Find the nth derivative of sqr(2x-1)

[Mindrust]

Find the nth derivative of sqr(2x-1).

 

That means we need to find a pattern. Here's what I've got so far, from 1st derivative to 5th:

 

 

The pattern for the exponent is simple, 1/2 - n

 

The pattern for the co-efficient is what I'm stuck on. Here's what it looks like. I'm not sure how to express it mathematically.

 

 

1 -> 1 = 1

2 -> -1 = 1(-1)

3 -> 3 = 1(-1)(-3)

4 -> -15 = 1(-1)(-3)(-5)

5 -> 105 = 1(-1)(-3)(-5)(-7)

 

and so on...

 

 

Anyone know how to make a function for the coefficient in terms of n?

Edited May 20, 2011 by Mindrust

[Bignose]

[math] \prod_{i=1}^n (-1)^{(i-1)}(2i-1) [/math]

[DrRocket]

Find the nth derivative of sqr(2x-1).

 

That means we need to find a pattern. Here's what I've got so far, from 1st derivative to 5th:

 

 

The pattern for the exponent is simple, 1/2 - n

 

The pattern for the co-efficient is what I'm stuck on. Here's what it looks like. I'm not sure how to express it mathematically.

 

 

1 -> 1 = 1

2 -> -1 = 1(-1)

3 -> 3 = 1(-1)(-3)

4 -> -15 = 1(-1)(-3)(-5)

5 -> 105 = 1(-1)(-3)(-5)(-7)

 

and so on...

 

 

Anyone know how to make a function for the coefficient in terms of n?

 

[math] \frac{d}{dx} (2x-1)^{\frac{1}{2}} = (2x-1)^{\frac {-1}{2}}[/math]

 

[math] \frac{d^2}{dx^2} (2x-1)^{\frac{1}{2}} = -(2x-1)^{\frac {-3}{2}}[/math]

 

[math]\frac{d^3}{dx^3} (2x-1)^{\frac{1}{2}} = 3(2x-1)^{\frac {-5}{2}}[/math]

 

.

.

.

[math]\frac{d^n}{dx^n} (2x-1)^{\frac{1}{2}} = ( (-1)^{n}\displaystyle\prod_{k=0}^{n-1} (2k-1) )(2x-1)^{\frac {-2n+1}{2}} [/math]

[Mindrust]

Thanks.

 

BTW, what does that big "pi" looking symbol mean?

Edited May 20, 2011 by Mindrust

[DJBruce]

Thanks.

 

BTW, what does that big "pi" looking symbol mean?

 

It is similar to summation notation, but instead of adding each terms you are going to multiply them together. So for example,

 

[math]\prod_{k=1}^n \frac{1}{x_{k}}=1*\frac{1}{2}*\frac{1}{3}*...*\frac{1}{n} [/math]

Edited May 20, 2011 by DJBruce