# Thread: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

1. ## Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Hello

So I tried solving this substituting x = 0 then I get f'(x) = 0.
But that seems to be wrong.
If I use the product law of derivatives then I get 0*(0-1)*(0-2)....=0
Why we can't substitute x=0 first, then take the derivative of the entire expression, instead we have to first take the derivative then substitute x=0 which gives another answer...
How to calculate this properly?

Thanks.

2. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by ryu1
Hello

So I tried solving this substituting x = 0 then I get f'(x) = 0.
But that seems to be wrong.
If I use the product law of derivatives then I get 0*(0-1)*(0-2)....=0
Why we can't substitute x=0 first, then take the derivative of the entire expression, instead we have to first take the derivative then substitute x=0 which gives another answer...
How to calculate this properly?

Thanks.
you did it wrong because you did it backward.

You have to take the derivative of a function, and after that to find f'(0) for the derivative you got (which has to be considered as a function),

just think about what you understand by x and f(x), or x and f'(x), what's the difference between those two?

btw, what did you obtain by blugging x=0 directly in your function f(x)?

3. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by dokrbb
you did it wrong because you did it backward.

You have to take the derivative of a function, and after that to find f'(0) for the derivative you got (which has to be considered as a function),

just think about what you understand by x and f(x), or x and f'(x), what's the difference between those two?

btw, what did you obtain by blugging x=0 directly in your function f(x)?
Oh I see, i got f'(0) = 0 which is like blugging(?) x=0 directly in the function f(x). but that is wrong, which is why we first take the derivative then plunging (is that the right word? :P) the value for x=0..
So how do you normally take the derivative of a polynomial this long
x(x-1)(x-2)....(x-96) ? theres obviously a trick here, no one would (normally) let you manually calculate it...but I can't figure it out.

4. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

You don't need to find the general form of f'(x), you only need to find f'(0). The function f(x) is a big polynomial, of order x^97:

f(x) = x(x-1)(x-2)...(x-96) = x^(97) + Ax^(96) + Bx^(95) + .... + Wx^3 + Yx^2 + Zx

where the A, B, etc terms are constants. Its derivative is also a polynomial, of order one less than the original polynomial:

f'(x) = 97x^(96) + 96Ax^(95) + 95Bx^(94) + ... + 3Wx^2 + 2Yx + Z

Note that at x=0 all these terms equal zero except for the last, so you have f'(0) = Z. Now the only thing left to do is figure out the value of Z, which is the coefficient of the x^1 term in the original polynomial. I'll leave that to you!

5. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by ebaines
You don't need to find the general form of f'(x), you only need to find f'(0). The function f(x) is a big polynomial, of order x^97:

f(x) = x(x-1)(x-2)...(x-96) = x^(97) + Ax^(96) + Bx^(95) + .... + Wx^3 + Yx^2 + Zx

where the A, B, etc terms are constants. Its derivative is also a polynomial, of order one less than the original polynomial:

f'(x) = 97x^(96) + 96Ax^(95) + 95Bx^(94) + ... + 3Wx^2 + 2Yx + Z

Note that at x=0 all these terms equal zero except for the last, so you have f'(0) = Z. Now the only thing left to do is figure out the value of Z, which is the coefficient of the x^1 term in the original polynomial. I'll leave that to you!
I get something crazy like -1*-2*-3...*-96
Is that right?
I think it's called factorial 96! ?
I read we can't calculate the factorial of negative numbers but (we can say it's positive because it's an even number therefore -96! = 96! ?)

Thanks.

6. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by ryu1
I get something crazy like -1*-2*-3...*-96
Is that right?
I think it's called factorial 96! ?
I read we can't calculate the factorial of negative numbers but (we can say it's positive because it's an even number therefore -96! = 96! ?)
This question can be done using logarithmic differentiation. But I suggest doing a simple example.

Let $f(x)=x(x-1)(x-2)$ then $f'(x)=(x-1)(x-2)+x(x-2)+x(x-1)$. Is that correct?

If so is it clear that $f'(0)=2~?$

Isn't easy to generalize?

Post Script

$f(x) = \prod\limits_{k = 0}^{96} {\left( {x - k} \right)}$

$f'(x) = \sum\limits_{j = 0}^{96} {\left( {\prod\limits_{k = 0,k \ne j}^{96} {\left( {x - k} \right)} } \right)}$

$f'(0)=~?$

7. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

$f'(x)=\prod_{i=1}^{96}(x-i)+x\left[\prod_{i=2}^{96}(x-i)+(x-1)\prod_{i=2}^{96}(x-i)+\cdots\right]\rightarrow f'(0)=\prod_{i=1}^{96}(-i)=96!$

8. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by math2009
$f'(x)=\prod_{i=1}^{96}(x-i)+x\left[\prod_{i=2}^{96}(x-i)+(x-1)\prod_{i=2}^{96}(x-i)+\cdots\right]\rightarrow f'(0)=\prod_{i=1}^{96}(-i)=96!$
So well, How does that help? Our object is to help? learning?

9. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by Plato
This question can be done using logarithmic differentiation. But I suggest doing a simple example.

Let $f(x)=x(x-1)(x-2)$ then $f'(x)=(x-1)(x-2)+x(x-2)+x(x-1)$. Is that correct?

If so is it clear that $f'(0)=2~?$

Isn't easy to generalize?

Post Script

$f(x) = \prod\limits_{k = 0}^{96} {\left( {x - k} \right)}$

$f'(x) = \sum\limits_{j = 0}^{96} {\left( {\prod\limits_{k = 0,k \ne j}^{96} {\left( {x - k} \right)} } \right)}$

$f'(0)=~?$
yes f'(0) = 2 in that case.
so if i have x(x-1)(x-2)...(x-n) f'(0) = n ? or n! ?
But I got 96! thing.

I have no idea what those big E and big pi thing operators mean yet.

10. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by ryu1
yes f'(0) = 2 in that case.
so if i have x(x-1)(x-2)...(x-n) f'(0) = n ? or n! ?
But I got 96! thing.

I have no idea what those big E and big pi thing operators mean yet.

In this case I must ask why in the world were yo you given this question? Frankly if I were you, I would complain loudly to your educational authority. It seems to be that your instructor is of bass on this one. You can use this post if you want.

11. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

You could accost somebody at your school, or you could take another look at posts 4 and 5. Do some simpler examples. Let $f(x)=x(x-1)$. Expand this and then take the derivative (as ebaines said, you really only need to know one term of $f(x)$ to find $f'(0)$). Now let $f(x)=x(x-1)(x-2)$. Expand this and then take the derivative. If you keep going and do this one more time, you should answer your $n$ or $n!$ question.

12. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by Plato
In this case I must ask why in the world were yo you given this question? Frankly if I were you, I would complain loudly to your educational authority. It seems to be that your instructor is of bass on this one. You can use this post if you want.
it is meant to practice only the derivative function.

someone offered to to do this

let g(x) = (x-1)(x-2)...(x-96)
let f(x) = xg(x)
f'(x) = xg'(x) + g(x)*1
f'(0) = 0g'(0) + 1g(0)
g(0) = (-1)(-2)...(-96)

does that work?

Thanks again.

13. ## Re: Derivative f'(0) for f(x) =x*(x-1)(x-2)....(x-96)

Originally Posted by ryu1
does that work?

Thanks again.
Yes that works - it's an interesting technique. So now you have two approaches that both give the same answer!

Originally Posted by Plato
In this case I must ask why in the world were yo you given this question? Frankly if I were you, I would complain loudly to your educational authority. It seems to be that your instructor is of bass on this one. You can use this post if you want.
I don't see the issue, as there are at least two ways to solve this without employing symbology that many college freshmen or seniors in highschool have not been exposed to. I'm referring here specifically to the use of the product operator; admittedly the OP should be familiar with the summation symbol - he must have been asleep in class when they covered that topic.