## frx Group Title 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.. one year ago one year ago

1. myininaya Group Title

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

2. myininaya Group Title

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

3. frx Group Title

Great!

4. myininaya Group Title

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

5. myininaya Group Title

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

6. frx Group Title

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

7. myininaya Group Title

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

8. myininaya Group Title

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

9. myininaya Group Title

You notice factorial is involved?

10. myininaya Group Title

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

11. frx Group Title

Oh, I didn't actually

12. myininaya Group Title

So you almost had it :)

13. frx Group Title

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

14. myininaya Group Title

Right! Totally.

15. myininaya Group Title

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

16. myininaya Group Title

That is when the pattern starts to occur

17. myininaya Group Title

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 :)

18. frx Group Title

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

19. frx Group Title

n>=2

20. myininaya Group Title

Yep that is exactly right :)

21. frx Group Title

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?

22. frx Group Title

thing*

23. frx Group Title

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

24. myininaya Group Title

I think it is weird it says find by induction.

25. myininaya Group Title

Usually I see show or prove by induction.

26. myininaya Group Title

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

27. frx Group Title

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

28. myininaya Group Title

29. frx Group Title

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

30. myininaya Group Title

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 $f^{(k+1)}(x)=\frac{(-1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}$

31. myininaya Group Title

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

32. myininaya Group Title

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

33. frx Group Title

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

34. frx Group Title

That makes perfect sense

35. myininaya Group Title

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

36. myininaya Group Title

Or for any known value k

37. frx Group Title

Exactly, I get it!

38. frx Group Title

A million thank you myininaya!

39. frx Group Title

Really great tutor!

40. myininaya Group Title

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

41. myininaya Group Title

too*

42. frx Group Title

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

43. frx Group Title

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

44. frx Group Title

now*

45. myininaya Group Title

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

46. myininaya Group Title

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

47. myininaya Group Title

Yeah, it works out nicely :)

48. frx Group Title

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

49. myininaya Group Title

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

50. frx Group Title

You did, really valueble!