# Calculus Class Bonus Question

• Apr 1st 2009, 01:23 PM
The0wn4g3
Calculus Class Bonus Question
My teacher gave us this bonus question (said getting help was fine as long as we understand the solution), but I think I'm stumped.

"Let f(x) = 1 / sqrt(x)
Determine a function of n for the nth derivative of this function."

I wrote out the first 5 derivatives to find a pattern. Here's the function I've come up with so far:
f(x) to the nth derivative = (-1)^n * ( ) * ( 1/ sqrt(x)^ (2n+1))

I know the (1)^n takes care of the alternating signs, and the ( 1/ sqrt(x)^ (2n+1)) takes care of the power of x. I just can't think of a function to describe the middle term which should be the multiplication of the previous coefficient with the coefficient(?) of the derivative.

Some help would be greatly appreciated :) Thanks!
• Apr 1st 2009, 01:38 PM
Jameson
Welcome to MHF. You have a good start.

I wouldn't think of the general derivative in terms of $\sqrt{x}$. You start out with $x^{-\frac{1}{2}}$ and each time you perform another derivative the initial numerator is increased by 2 and the denominator stays the same. For example the first derivative has a power of (-3/2) and a coefficient of (-1/2). The second derivative has a power of (-5/2) and a coefficient of (-1/2)*(-3/2). So how does the power of the derivative relate to the last coefficient? The coefficients follow a pattern and I think the best way to express them generally is using factorials.

Are you familiar with factorials and can you do some more now or are you still stuck?
• Apr 1st 2009, 02:11 PM
The0wn4g3
Hmmm, Ok. So seeing as the coefficients for the first 6 derivatives are 1, 1/2, 3/4, 15/8, 105/6, 945/32 I would think the easiest way to express the denominator would just be 1/(2^n). As far as factorials are concerned, we haven't done much beyond having them in infinite series problems... I was thinking a factorial was the only way to gain larger and larger increases in the numerator, but I have no idea as to how to express that.

Hopefully I'm thinking in the right direction?

Thanks!
• Apr 1st 2009, 02:46 PM
Jameson
Quote:

Originally Posted by The0wn4g3
Hmmm, Ok. So seeing as the coefficients for the first 6 derivatives are 1, 1/2, 3/4, 15/8, 105/6, 945/32 I would think the easiest way to express the denominator would just be 1/(2^n). As far as factorials are concerned, we haven't done much beyond having them in infinite series problems... I was thinking a factorial was the only way to gain larger and larger increases in the numerator, but I have no idea as to how to express that.

Hopefully I'm thinking in the right direction?

Thanks!

Don't simplify the coefficients. Think of them in terms of multiplication. So for n=1, meaning the first derivative, we get $f' = -\frac{1}{2} x^{-\frac{3}{2}}$. For the second we get $f'' = \left( -\frac{3}{2}*-\frac{1}{2} \right) x^{-\frac{5}{2}}$. So you should see the pattern being that the last coefficient is going to be $\frac{2n-1}{2}$, and the next term will be $\frac{(2n-1)-2}{2}$ and so on until you reach 1/2. The sign of the total multiplication result will alternate like you said.

Look at the expression $\left( \frac{(2n-1)}{2} \right) !$ This is going to look like $\frac{2n-1}{2} * \frac{2(n-1)-1}{2} * \frac{2(n-2)-1}{2}...$. Take note that when you distribute the 2 to the parentheses in the numerators you get the pattern $\frac{2n-1}{2}*\frac{2n-3}{2}*\frac{2n-5}{2}...$ The factorial is defined for how you'll use it to only work with positive numbers. So after the numerator goes below 0, the process will stop.

Make sense so far?
• Apr 1st 2009, 03:26 PM
The0wn4g3
Oh wow, that's cool. I don't think I ever would have thought of it that way. I was trying to calculate the numerator and the denominator separately, but that makes a lot more sense!

Thanks a ton!