# nth Derivative notation question

• Aug 7th 2011, 08:59 PM
NeedDirection
nth Derivative notation question
Find an expression for the nth derivative of f(x)=e^(x) * sin(x)

so what would the nth derivative of f be: f^(n) (x)

You can make a different expression for odd or even numbers, but I still can't quiet get one that works.
Please show me one that would work!!!
THANKS (Happy)
• Aug 7th 2011, 09:05 PM
Prove It
Re: nth Derivative notation question
\displaystyle \displaystyle \begin{align*} f(x) &= e^x\sin{x}\\ \\ f'(x) &= e^x\sin{x} + e^x\cos{x}\\ &= f(x) + e^x\cos{x}\\ \\ f''(x) &= f'(x) + e^x\cos{x} - e^x\sin{x} \\ &= f'(x) - f(x) + e^x\cos{x}\\ \\ f'''(x) &= f''(x) - f'(x) + e^x\cos{x} - e^x\sin{x} \\ &= f''(x) - f'(x) - f(x) + e^x\cos{x}\\ \\ f^{(\textrm{iv})}(x) &= f'''(x) - f''(x) - f'(x) + e^x\cos{x} - e^x\sin{x} \\ &= f'''(x) - f''(x) - f'(x) - f(x) + e^x\cos{x} \end{align*}

I think you can see a pattern here...
• Aug 7th 2011, 09:21 PM
NeedDirection
Re: nth Derivative notation question
Yeah I know the pattern but how could you put it all into only a few expressions?
• Aug 7th 2011, 11:07 PM
Bacterius
Re: nth Derivative notation question
You could use the sum notation. For the nth derivative, you start with (n-1)th derivative + $\displaystyle e^x \cos x$, and then do the sum of all the lower order derivatives and the original function.. don't forget the minus sign... It should look something like this:

$\displaystyle y = e^x \sin x$

$\displaystyle \frac{\mathrm{d}^n y}{\mathrm{d} x^n} = e^x \cos x + \frac{\mathrm{d}^{n - 1} y}{\mathrm{d} x^{n - 1}} - \sum_{i=0}^{n-2} \left [\frac{\mathrm{d}^i y}{\mathrm{d} x^i} \right \]$

This gives the expression for all the $\displaystyle n$th derivatives of the function (so $\displaystyle n > 0$), assuming the sum term is ignored if $\displaystyle n -2 < 0$.

That being said, I don't really see the point in doing this. It's not like that new expression is easier to use or something. Do you have any other information associated with this, perhaps the context would make it easier to know what you are really asking for (because I'm not sure this is what you want).
• Aug 8th 2011, 02:12 AM
Ackbeet
Re: nth Derivative notation question
Quote:

Originally Posted by NeedDirection
Find an expression for the nth derivative of f(x)=e^(x) * sin(x)

so what would the nth derivative of f be: f^(n) (x)

You can make a different expression for odd or even numbers, but I still can't quiet get one that works.
Please show me one that would work!!!
THANKS (Happy)

Try the Leibniz Rule for the nth derivative of a product.
• Aug 8th 2011, 02:20 PM
NickTaken
Re: nth Derivative notation question
Quote:

Originally Posted by NeedDirection
Find an expression for the nth derivative of f(x)=e^(x) * sin(x)

so what would the nth derivative of f be: f^(n) (x)

You can make a different expression for odd or even numbers, but I still can't quiet get one that works.
Please show me one that would work!!!
THANKS (Happy)

\displaystyle \begin{aligned} f^{(n)}(x) & = \sum_{k=0}^{n}\binom{n}{k} e^{x}\sin\left(\frac{k\pi}{2}+x\right) \\& = \Im \bigg[ e^{x} \sum_{k=0}^{n} \binom{n}{k}e^{i(\frac{k\pi}{2}+x)}\bigg] \\ & = \Im\bigg[e^{x}e^{ix}\sum_{k=0}^{n}\binom{n}{k}(e^\frac{i\pi }{2})^k}\bigg] \\ & =\Im\bigg[e^{x}e^{ix}(1+e^\frac{\i\pi}{2})^n\bigg] \\ & = e^{x}~2^{\frac{n}{2}}~\sin\left(\frac{n\pi}{4}+x \right). \end{aligned}
• Aug 8th 2011, 04:52 PM
NeedDirection
Re: nth Derivative notation question
Wow, thanks a lot everyone! This makes perfect sense now!