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.

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).

Re: Calculating the fifth derivative

Re: Calculating the fifth derivative

Actually, not all that messy.....

- Hollywood

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.

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

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.. :(

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

Re: Calculating the fifth derivative

Re: Calculating the fifth derivative

its easy the nth derivative of xe^x is e^x[x+n] as simple as such.