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swanny Group Title

derivative of (x^(2))(e^(1/x))

  • 2 years ago
  • 2 years ago

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  1. swanny Group Title
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    @Callisto

    • 2 years ago
  2. swanny Group Title
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    i got (2x)(e^(1/x))+x^(2)(e^(1/x) which is wrong

    • 2 years ago
  3. satellite73 Group Title
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    yeah it is

    • 2 years ago
  4. satellite73 Group Title
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    \[g(x)=e^{\frac{1}{x}}\] \[g'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}\] by the chain rule

    • 2 years ago
  5. satellite73 Group Title
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    so the second part of your product rule is the problem. first part is good

    • 2 years ago
  6. swanny Group Title
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    but how is it (-1/x^2) e^(1/x)?

    • 2 years ago
  7. swanny Group Title
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    shouldn't the derivative e^(1/x)

    • 2 years ago
  8. Callisto Group Title
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    \[y=e^{\frac{1}{x}}\] Let u = 1/x \[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\]

    • 2 years ago
  9. swanny Group Title
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    1?

    • 2 years ago
  10. swanny Group Title
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    or 0

    • 2 years ago
  11. swanny Group Title
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    by quotient rule i did (1)'(x)-(1)(x)'/(x)^@

    • 2 years ago
  12. Callisto Group Title
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    You can use power for d/dx (1/x) \[\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} (x^{-1}) = (-1)x^{-1-1} =...?\]

    • 2 years ago
  13. swanny Group Title
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    -x^(-2)?

    • 2 years ago
  14. Callisto Group Title
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    Yes.

    • 2 years ago
  15. swanny Group Title
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    ok so it will be e^(-x^(-2))?

    • 2 years ago
  16. Callisto Group Title
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    The power of e would not change when you differentiate e^(something)!

    • 2 years ago
  17. Callisto Group Title
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    \[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\] u = 1/x and you found d/dx(1/x). So......

    • 2 years ago
  18. swanny Group Title
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    i'm lost with this one

    • 2 years ago
  19. Callisto Group Title
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    Where?

    • 2 years ago
  20. swanny Group Title
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    are you asking for the derivative of (1/x) above? i dont understand what exactly you are asking

    • 2 years ago
  21. Callisto Group Title
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    I asked derivative of 1/x because you didn't get that right. To get the derivative of derivative of (x^(2))(e^(1/x)), it should be like this: \[y = x^2e^{\frac{1}{x}}\]\[y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)\] Then, then you need to work out what d/dx (e^(1/x)) is and d/dx (x^2) are. The later one is easy, and you got that right. The problem is to find d/dx (e^(1/x)) So, let u = 1/x \[\frac{d}{dx}e^{\frac{1}{x}} = \frac{d}{du}(e^u) \times \frac{d}{dx} (\frac{1}{x}) = ...?\]

    • 2 years ago
  22. Callisto Group Title
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    Is that clear? Do you understand what we are working on?

    • 2 years ago
  23. swanny Group Title
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    but isn't derivative of (1/x) = -x^(-2) ? i thought we got that right

    • 2 years ago
  24. swanny Group Title
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    wait. give me second

    • 2 years ago
  25. Callisto Group Title
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    Yes. d/dx (1/x) = -x^(-2) = -1/x^2

    • 2 years ago
  26. swanny Group Title
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    so you are asking derivative of e^(1/x) * -1/x^2?

    • 2 years ago
  27. Callisto Group Title
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    No. Instead, derivative of e^(1/x), which is equal to e^(1/x) * -1/x^2. Do you understand how to get it?

    • 2 years ago
  28. swanny Group Title
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    so looking at the original question question my answer should be (2x)(e^1/x)-(x^2)(e^(1/x))?

    • 2 years ago
  29. swanny Group Title
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    derivative of (x^(2))(e^(1/x)) is the original question

    • 2 years ago
  30. Callisto Group Title
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    No... The second part is not correct. For the second part, you need to find the derivative of e^(1/x). What is it?

    • 2 years ago
  31. swanny Group Title
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    isn't it e^(1/x)

    • 2 years ago
  32. swanny Group Title
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    e^(1/x) * -1/x^2?

    • 2 years ago
  33. Callisto Group Title
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    It is *NOT* e^(1/x) e^(1/x) * -1/x^2 ^How do you get it?

    • 2 years ago
  34. swanny Group Title
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    because derivative of 1/x = -1/x^2

    • 2 years ago
  35. Callisto Group Title
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    Yup.. So, can you solve \(y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)\) now?

    • 2 years ago
  36. Callisto Group Title
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    ^That is your derivative btw.

    • 2 years ago
  37. swanny Group Title
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    yes i had already gotten the first part. just the second was the problem

    • 2 years ago
  38. Callisto Group Title
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    Just show us what you've got.

    • 2 years ago
  39. swanny Group Title
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    final answer for the original question is (2x)(e^(1/x))-(x^2)(1/x^2)(e^(1/x))

    • 2 years ago
  40. swanny Group Title
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    this is what i got

    • 2 years ago
  41. Callisto Group Title
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    That looks pretty cool~ But you can simplify the last term.

    • 2 years ago
  42. swanny Group Title
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    i don't think we have to. but you can show me if you want

    • 2 years ago
  43. Callisto Group Title
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    \[(2x)(e^{\frac{1}{x}})-(x^2)\frac{1}{x^2}(e^{\frac{1}{x}}) = (2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =...?\]

    • 2 years ago
  44. swanny Group Title
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    x^2 / x^2 gets cancelled out = 1

    • 2 years ago
  45. Callisto Group Title
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    Yes.

    • 2 years ago
  46. swanny Group Title
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    perfect.

    • 2 years ago
  47. swanny Group Title
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    Thank you very much for your help

    • 2 years ago
  48. swanny Group Title
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    you explain concepts very well

    • 2 years ago
  49. Callisto Group Title
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    You're welcome. I hope I didn't confuse you :S \[(2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =(2x)(e^{\frac{1}{x}})-e^{\frac{1}{x}}=e^{\frac{1}{x}}(2x-1)\]That looks nice :)

    • 2 years ago
  50. swanny Group Title
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    No, not confusing at all

    • 2 years ago
  51. swanny Group Title
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    yes that looks better

    • 2 years ago
  52. Callisto Group Title
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    :)

    • 2 years ago
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