Which Derivative of f(x) = x^n Is Zero?

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The discussion revolves around determining which derivative of the function f(x) = x^n, where n is a positive integer, is identically zero. The correct answer is that the (n+1)th derivative is zero, as differentiating n times yields a constant, and the subsequent derivative of that constant results in zero. Participants emphasize understanding the process of differentiation and suggest testing simple examples like x^2 to observe the pattern. The question's wording is noted as tricky, but the underlying concept is straightforward. Ultimately, the key takeaway is that it takes n+1 differentiations to reach zero.
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i got this question from a friend and its bugging me because i cannot understand it. i just cannot understand what it means... here is it, word for word what i have on the assigned paper

"if f(x) = x^n , "n" is a positive interger, the first derivative of f(x) which is identically zero is "

A) the nth
B) the (n-1)st
C) the (n+2)nd
D) the first
E) the (n+1)st

those are the options...am I to assume that its so easy that its B? i am hesitant to pick B tho because this teacher is known for his tricks and it seemed a little too easy...a little help would be great for my friend and myself
 
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I suggest trying a few simple example like x^2 and so forth - you should see a pattern emerge! :)
 
Basically, it takes n iterations to get to a constant, and then one more.
 
Not sure if you got this already, but the question is asking "How many times do you need to differentiate this thing to get 0?"

It's worded in a tricky manner, but essentially, what do you know about differentiating a constant? How many times will you need to differentiate to get a constant? Then, how many times will you have to differentiate that to get 0?

PS - please use more descriptive thread titles. I've noticed a few threads by you with no indication as to what lies within. It makes it very difficult to get help when you need it if people skip over it!
 
The answer is n+1 times

The hint lies in the fact
The nth differential of f(X)=x^n gives a constant
And n+1th diffrential i.e of a constant gives us Zero
 
Remember that each time you take de derivative, the exponent reduces by one.
And the derivative of a constant is zero.
Ciao
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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