Just couldnt figure this out =

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The discussion centers on the application of the derivative operator \(\frac{\partial^j}{\partial x^j}\) in calculus, specifically regarding its action on the variable \(x\). The user clarifies that the zeroth-order derivative of \(x\) is \(x\) itself, while the first-order derivative results in a constant value of 1. Consequently, all higher-order derivatives (for \(j > 1\)) yield zero, simplifying the expression to only include the zeroth and first-order terms. This understanding resolves the user's confusion regarding the equivalence in the derivative calculations.

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toni
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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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The derivative operator [itex]\frac{\partial^j}{\partial x^j}[/itex] 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 [itex]\frac{\partial}{\partial x}x=1[/itex]. 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.
 
THANK YOU SOOOOOOOOOOO MUCH!

Now i understand why it goes like that!
 

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