## IsTim 3 years ago Differentiate, expressing each answer using positive exponents. d) y=(4x^2+3x)^-2

1. lgbasallote

you know how to use chain rule right? let u = (4x^2+3x) for a while...to solve for the derivative... $$u^{-2} du$$ can you do that? use power rule on u..then differentiate the value of u

2. IsTim

What's U? I see it in my textbook, but I don't understand.

3. lgbasallote

u is (4x^2 + 3x) i substituted it...we're going to solve the derivative this way (u^-2)(du) first..we'll solve for the derivative of u by power rule...

4. Mimi_x3

@lgbasallote: this is not integration..

5. lgbasallote

yeah...but i find it easier using u :C

6. IsTim

Here's what I wrote: h(x)=4x^2+3x, h'(x)=8x+3, g(x)=x^-2 and g'(x)=-2x. For some reason I put f'(x)=-2(8x+3) as the final answer.

7. IsTim

Wait. Where'd that comment go. I was working on that...

8. lgbasallote

derivative of x^-2 is -2x^-3

9. lgbasallote

that's the only problem with your solution...though i do not know how you got the final answer..remember that it's [g'(h(x))][h'(x)]

10. lgbasallote

raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)

11. IsTim

Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)

12. IsTim

$y'=-2(8x+3)/(4x^2+3x)^3$

13. lgbasallote

14. IsTim

I don't know how to get that...

15. lgbasallote

(4x^2 +3x)^-2..use power rule -2(4x^2 +3x)^-3 now take the derivative of 4x^2 + 3x...you get 8x +3 -2(4x^2 + 3x)^-3 (8x+3) put 4x^2 + 3x in denom... -2(8x+3)/(4x^2+3x)^3 make sense?

16. lgbasallote

LOL you forgot the -2 this time mimi :PPPP

17. Mimi_x3

$y=(4x^2+3x)^{-2}$ $y' = -2(4x^2+3x)^{-3}*(8x+3)$ $y' = \frac{-2}{4x^{2}+3x} *(8x+3)=>\frac{-2(8x+3)}{4x^{2}+3x}$

18. lgbasallote

ahh that's about right :) do you get it now @IsTim ?

19. Mimi_x3

And sorry for the various typos..haven't touched derivivatives in a while..

20. lgbasallote

for the benefit of the doubt...i'd like to write what i was starting with u... u^-2 (du) derivative of u^-2 is -2u^-3 du = derivative of (4x^2 + 3x) = 8x + 3 substitute back u to 4x^2 + 3x -2(4x^2 + 3x)^-3 (8X + 3) @Mimi_x3

21. Mimi_x3

lol, compare yoursteps to mine; which one is easier? :P

22. lgbasallote

23. lgbasallote

just kdding dont kill me

24. Mimi_x3

lol, pity a poor kid who has not touched derivatives in a LONG time! :P

25. lgbasallote

hahaha well considering as you have more medals..ill assume your method was easier :p

26. Mimi_x3

lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p

27. lgbasallote

that's the worst part mine was a pity medal :P

28. Mimi_x3

aww..i take sympathy to those who uses strange methods. :P

29. lgbasallote

=))) it's innovation i tell you :p haha

30. IsTim

Oh no. Mimi, there's a problem!

31. Mimi_x3

What's the problem?

32. IsTim

The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?

33. Mimi_x3

Woops, typo, lol..so much typos today sorry

34. lgbasallote

i told you if you'd just use my method you wont miss that :P

35. Mimi_x3

$y' = -2(4x^2+3x)^{-3}*(8x+3)=>-2*\frac{1}{(4x^{2}+3x)^{3}} *(8x+3)$ $=>\frac{-2}{(4x^{2}+3x)^{3}} *(8x+3) =>\frac{-2(8x+3)}{4x^{2}+3x^{3}}$

36. lgbasallote

haha missed it again :P

37. Mimi_x3

Man this site should have an editing option!!!!! the last one is: $\large \frac{-2(8x+3)}{(4x^{2}+3x)^{3}}$

38. lgbasallote

HAHAHAHHAA =))))

39. Mimi_x3

& igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P

40. lgbasallote

okay :C i just wanted to share my innovations

41. IsTim

I don't understand. How?

42. Mimi_x3

Which step you do not understand?

43. IsTim

When you multiply 8x+3 into the fraction thing, how does the denominator stay the same, but have a exponent of 3? I only have the vaguest idea why.

44. IsTim

I just know that the denominator is the original function, and it is being multiplied to its deriative.

45. lgbasallote

remember how $$\large a^{-n} = \frac{1}{a^n}$$?

46. Mimi_x3

that is simple algebra..which i do not know how to explain..

47. IsTim

No...

48. IsTim

I don't remember.

49. Mimi_x3

Just think as though it's simple algebra..when you go to the last step..and do it normally as you did in algebra..forget it's the derivative.

50. lgbasallote

the rules on exponent says that if you have a variable raised to a negative exponent you take it's reciprocal and change the exponent to positive

51. IsTim

Ok.

52. lgbasallote

i.e. $$\LARGE a^{-n} = \frac{1}{a^n}$$ and $$\Large \frac{1}{a^{-n}} = a^n$$