# Just couldnt figure this out =

1. Oct 31, 2008

### toni

why the equivalence??????

there must be some formula for this...just i donno.
i really need some hints!!! thank you guys!

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2. Oct 31, 2008

### gabbagabbahey

The derivative operator $\frac{\partial^j}{\partial x^j}$ is acting only on x...the j=0 derivative is just the zeroith order derivative of x, which is x. The j=1 derivative is $\frac{\partial}{\partial x}x=1$. Since this is a constant, all higher order derivatives are zero and so all the j>1 terms in each of the sums is zero. You are left with only the j=0 and j=1 terms which you can write out explicitly to obtain the expression on the right.

3. Oct 31, 2008

### toni

THANK YOU SOOOOOOOOOOO MUCH!

Now i understand why it goes like that!!!