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January 25th, 2019, 02:22 PM   #1
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n-th derivative

Find n-th derivative of function $\displaystyle f(x)=\sin^{n}(x)$ .
n-natural number
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January 25th, 2019, 02:54 PM   #2
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Originally Posted by idontknow View Post
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
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January 25th, 2019, 03:19 PM   #3
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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
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January 25th, 2019, 03:29 PM   #4
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Originally Posted by idontknow View Post
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
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Last edited by skipjack; January 26th, 2019 at 02:54 AM.
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January 26th, 2019, 06:33 AM   #5
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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.
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Last edited by skipjack; January 26th, 2019 at 03:22 PM.
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January 26th, 2019, 11:10 AM   #6
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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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