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, 201114 yr by Mindrust

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

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]

Thanks.

 

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

Edited May 20, 201114 yr by Mindrust

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, 201114 yr by DJBruce