Proving the n-th Derivative of f(x)=\sqrt{1-x} Using Induction

[tjr39]

Homework Statement

Prove by induction that the $$n^{th}$$ derivative of f(x)=$$\sqrt{1-x}$$ is

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

for all n geater or equal to 1

Homework Equations

The Attempt at a Solution

To start I showed that it is true for n=1.

Then I assumed true for all n=k. Now test for n=k+1.

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

From here I rearranged and multiplyed by 4/4 and (k+1)/(k+1) to obtain

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

This is where i got stuck.
Was wondering if someone could tell me if I'm on the right track and/or point me in the right direction.

I know I'm trying to get to

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

but can't quite make the leap to get there. Any help/advise would be appreciated.
Thanks

 

[Redbelly98]

You are on the right track.

First, just fixing a couple of typos (for the benefit of other helpers):
You multiplied by 4/4, not 2/2
"(2-4n)" should be (2-4k)

Next, a couple of observations:
What is the derivative of (1-x)^m? What factor did you miss when you took the derivative before? If you're stumped by that question, I'll instead ask: what is the derivative of (1-x)?

What extra terms are needed for (2k)! to become (2(k+1))! ?

 

[tjr39]

Cool thanks I got it now.