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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..

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\[\text{ Let } f^{(0)}(x)=f(x)\] \[\text{ Let } f^{(i)}(x) \text{ be the ith derivative of } f \] So we have \[f^{(0)}(x)=\frac{x^2+2x-3}{4x+4}=\frac{x^2+2x-3}{4(x+1)}\] \[f^{(0)}(x)=\frac{1}{4} \cdot \frac{x^2+2x-3}{x+1}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{(2x+2)(x+1)-(x^2+2x-3)(1)}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{2x^2+2x+2x+2-x^2-2x+3}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{x^2+2x+5}{(x+1)^2}\]
Ok so good so have the 1st derivative right... Okay...checking your second

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\[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x+1)^2-(x^2+2x+5) \cdot 2(x+1)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x^2+2x+1)-(2x^3+2x^2+4x^2+4x+10x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x^3+4x^2+2x+2x^2+4x+2)-(2x^3+6x^2+14x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{(2x^3+6x^2+6x+2)-2x^3-6x^2-14x-10}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{-8x-8}{(x+1)^4}=\frac{1}{4} \frac{-8(x+1)}{(x+4)^4}\] \[f^{(2)}(x)=\frac{-8}{4} \frac{1}{(x+4)^3}\]
Ok that one is fine...I'm going to assume your others one are correct then lol
Haha, yeah I think they are, a lot of work you're doing ;)
Ok..I see part of sec.
\[f^{(2)}=\frac{2 \cdot 3}{(x+1)^3} , f^{(3)}= - \frac{2 \cdot 3 \cdot 4}{(x+1)^4}\]
You notice factorial is involved?
I just notice the exponents and how the top was changing.
Oh, I didn't actually
So you almost had it :)
But the (-1)^n+1 should be there to control the signs right?
Right! Totally.
Your only expression that doesn't work is n=1 So I would say for your expression n>=2
That is when the pattern starts to occur
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 :)
So the it's \[\frac{ (-1)^{n+1} n! }{ (x+1)^{n+1} }\] \[n\]
Yep that is exactly right :)
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?
Find by induction it says, so isn't this the wrong way?
I think it is weird it says find by induction.
Usually I see show or prove by induction.
And I know what that means when it is written but find by induction I have trouble understanding what they mean
Ok, maybe they just mean proove the n:th derivitive by induction
If that is the case, I can help you with that
I assume that's it. We've never talked about finding the derivitive by induction anyway
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 \[f^{(k+1)}(x)=\frac{(-1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}\]
Recall, \[f^{(k+1)}(x)=(f^{(k)}(x))'\]
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
Oh, so proving k+1 is just the derivitive of n=k ?
That makes perfect sense
yep just like \[f^{(3)}(x)=(f^{(2)}(x))'\]
Or for any known value k
Exactly, I get it!
A million thank you myininaya!
Really great tutor!
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
One last question, the derivitive of the factorial part, how does that work?
Ohh I see it know, it nothing but a constant to :)
You know what I don't think you actually need to do an even k and odd k
It doesn't matter if k is even or odd, you can still do it as one case :)
Yeah, it works out nicely :)
Awesome! I tag you later on if i get any trouble but i doubt it :) Once again, thank you so much!:D
No problem. I like your questions. I think I remember helping you with some other problems you presented.
You did, really valueble!

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