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..

- anonymous

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- myininaya

\[\text{ Let } f^{(0)}(x)=f(x)\]
\[\text{ Let } f^{(i)}(x) \text{ be the ith derivative of } f \]
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

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

- anonymous

Great!

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## More answers

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

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

- myininaya

You notice factorial is involved?

- myininaya

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

- anonymous

Oh, I didn't actually

- myininaya

So you almost had it :)

- anonymous

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

- myininaya

Right! Totally.

- myininaya

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

- myininaya

That is when the pattern starts to occur

- myininaya

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

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

- anonymous

n>=2

- myininaya

Yep that is exactly right :)

- anonymous

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

thing*

- anonymous

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

- myininaya

I think it is weird it says find by induction.

- myininaya

Usually I see show or prove by induction.

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

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

Recall,
\[f^{(k+1)}(x)=(f^{(k)}(x))'\]

- myininaya

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

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

- anonymous

That makes perfect sense

- myininaya

yep just like
\[f^{(3)}(x)=(f^{(2)}(x))'\]

- myininaya

Or for any known value k

- anonymous

Exactly, I get it!

- anonymous

A million thank you myininaya!

- anonymous

Really great tutor!

- myininaya

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

too*

- anonymous

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

- anonymous

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

- anonymous

now*

- myininaya

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

- myininaya

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

- myininaya

Yeah, it works out nicely :)

- anonymous

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

- myininaya

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

- anonymous

You did, really valueble!

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