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frx
Group Title
Find, by induction, the n:th derivitive of
\[\frac{ x ^{2}+2x3 }{ 4x+4 }\]
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..
 one year ago
 one year ago
frx Group Title
Find, by induction, the n:th derivitive of \[\frac{ x ^{2}+2x3 }{ 4x+4 }\] 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..
 one year ago
 one year ago

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myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\text{ Let } f^{(0)}(x)=f(x)\] \[\text{ Let } f^{(i)}(x) \text{ be the ith derivative of } f \] So we have \[f^{(0)}(x)=\frac{x^2+2x3}{4x+4}=\frac{x^2+2x3}{4(x+1)}\] \[f^{(0)}(x)=\frac{1}{4} \cdot \frac{x^2+2x3}{x+1}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{(2x+2)(x+1)(x^2+2x3)(1)}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{2x^2+2x+2x+2x^22x+3}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{x^2+2x+5}{(x+1)^2}\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Ok so good so far...you have the 1st derivative right... Okay...checking your second
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x+1)^2(x^2+2x+5) \cdot 2(x+1)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x^2+2x+1)(2x^3+2x^2+4x^2+4x+10x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x^3+4x^2+2x+2x^2+4x+2)(2x^3+6x^2+14x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{(2x^3+6x^2+6x+2)2x^36x^214x10}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{8x8}{(x+1)^4}=\frac{1}{4} \frac{8(x+1)}{(x+4)^4}\] \[f^{(2)}(x)=\frac{8}{4} \frac{1}{(x+4)^3}\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Ok that one is fine...I'm going to assume your others one are correct then lol
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Haha, yeah I think they are, a lot of work you're doing ;)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Ok..I see part of it...one sec.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[f^{(2)}=\frac{2 \cdot 3}{(x+1)^3} , f^{(3)}=  \frac{2 \cdot 3 \cdot 4}{(x+1)^4}\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You notice factorial is involved?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I just notice the exponents and how the top was changing.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Oh, I didn't actually
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So you almost had it :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
But the (1)^n+1 should be there to control the signs right?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Right! Totally.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Your only expression that doesn't work is n=1 So I would say for your expression n>=2
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
That is when the pattern starts to occur
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
So the it's \[\frac{ (1)^{n+1} n! }{ (x+1)^{n+1} }\] \[n\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Yep that is exactly right :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Find by induction it says, so isn't this the wrong way?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I think it is weird it says find by induction.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Usually I see show or prove by induction.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
And I know what that means when it is written but find by induction I have trouble understanding what they mean
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Ok, maybe they just mean proove the n:th derivitive by induction
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
If that is the case, I can help you with that
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
I assume that's it. We've never talked about finding the derivitive by induction anyway
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 \[f^{(k+1)}(x)=\frac{(1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Recall, \[f^{(k+1)}(x)=(f^{(k)}(x))'\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Oh, so proving k+1 is just the derivitive of n=k ?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
That makes perfect sense
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
yep just like \[f^{(3)}(x)=(f^{(2)}(x))'\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Or for any known value k
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Exactly, I get it!
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
A million thank you myininaya!
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Really great tutor!
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
One last question, the derivitive of the factorial part, how does that work?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Ohh I see it know, it nothing but a constant to :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You know what I don't think you actually need to do an even k and odd k
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
It doesn't matter if k is even or odd, you can still do it as one case :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Yeah, it works out nicely :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
Awesome! I tag you later on if i get any trouble but i doubt it :) Once again, thank you so much!:D
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
No problem. I like your questions. I think I remember helping you with some other problems you presented.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.1
You did, really valueble!
 one year ago
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