January 25th, 2019, 02:22 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92  nth derivative
Find nth derivative of function $\displaystyle f(x)=\sin^{n}(x)$ . nnatural number 
January 25th, 2019, 02:54 PM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,273 Thanks: 943 Math Focus: Wibbly wobbly timeywimey stuff.  
January 25th, 2019, 03:19 PM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92 
Worked four and got it. $\displaystyle y^{(1)}=n\sin^{n1}(x)\cos(x)$ . Last edited by idontknow; January 25th, 2019 at 03:36 PM. Reason: Wrong 
January 25th, 2019, 03:29 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,273 Thanks: 943 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle y^{(0)}(x) = \sin^n(x)$ $\displaystyle y^{(1)}(x) = n ~ \sin^{n  1}(x) \cdot \cos(x)$ 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: 20,975 Thanks: 2226 
I'll assume $n$ is a nonnegative integer. If $n$ is odd, $\displaystyle \sin^n(x) = \frac{2}{2^n} \sum_{k=0}^{\frac{n1}{2}} (1)^{\left(\frac{n1}{2}k\right)} \binom{n}{k} \sin{\left((n2k)x\right)}$ If $n$ is even, $\displaystyle \sin^n(x) = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}1} (1)^{\left(\frac{n}{2}k\right)} \binom{n}{k} \cos{\left((n2k)x\right)}$ The individual terms in the above sums are easy to differentiate $n$ times. Last edited by skipjack; January 26th, 2019 at 03:22 PM. 
January 26th, 2019, 11:10 AM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,273 Thanks: 943 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; January 26th, 2019 at 03:22 PM.  

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