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IsTim

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

  • one year ago
  • one year ago

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  1. lgbasallote
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    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

    • one year ago
  2. IsTim
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    What's U? I see it in my textbook, but I don't understand.

    • one year ago
  3. lgbasallote
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    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...

    • one year ago
  4. Mimi_x3
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    @lgbasallote: this is not integration..

    • one year ago
  5. lgbasallote
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    yeah...but i find it easier using u :C

    • one year ago
  6. IsTim
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    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.

    • one year ago
  7. IsTim
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    Wait. Where'd that comment go. I was working on that...

    • one year ago
  8. lgbasallote
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    derivative of x^-2 is -2x^-3

    • one year ago
  9. lgbasallote
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    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)]

    • one year ago
  10. lgbasallote
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    raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)

    • one year ago
  11. IsTim
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    Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)

    • one year ago
  12. IsTim
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    \[y'=-2(8x+3)/(4x^2+3x)^3\]

    • one year ago
  13. lgbasallote
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    yup seems about right

    • one year ago
  14. IsTim
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    I don't know how to get that...

    • one year ago
  15. lgbasallote
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    (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?

    • one year ago
  16. lgbasallote
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    LOL you forgot the -2 this time mimi :PPPP

    • one year ago
  17. Mimi_x3
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    \[y=(4x^2+3x)^{-2}\] \[y' = -2(4x^2+3x)^{-3}*(8x+3)\] \[y' = \frac{-2}{4x^{2}+3x} *(8x+3)=>\frac{-2(8x+3)}{4x^{2}+3x}\]

    • one year ago
  18. lgbasallote
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    ahh that's about right :) do you get it now @IsTim ?

    • one year ago
  19. Mimi_x3
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    And sorry for the various typos..haven't touched derivivatives in a while..

    • one year ago
  20. lgbasallote
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    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

    • one year ago
  21. Mimi_x3
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    lol, compare yoursteps to mine; which one is easier? :P

    • one year ago
  22. lgbasallote
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    mine had 3 steps..yours had 3 reposts :p

    • one year ago
  23. lgbasallote
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    just kdding dont kill me

    • one year ago
  24. Mimi_x3
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    lol, pity a poor kid who has not touched derivatives in a LONG time! :P

    • one year ago
  25. lgbasallote
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    hahaha well considering as you have more medals..ill assume your method was easier :p

    • one year ago
  26. Mimi_x3
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    lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p

    • one year ago
  27. lgbasallote
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    that's the worst part mine was a pity medal :P

    • one year ago
  28. Mimi_x3
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    aww..i take sympathy to those who uses strange methods. :P

    • one year ago
  29. lgbasallote
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    =))) it's innovation i tell you :p haha

    • one year ago
  30. IsTim
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    Oh no. Mimi, there's a problem!

    • one year ago
  31. Mimi_x3
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    What's the problem?

    • one year ago
  32. IsTim
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    The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?

    • one year ago
  33. Mimi_x3
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    Woops, typo, lol..so much typos today sorry

    • one year ago
  34. lgbasallote
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    i told you if you'd just use my method you wont miss that :P

    • one year ago
  35. Mimi_x3
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    \[y' = -2(4x^2+3x)^{-3}*(8x+3)=>-2*\frac{1}{(4x^{2}+3x)^{3}} *(8x+3)\] \[=>\frac{-2}{(4x^{2}+3x)^{3}} *(8x+3) =>\frac{-2(8x+3)}{4x^{2}+3x^{3}} \]

    • one year ago
  36. lgbasallote
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    haha missed it again :P

    • one year ago
  37. Mimi_x3
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    Man this site should have an editing option!!!!! the last one is: \[\large \frac{-2(8x+3)}{(4x^{2}+3x)^{3}} \]

    • one year ago
  38. lgbasallote
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    HAHAHAHHAA =))))

    • one year ago
  39. Mimi_x3
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    & igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P

    • one year ago
  40. lgbasallote
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    okay :C i just wanted to share my innovations

    • one year ago
  41. IsTim
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    I don't understand. How?

    • one year ago
  42. Mimi_x3
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    Which step you do not understand?

    • one year ago
  43. IsTim
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    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.

    • one year ago
  44. IsTim
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    I just know that the denominator is the original function, and it is being multiplied to its deriative.

    • one year ago
  45. lgbasallote
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    remember how \(\large a^{-n} = \frac{1}{a^n}\)?

    • one year ago
  46. Mimi_x3
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    that is simple algebra..which i do not know how to explain..

    • one year ago
  47. IsTim
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    No...

    • one year ago
  48. IsTim
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    I don't remember.

    • one year ago
  49. Mimi_x3
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    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.

    • one year ago
  50. lgbasallote
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    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

    • one year ago
  51. IsTim
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    Ok.

    • one year ago
  52. lgbasallote
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    i.e. \(\LARGE a^{-n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{-n}} = a^n\)

    • one year ago
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