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 January 25th, 2019, 02:22 PM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 n-th derivative Find n-th derivative of function $\displaystyle f(x)=\sin^{n}(x)$ . n-natural number
January 25th, 2019, 02:54 PM   #2
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 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, 03:19 PM #3 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 Worked four and got it. $\displaystyle y^{(1)}=n\sin^{n-1}(x)\cos(x)$ . Last edited by idontknow; January 25th, 2019 at 03:36 PM. Reason: Wrong
January 25th, 2019, 03:29 PM   #4
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 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 02:54 AM.

 January 26th, 2019, 06:33 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 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 03:22 PM.
January 26th, 2019, 11:10 AM   #6
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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 03:22 PM.

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