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Mertsj

  • 3 years ago

Derivative question

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  1. myininaya
    • 3 years ago
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    yes?

  2. Mertsj
    • 3 years ago
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    |dw:1328147554735:dw|

  3. satellite73
    • 3 years ago
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    ho ho ho

  4. Mertsj
    • 3 years ago
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    Hi Santa

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

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

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

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

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

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

  11. myininaya
    • 3 years ago
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    yes it did satellite :)

  12. satellite73
    • 3 years ago
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    i am tired, time for bed. gnight

  13. myininaya
    • 3 years ago
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    me too we are going to bed

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

  15. Mertsj
    • 3 years ago
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    Not yet, I hope.

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

  17. Mertsj
    • 3 years ago
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    Maybe if I study it.

  18. myininaya
    • 3 years ago
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    :do you understand what I did?

  19. Mertsj
    • 3 years ago
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    Not quite.

  20. Mertsj
    • 3 years ago
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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.

  21. myininaya
    • 3 years ago
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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?

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

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

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

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

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

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

  28. myininaya
    • 3 years ago
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    i used chain rule for this

  29. Mertsj
    • 3 years ago
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    One second.

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

  31. myininaya
    • 3 years ago
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    ok one sec granted

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

  33. Mertsj
    • 3 years ago
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    And what happened to F(x)?

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

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

  36. Mertsj
    • 3 years ago
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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?

  37. myininaya
    • 3 years ago
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    i'm a little smart not much

  38. myininaya
    • 3 years ago
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    well i'm not really smart lol

  39. myininaya
    • 3 years ago
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    but thanks mert

  40. Mertsj
    • 3 years ago
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    thanks to you and sleep well...

  41. myininaya
    • 3 years ago
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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

  42. myininaya
    • 3 years ago
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    he likes to split the integral at a constant

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

  44. myininaya
    • 3 years ago
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    goodnight! :)

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

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