anonymous
  • anonymous
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..
Geometry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • 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
  • myininaya
Ok so good so far...you have the 1st derivative right... Okay...checking your second
anonymous
  • anonymous
Great!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.