1. ## n+1 th derivative

Hi,

f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

I kind of know the answer for this but just don't know how to get there.(need steps).

f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}

2. Originally Posted by ammayi
Hi,

f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

I kind of know the answer for this but just don't know how to get there.(need steps).
Well, there are two ways to do this. Firstly, you can substitute $(n+1)$ for $n$, which is what you've done here:

f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}
And which we can simplify:

$f^{(n+1)}(x)=-\frac{(2n+2)!}{(4n+4)(n+1)!(2n+1)}(1-x)^{(\frac{1}{2}-n-1)}$

$=-\frac{(2n)!(2n+1)(2n+2)}{2(2n+2)(n+1)!(2n+1)}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{2(n+1)!}(1-x)^{(\frac{1}{2}-n-1)}$

$=-\frac{(2n)!}{n!2(n+1)}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{n!(2n+2)}(1-x)^{(\frac{1}{2}-n-1)}$

To check our work, we can take the derivative of $f^n(x)$, like so:

$f^{(n+1)}(x)=\frac{d}{dx}[-\frac{(2n)!}{4n!n(2n-1)}(1-x)^{(\frac{1}{2}-n)}]=-\frac{(2n)!}{4n!n(2n-1)}\frac{d}{dx}[(1-x)^{(\frac{1}{2}-n)}]$

$=\frac{(2n)!}{4n!n(2n-1)}(\frac{1}{2}-n)(1-x)^{(\frac{1}{2}-n-1)}=[\frac{(2n)!}{4n!n(2n-1)2}-\frac{(2n)!n}{4n!n(2n-1)}](1-x)^{(\frac{1}{2}-n-1)}$

$=[\frac{(2n)!}{4n!n(2n-1)2}-\frac{2(2n)!n}{4n!n(2n-1)2}](1-x)^{(\frac{1}{2}-n-1)}=\frac{(2n)!-2(2n)!n}{4n!n(2n-1)2}(1-x)^{(\frac{1}{2}-n-1)}$

$=\frac{(2n)!(1-2n)}{4n!n(2n-1)2}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{n!8n}(1-x)^{(\frac{1}{2}-n-1)}$

And we can see that this does not check out. So, something is either wrong with my work or with the original functions. I believe it is the latter.

3. Originally Posted by ammayi
Hi,

f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

I kind of know the answer for this but just don't know how to get there.(need steps).

f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}

To differentiate, multiply by the power, $\frac{1}{2} - n$, and substract 1 off the power. Don't forget to also multiply by -1:
f^(n+1).(x) = $\left(\frac{1}{2} - n\right)$ [(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2− n - 1}