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diamondboy

  • one year ago

Just a quick question

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  1. diamondboy
    • one year ago
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    |dw:1442283012866:dw|

  2. myininaya
    • one year ago
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    yes for x>0

  3. diamondboy
    • one year ago
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    alright

  4. diamondboy
    • one year ago
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    could you help with one more?

  5. myininaya
    • one year ago
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    k

  6. diamondboy
    • one year ago
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    or maybe I should say more?

  7. diamondboy
    • one year ago
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    |dw:1442283174794:dw|

  8. diamondboy
    • one year ago
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    find y prime

  9. diamondboy
    • one year ago
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    is it 3/x?

  10. myininaya
    • one year ago
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    \[y=\ln(5x^3) =\ln(5)+3 \ln(x) \\ y'=0+3 \frac{1}{x}=\frac{3}{x} \\ \text{ yep or you could have done } \\ y=\ln(5x^3) \implies y'=(5x^3)' \frac{1}{5x^3}=15x^2 \frac{1}{5x^3} =\frac{3}{x}\]

  11. diamondboy
    • one year ago
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    yes... I did it d second way

  12. diamondboy
    • one year ago
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    so I need to understand something about this chain rule

  13. myininaya
    • one year ago
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    \[\text{ Let } f(x)>0 \\ y=\ln(f(x)) \implies y'=f'(x) \cdot \frac{1}{f(x)} \]

  14. diamondboy
    • one year ago
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    Oh ok...now I see it

  15. myininaya
    • one year ago
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    there was another way... to find y' for the earlier problem \[y=\ln(5x^3) \\ e^y=5x^3 \\ \text{ now differentiate both sides } \\ y'e^{y}=15x^2 \\ \text{ solve for } y' \\ y'=15x^2 \cdot \frac{1}{e^y} \\ \text{ but } e^y=5x^3 \\ \text{ so } y'=15x^2 \cdot \frac{1}{5x^3}\]

  16. diamondboy
    • one year ago
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    ok

  17. diamondboy
    • one year ago
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    |dw:1442283585590:dw| is that true?

  18. myininaya
    • one year ago
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    well what did you get for f'(x)?

  19. diamondboy
    • one year ago
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    e^(-2x)*-2

  20. myininaya
    • one year ago
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    ok so you differentiate -2*(-2)=4 not -4

  21. diamondboy
    • one year ago
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    yes thats true

  22. myininaya
    • one year ago
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    \[y=e^{-2x} \\ y'=(-2x)'e^{-2x}=-2 e^{-2x} \\ y''=-2(-2x)'e^{-2x}\]

  23. myininaya
    • one year ago
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    ok so you differentiate again but the coefficient should be -2*(-2)=4 not -4 *

  24. diamondboy
    • one year ago
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    ok so I used the product rule to solve for d second derivative

  25. myininaya
    • one year ago
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    product rule is not needed

  26. diamondboy
    • one year ago
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    yea I get that

  27. myininaya
    • one year ago
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    constant multiple rule and chain rule

  28. myininaya
    • one year ago
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    though you could use product rule

  29. diamondboy
    • one year ago
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    ok

  30. diamondboy
    • one year ago
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    well what do you think?

  31. diamondboy
    • one year ago
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    o I leave you without a medal?....anyways thanks

  32. myininaya
    • one year ago
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    had internet connection failure

  33. myininaya
    • one year ago
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    or at least openstudy internet connection failure :p

  34. myininaya
    • one year ago
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    what do I think about what?

  35. myininaya
    • one year ago
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    you could use product rule but I think it is shorter to use constant multiple rule is that what you are talking about?

  36. myininaya
    • one year ago
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    \[y'=-2e^{-2x} \\ y''=-2 (e^{-2x})' \text{ by constant multiple rule } \\ \text{ or as I said you could have use product rule to get this same result } \\ y'' =(-2)'e^{-2x}+-2(e^{-2x})'=0e^{-2x}-2(e^{-2x})'=-2(e^{-2x})'\]

  37. myininaya
    • one year ago
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    where you already found the derivative of exp(-2x) w.r.t . x

  38. diamondboy
    • one year ago
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    yep

  39. diamondboy
    • one year ago
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    wait wat do u mean?

  40. myininaya
    • one year ago
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    are you asking what I mean by finding the derivative of exp(-2x) w.r.t x?

  41. myininaya
    • one year ago
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    \[\frac{d}{dx}e^{-2x}=(-2)e^{-2x}=-2e^{-2x}\]

  42. diamondboy
    • one year ago
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    oh noo

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