## frx 3 years ago 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..

1. myininaya

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

2. myininaya

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

3. frx

Great!

4. myininaya

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

5. myininaya

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

6. frx

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

7. myininaya

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

8. myininaya

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

9. myininaya

You notice factorial is involved?

10. myininaya

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

11. frx

Oh, I didn't actually

12. myininaya

So you almost had it :)

13. frx

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

14. myininaya

Right! Totally.

15. myininaya

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

16. myininaya

That is when the pattern starts to occur

17. 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 :)

18. frx

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

19. frx

n>=2

20. myininaya

Yep that is exactly right :)

21. frx

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

thing*

23. frx

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

24. myininaya

I think it is weird it says find by induction.

25. myininaya

Usually I see show or prove by induction.

26. myininaya

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

27. frx

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

28. myininaya

29. frx

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

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

31. myininaya

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

32. 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

33. frx

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

34. frx

That makes perfect sense

35. myininaya

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

36. myininaya

Or for any known value k

37. frx

Exactly, I get it!

38. frx

A million thank you myininaya!

39. frx

Really great tutor!

40. 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

41. myininaya

too*

42. frx

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

43. frx

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

44. frx

now*

45. myininaya

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

46. myininaya

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

47. myininaya

Yeah, it works out nicely :)

48. frx

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

49. myininaya

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

50. frx

You did, really valueble!