# Calculating the fifth derivative

• Feb 24th 2013, 05:05 PM
Paze
Calculating the fifth derivative
So I've got this problem:

Calculate the fifth derivative of the function \$\displaystyle f(x)=xe^x\$

I tried going through the hassle of doing product rule again and again but it became so messy by the third iteration that I have to believe that there is an easier way to solve this problem (and perhaps other problems like it?).

Thanks.
• Feb 24th 2013, 05:07 PM
chiro
Re: Calculating the fifth derivative
Hey Paze.

The way you are doing it is the way to do it (and yes it will probably be very messy).
• Feb 24th 2013, 05:11 PM
Paze
Re: Calculating the fifth derivative
Oh, I see. Thanks!
• Feb 25th 2013, 12:06 AM
hollywood
Re: Calculating the fifth derivative
Actually, not all that messy.....

- Hollywood
• Feb 25th 2013, 12:07 AM
Paze
Re: Calculating the fifth derivative
Quote:

Originally Posted by hollywood
Actually, not all that messy.....

- Hollywood

Care to explain? It becomes pretty messy for me.
• Feb 25th 2013, 12:21 AM
hollywood
Re: Calculating the fifth derivative
If you apply the product rule to \$\displaystyle (x+n)e^x\$, you get \$\displaystyle (x+n)(e^x)'+(x+n)'(e^x) = (x+n)e^x + e^x = (x+n+1)e^x\$. So if you start with \$\displaystyle (x+0)e^x\$ and apply it 5 times, you get \$\displaystyle (x+5)e^x\$.

- Hollywood
• Feb 25th 2013, 12:24 AM
Paze
Re: Calculating the fifth derivative
Quote:

Originally Posted by hollywood
If you apply the product rule to \$\displaystyle (x+n)e^x\$, you get \$\displaystyle (x+n)(e^x)'+(x+n)'(e^x) = (x+n)e^x + e^x = (x+n+1)e^x\$. So if you start with \$\displaystyle (x+0)e^x\$ and apply it 5 times, you get \$\displaystyle (x+5)e^x\$.

- Hollywood

Where does that n come from? I just did product rule on \$\displaystyle xe^x\$ and somehow, natural logarithms started getting mixed and what not.. :(
• Feb 25th 2013, 07:12 AM
hollywood
Re: Calculating the fifth derivative
n is any integer.

So when you take the first derivative, you get \$\displaystyle xe^x+e^x\$. You can write that \$\displaystyle (x+1)e^x\$. Now try taking the derivative again.

- Hollywood
• Feb 25th 2013, 07:31 AM
Paze
Re: Calculating the fifth derivative
Ah, I see. Thanks.
• Feb 25th 2013, 07:32 AM
MINOANMAN
Re: Calculating the fifth derivative
its easy the nth derivative of xe^x is e^x[x+n] as simple as such.