Find Nth Derivative of Square Root of X

In summary, you find the general formula for the nth derivative of x^1/2 (square root of x) by using the following steps:1) figure out the coefficients for each term in the derivative equation2) find a formula that works for the terms you have (factorials)3) prove by induction that the formula works for all n.
  • #36
0rthodontist said:
Data's expression is the same as HallsofIvy's.

your right...why didnt it work the first time i tried it? :confused:

dont know what i did. i think i substituted wrong or something.
 
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  • #37
Halls is HallsofIvy
 
  • #38
lol. i was thinking halls was some calc theorem :)
 
  • #39
occasionally an apparently useless thing (like, say, multiplying by 1, which is all I did) can allow you to avoid splitting things up into ugly "special cases"
 
  • #40
Data said:
yes, it's the same except that you can still evaluate it for [itex]n= -1[/itex] (even though the "product of the first 0 odd integers" doesn't make any sense). It happens to give dnt exactly what he needs though~
Ah, good idea.
I do think it would have been easier to generalize from example (continuing from before, at the fourth derivative):
[tex](-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}[/tex]
Now you know that you're going to change the sign every time, so the exponent on -1 is going to increase by 1 every time. The 6! came from 5 * 3 * 1 so that is going to increase by 2 every time. The 3! will increase by 1 every time, and the exponent on the 2 will increase by 1 every time from 5 * 3 * 1 and by another 1 every time from the denominator. So you can then write
[tex](-1)^{n - 1} \frac{(2 * (n - 1))!}{((n-1)!)2^{2n - 1}} \cdot x^{\frac{1}{2} - n}[/tex]
where all I did was make sure the various terms increase as they should, and otherwise adjust them by constants so they match the example. This saves you some fiddling with off-by-one errors and then you can verify it with more general notation.
 
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  • #41
HallsofIvy said:
Also, your statement that "the numerator part goes 1,-1,-3,-5" is wrong.

What you put in your very first post was:
n=1 (1st der): (1/2)(x^-1/2) = (1/2)(x^(-1/2))
n=2 (2nd der): (1/2)(-1/2)(x^-3/2) = (-1/4)(x^(-3/2))
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2) = (+3/8)(x^(-5/2))
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2) = (-15/16)(x^(-7/2))
(I've added the last in each line.)
so the numerators are 1, -1, 3= 1*3, -15= -1*3*15.

It might help you to realize this:

A product of even integers is 2*4*6*...*2n= (2*1)(2*2)(2*3)...(2*n)= (2*2*2...2)(1*2*3*...n)= 2nn!.

A product of odd integer, 1*3*5*7*...*(2n+1) is missing the even integers: multiply and divide by them:
[tex]\frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*6*...*(2n)}= \frac{(2n+1)!}{2^n(n!)}[/tex]


i konw the problem has been solved but i wanted to make sure i really understood it and one more question about the above equation popped into my head.

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

eg. if n=5 (product of first 5 integers = 1*2*3*4*5), well that should be equal to 120 which is n!

if you use (2n+1)! you get 11!, a much bigger number. can someone reexplain that? thanks.
 
  • #42
dnt said:
i konw the problem has been solved but i wanted to make sure i really understood it and one more question about the above equation popped into my head.

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

eg. if n=5 (product of first 5 integers = 1*2*3*4*5), well that should be equal to 120 which is n!

if you use (2n+1)! you get 11!, a much bigger number. can someone reexplain that? thanks.

can anyone explain it?
thanks.
 
  • #43
dnt said:
in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

it doesn't. (2n+1)! is the product of the first 2n+1 integers, of course, not the first n integers.

But [itex](2n+1)!/(n!2^n)[/itex] is the product of the first [itex]n+1[/itex] odd integers, because you're multiplying together the first 2n+1 integers and then dividing out the even ones. Here [itex]n!2^n[/itex] is the product of the first n even integers:

[tex]n!2^n = n(n-1)...(1)*2^n = (2n)(2n-2)...(2).[/tex]

So in [itex](2n+1)!/(n!2^n)[/itex], you're multiplying together the first [itex]2n+1[/itex] integers in the top, then you divide out the first [itex]n[/itex] even integers in the bottom, leaving you with the first [itex](2n+1)-n = n+1[/itex] odd integers!
 
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<h2>1. What is the formula for finding the Nth derivative of the square root of x?</h2><p>The formula for finding the Nth derivative of the square root of x is: <br>d<sup>N</sup>/dx<sup>N</sup> √x = (-1)<sup>N</sup> (2N-1)!!/2<sup>N</sup> x<sup>-(2N-1)/2</sup></p><h2>2. How do you find the first derivative of the square root of x?</h2><p>To find the first derivative of the square root of x, you can use the power rule for derivatives. This means taking the exponent (1/2) and multiplying it by the coefficient (1), and then subtracting 1 from the exponent. This results in 1/2x<sup>-1/2</sup>, which simplifies to 1/2√x.</p><h2>3. Can the Nth derivative of the square root of x be negative?</h2><p>Yes, the Nth derivative of the square root of x can be negative. This is because the (-1)<sup>N</sup> term in the formula alternates between positive and negative values for even and odd values of N. So, for example, the 2nd derivative would be positive, but the 3rd derivative would be negative.</p><h2>4. Is there a shortcut or easier way to find the Nth derivative of the square root of x?</h2><p>There is no known shortcut or easier way to find the Nth derivative of the square root of x. It is recommended to use the formula mentioned in question 1 to find the Nth derivative.</p><h2>5. Can the Nth derivative of the square root of x be simplified further?</h2><p>No, the Nth derivative of the square root of x cannot be simplified further. The formula mentioned in question 1 is the most simplified form for finding the Nth derivative of the square root of x.</p>

1. What is the formula for finding the Nth derivative of the square root of x?

The formula for finding the Nth derivative of the square root of x is:
dN/dxN √x = (-1)N (2N-1)!!/2N x-(2N-1)/2

2. How do you find the first derivative of the square root of x?

To find the first derivative of the square root of x, you can use the power rule for derivatives. This means taking the exponent (1/2) and multiplying it by the coefficient (1), and then subtracting 1 from the exponent. This results in 1/2x-1/2, which simplifies to 1/2√x.

3. Can the Nth derivative of the square root of x be negative?

Yes, the Nth derivative of the square root of x can be negative. This is because the (-1)N term in the formula alternates between positive and negative values for even and odd values of N. So, for example, the 2nd derivative would be positive, but the 3rd derivative would be negative.

4. Is there a shortcut or easier way to find the Nth derivative of the square root of x?

There is no known shortcut or easier way to find the Nth derivative of the square root of x. It is recommended to use the formula mentioned in question 1 to find the Nth derivative.

5. Can the Nth derivative of the square root of x be simplified further?

No, the Nth derivative of the square root of x cannot be simplified further. The formula mentioned in question 1 is the most simplified form for finding the Nth derivative of the square root of x.

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