Find, by induction, the n:th derivitive of
$$\frac{ x ^{2}+2x-3 }{ 4x+4 }$$

I've counted the 4 first derivitives which are:
1. $$\frac{ x ^{2}+2x+5 }{ 4(x+1)^{2} }$$
2. $$\frac{ -2 }{ (x+1)^{3} }$$
3. $$\frac{ 6 }{ (x+1)^{4} }$$
4. $$\frac{ -24 }{ (x+1)^{5} }$$
5. $$\frac{ (-1)^{n+1}*?*n }{ (x+1)^{n+1} } $$

I don't really know if i'm on the right track cause i didn't find the n:th derivitive by induction..

Question

Find, by induction, the n:th derivitive of
$$\frac{ x ^{2}+2x-3 }{ 4x+4 }$$

I've counted the 4 first derivitives which are:
1. $$\frac{ x ^{2}+2x+5 }{ 4(x+1)^{2} }$$
2. $$\frac{ -2 }{ (x+1)^{3} }$$
3. $$\frac{ 6 }{ (x+1)^{4} }$$
4. $$\frac{ -24 }{ (x+1)^{5} }$$
5. $$\frac{ (-1)^{n+1}*?*n }{ (x+1)^{n+1} } $$

I don't really know if i'm on the right track cause i didn't find the n:th derivitive by induction..

myininaya · Answer

$$\text{ Let } f^{(0)}(x)=f(x)$$
$$\text{ Let } f^{(i)}(x) \text{ be the ith derivative of } f  $$
So we have 
$$f^{(0)}(x)=\frac{x^2+2x-3}{4x+4}=\frac{x^2+2x-3}{4(x+1)}$$
$$f^{(0)}(x)=\frac{1}{4} \cdot \frac{x^2+2x-3}{x+1}$$
$$f^{(1)}(x)=\frac{1}{4} \cdot \frac{(2x+2)(x+1)-(x^2+2x-3)(1)}{(x+1)^2}$$
$$f^{(1)}(x)=\frac{1}{4} \cdot \frac{2x^2+2x+2x+2-x^2-2x+3}{(x+1)^2}$$
$$f^{(1)}(x)=\frac{1}{4} \cdot \frac{x^2+2x+5}{(x+1)^2}$$

myininaya · Answer

Ok so good so far...you have the 1st derivative right...
Okay...checking your second

anonymous · Answer

Great!

myininaya · Answer

$$f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x+1)^2-(x^2+2x+5) \cdot 2(x+1)}{(x+1)^4}$$
$$f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x^2+2x+1)-(2x^3+2x^2+4x^2+4x+10x+10)}{(x+1)^4}$$
$$f^{(2)}(x)=\frac{1}{4} \frac{(2x^3+4x^2+2x+2x^2+4x+2)-(2x^3+6x^2+14x+10)}{(x+1)^4}$$
$$f^{(2)}(x)=\frac{1}{4}\frac{(2x^3+6x^2+6x+2)-2x^3-6x^2-14x-10}{(x+1)^4}$$
$$f^{(2)}(x)=\frac{1}{4}\frac{-8x-8}{(x+1)^4}=\frac{1}{4} \frac{-8(x+1)}{(x+4)^4}$$
$$f^{(2)}(x)=\frac{-8}{4} \frac{1}{(x+4)^3}$$

myininaya · Answer

Ok that one is fine...I'm going to assume your others one are correct then lol

anonymous · Answer

Haha, yeah I think they are, a lot of work you're doing ;)

myininaya · Answer

Ok..I see part of it...one sec.

myininaya · Answer

$$f^{(2)}=\frac{2 \cdot 3}{(x+1)^3} , f^{(3)}= - \frac{2 \cdot 3 \cdot 4}{(x+1)^4}$$

myininaya · Answer

You notice factorial is involved?

myininaya · Answer

I just notice the exponents and how the top was changing.

anonymous · Answer

Oh, I didn't actually

myininaya · Answer

So you almost had it :)

anonymous · Answer

But the (-1)^n+1 should be there to control the signs right?

myininaya · Answer

Right! Totally.

myininaya · Answer

Your only expression that doesn't work is n=1
So I would say for your expression n>=2

myininaya · Answer

That is when the pattern starts to occur

myininaya · Answer

And you want for n=2 for the expression to be negative so yeah n+1 works
(-1)^{n+1} is -1 for n=2 :)

anonymous · Answer

So the it's 
$$\frac{ (-1)^{n+1} n! }{ (x+1)^{n+1} }$$
$$n$$

anonymous · Answer

n>=2

myininaya · Answer

Yep that is exactly right :)

anonymous · Answer

Wonderful! But what really made me confused was the induction think, so is the task now to prove this for n>=2 by induction or are we done?

anonymous · Answer

thing*

anonymous · Answer

Find by induction it says, so isn't this the wrong way?

myininaya · Answer

I think it is weird it says find by induction.

myininaya · Answer

Usually I see show or prove by induction.

myininaya · Answer

And I know what that means when it is written but find by induction I have trouble understanding what they mean

anonymous · Answer

Ok, maybe they just mean proove the n:th derivitive by induction

myininaya · Answer

If that is the case, I can help you with that

anonymous · Answer

I assume that's it. We've never talked about finding the derivitive by induction anyway

myininaya · Answer

So I will not be too fancy in helping you with this proof...
I will just give you what you need and you can make it all fancy if you choose 
So you need to show it is true for n=2
Then assume it is true for some integer n=k
That is, that we are assuming 
$$f^{(k)}(x)=\frac{(-1)^{k+1} k!}{(x+1)^{k+1}}$$
Now you want to show that it is true for n=k+1
That is, you want to show that
$$f^{(k+1)}(x)=\frac{(-1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}$$

myininaya · Answer

Recall,
$$f^{(k+1)}(x)=(f^{(k)}(x))'$$

myininaya · Answer

If you want it might make it easier if you do an odd k version and an even k version
I think I would do that
I see nothing wrong with it

anonymous · Answer

Oh, so proving k+1 is just the derivitive of  n=k ?

anonymous · Answer

That makes perfect sense

myininaya · Answer

yep just like
$$f^{(3)}(x)=(f^{(2)}(x))'$$

myininaya · Answer

Or for any known value k

anonymous · Answer

Exactly, I get it!

anonymous · Answer

A million thank you myininaya!

anonymous · Answer

Really great tutor!

myininaya · Answer

Aww...Thanks.
But let me know if you run into any trouble with the derivative part
Don't forget k is just a constant 
So don't look to scared when differentiating

myininaya · Answer

too*

anonymous · Answer

One last question, the derivitive of the factorial part, how does that work?

anonymous · Answer

Ohh I see it know, it nothing but a constant to :)

anonymous · Answer

now*

myininaya · Answer

You know what I don't think you actually need to do an even k and odd k

myininaya · Answer

It doesn't matter if k is even or odd, you can still do it as one case :)

myininaya · Answer

Yeah, it works out nicely :)

anonymous · Answer

Awesome! I tag you later on if i get any trouble but i doubt it :) Once again, thank you so much!:D

myininaya · Answer

No problem. I like your questions. I think I remember helping you with some other problems you presented.

anonymous · Answer

You did, really valueble!