Expert: Paul Klarreich - 10/24/2006

Question
Hello again,
my name is David and I'm taking cal.1 right now.
The question is : What function f(x) has a fourth derivative equal to 1?

Well, I was thinking of working backward:
Original Function:
1st d/dx:
2nd d/dx:?
3rd d/dx: x
4th d/dx: 1
Now im stuck becuase I dont know any derivative that equals to x.

Answer
Questioner:  David S.
Category:  Calculus
 
Subject:  Derivative
Question:  Hello again,
my name is David and I'm taking cal.1 right now.
The question is : What function f(x) has a fourth derivative equal to 1?

Well, I was thinking of working backward:
Original Function:
1st d/dx:
2nd d/dx:?
3rd d/dx: x
4th d/dx: 1
Now im stuck becuase I dont know any derivative that equals to x.
-----------------------------------------------------------
Hi, David,

Good news -- it isn't really hard.  Bad news -- the answer is NOT UNIQUE.  There are many answers.

So, you are on the right track asking:

What function has a derivative that is 1?

Well, f(x) = x is one answer, but how about  f(x) = x + 7?  Isn't it's derivative equal to 1?  And what about  f(x) = x - 29?

Or in general,  f(x) = x + C1, where C1 is any constant term, which can be positive or negative or zero.

We'll deal with that as we go along.

Now can you think of a function whose derivative is  x?   You do recall one of the derivative formulas that says:

If  f(x) = x^n,  then  f'(x) = nx^(n-1).

Put another way, if you differentiate a power of x, the exponent decreases by 1.  So if you differentiated something, and got  x, which is  x^1, and it went down by 1, what was it before that?

Now you have the answer.  It must have had  x^2 in it.  But was f(x) = x^2  the right answer?  Not exactly.

If f(x) = x^2, then  f'(x) = 2x, not x -- it's too big by a factor of 2.  Solution:

Cut the original by a factor of 2.

Try  f(x) = x^2/2.  f'(x) = (2x)/2 = x.

But again, if x^2/2 is correct, then so is  x^2/2 + 17, or - 8,  or + 5.3, or, in general, any constant.

I think you see the scheme now.  As you work your way back, the powers of x increase by 1 at each step, but you must divide by the new exponent as a compensation.

AND you have to add in a constant term, which counts in the process, too.  So the answers you get look like:

Fourth derivative was  1
Third ..........  was  x + C1
Second ...........was  x^2/2 + C1x + C2
First ........... was  x^3/6 + C1x^2/2 + C2x + C3
Original function was  x^4/24 + C1x^3/6 + C2x^2/2 + C3x + C4

Easy, right?  Just what you expected.