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Mertsj Group Title

Derivative question

  • 2 years ago
  • 2 years ago

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  1. myininaya Group Title
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    yes?

    • 2 years ago
  2. Mertsj Group Title
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    |dw:1328147554735:dw|

    • 2 years ago
  3. satellite73 Group Title
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    ho ho ho

    • 2 years ago
  4. Mertsj Group Title
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    Hi Santa

    • 2 years ago
  5. myininaya Group Title
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    Let F be the antiderivative of \[f(x)=\frac{u^2-1}{u^2+1}\] Now we have \[[g(x)]'=[F(x)|_{2x}^{3x}]=[F(3x)-F(2x)]'=(3x)'f(3x)-(2x)'f(2x)\]

    • 2 years ago
  6. myininaya Group Title
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    that chain rule stuff is the best! :)

    • 2 years ago
  7. myininaya Group Title
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    so we have \[3 \cdot \frac{(3x)^2-1}{(3x)^2+1}-2 \cdot \frac{(2x)^2-1}{(2x)^2+1}\]

    • 2 years ago
  8. myininaya Group Title
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    \[3 \cdot \frac{9x^2-1}{9x^2+1}-2 \cdot \frac{4x^2-1}{4x^2+1}\]

    • 2 years ago
  9. satellite73 Group Title
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    \[-\int _a^{2x}\frac{u^2-1}{u^2+1}du+\int_a^{3x}\frac{u^2-1}{u^2+1}du\]

    • 2 years ago
  10. satellite73 Group Title
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    dang did that take me a long time to type now chain rule

    • 2 years ago
  11. myininaya Group Title
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    yes it did satellite :)

    • 2 years ago
  12. satellite73 Group Title
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    i am tired, time for bed. gnight

    • 2 years ago
  13. myininaya Group Title
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    me too we are going to bed

    • 2 years ago
  14. Mertsj Group Title
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    \[g'(x)=\frac{-2(4x^2-1)}{4x^2+1}+\frac{3(9x^2-1)}{9x^2+1}\]

    • 2 years ago
  15. Mertsj Group Title
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    Not yet, I hope.

    • 2 years ago
  16. Mertsj Group Title
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    I don't get the -2 and the 3

    • 2 years ago
  17. Mertsj Group Title
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    Maybe if I study it.

    • 2 years ago
  18. myininaya Group Title
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    :do you understand what I did?

    • 2 years ago
  19. Mertsj Group Title
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    Not quite.

    • 2 years ago
  20. Mertsj Group Title
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    I thought the u^2-1/u^2+1 would be the derivative and all I had to do was plug in the interval endpoints.

    • 2 years ago
  21. myininaya Group Title
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    \[g(x)=\int\limits_{2x}^{3x}f(u) du=F(u)|_{2x}^{3x}=F(3x)-F(2x)\] Do you understand this part?

    • 2 years ago
  22. Mertsj Group Title
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    Why isn't f(u) the g'(x)?

    • 2 years ago
  23. myininaya Group Title
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    no \[g(x)=F(3x)-F(2x)\]

    • 2 years ago
  24. myininaya Group Title
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    g is the integral from 2x to 3x of f(u) with respect to u

    • 2 years ago
  25. myininaya Group Title
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    now since we know that \[g(x)=F(3x)-F(2x)\] we can differentiate both sides

    • 2 years ago
  26. Mertsj Group Title
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    Ok. I get that and I understand what you wrote about F(3x)-F(2x)

    • 2 years ago
  27. myininaya Group Title
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    \[g'(x)=(3x)'f(3x)-(2x)'f(2x)\]

    • 2 years ago
  28. myininaya Group Title
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    i used chain rule for this

    • 2 years ago
  29. Mertsj Group Title
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    One second.

    • 2 years ago
  30. myininaya Group Title
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    \[g'(x)=3 f(3x)-2 f(2x)\] since (3x)'=3 and (2x)'=2

    • 2 years ago
  31. myininaya Group Title
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    ok one sec granted

    • 2 years ago
  32. Mertsj Group Title
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    so F(3x)=(3x)'f(x)?

    • 2 years ago
  33. Mertsj Group Title
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    And what happened to F(x)?

    • 2 years ago
  34. myininaya Group Title
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    well (F(3x))'=(3x)'f(3x) by chain rule

    • 2 years ago
  35. myininaya Group Title
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    remember we differentiate both sides so we could find g'

    • 2 years ago
  36. Mertsj Group Title
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    Yep. Ok. I think i can figure it out now. thank you very much and how did you get so smart?

    • 2 years ago
  37. myininaya Group Title
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    i'm a little smart not much

    • 2 years ago
  38. myininaya Group Title
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    well i'm not really smart lol

    • 2 years ago
  39. myininaya Group Title
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    but thanks mert

    • 2 years ago
  40. Mertsj Group Title
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    thanks to you and sleep well...

    • 2 years ago
  41. myininaya Group Title
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    thanks i'm going to sleep for reals i'm so tired by the way satellite likes to do these types of problems another way but i love my way

    • 2 years ago
  42. myininaya Group Title
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    he likes to split the integral at a constant

    • 2 years ago
  43. myininaya Group Title
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    and then use the other part of the fundamental thm of calculus

    • 2 years ago
  44. myininaya Group Title
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    goodnight! :)

    • 2 years ago
  45. Mertsj Group Title
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    Satellite is very smart also. Do you know him personally?

    • 2 years ago
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