, 2011 06:07 PM

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