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ta123

  • 3 years ago

need help with the attachment

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  1. ta123
    • 3 years ago
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  2. YTheManifold
    • 3 years ago
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    a) Just plug in 1 for the numerator and g(x) for the denominator in the quotient rule. b) \[ -{\frac {1+k{{\rm e}^{s}}}{ \left( s+k{{\rm e}^{s}} \right) ^{2}}} \] assuming y is a function of s. c) Follows from \[ x^{-n} = \frac{1}{x^n} \] and \[ \frac{d}{dx} x^n = nx^{n-1} \]

  3. ta123
    • 3 years ago
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    is the quotient rule the same thing as the reciprocal part b?

  4. YTheManifold
    • 3 years ago
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    What do you mean? The reciprocal rule is a special case of the quotient rule. You can use either one for part b, but it's easier to use the reciprocal rule.

  5. ta123
    • 3 years ago
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    The quotient rule f'g-fg'/g^(2)

  6. ta123
    • 3 years ago
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    but g(x) is just a variable

  7. ta123
    • 3 years ago
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    ok I got it

  8. YTheManifold
    • 3 years ago
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    What is your question? g is not just a variable, it's a function.

  9. YTheManifold
    • 3 years ago
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    Ok, hit good answer.

  10. ta123
    • 3 years ago
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    OK I need help with differentiating the bottom on part could you show me step by step how to differentiate that one?

  11. ta123
    • 3 years ago
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    part b

  12. ta123
    • 3 years ago
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    and also the numerator of the function

  13. YTheManifold
    • 3 years ago
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    You need to differentiate\[ \Large s+k\cdot e^s \] with respect to s.

  14. YTheManifold
    • 3 years ago
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    The derivative of a sum is the sum of derivatives, so differentiate s and then differentiate k*e^s.

  15. YTheManifold
    • 3 years ago
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    What is the derivative of s?

  16. raheen
    • 3 years ago
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  17. ta123
    • 3 years ago
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    1

  18. YTheManifold
    • 3 years ago
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    Yes. And what is the derivative of e^s ?

  19. raheen
    • 3 years ago
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    @ta, see the attachee

  20. ta123
    • 3 years ago
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    e^(s)

  21. YTheManifold
    • 3 years ago
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    Your attachment doesn't include part b) explanation, thats what he asked about.

  22. YTheManifold
    • 3 years ago
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    Yes, so now you can add them together and you get the derivative

  23. YTheManifold
    • 3 years ago
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    And then just plug that into the formula in part a)

  24. ta123
    • 3 years ago
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    1+e^(s)

  25. YTheManifold
    • 3 years ago
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    No! Don't forget the factor k!

  26. YTheManifold
    • 3 years ago
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    k does not depend on s, so you can just leave it unchanged

  27. YTheManifold
    • 3 years ago
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    But you must not forget it :D

  28. raheen
    • 3 years ago
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  29. raheen
    • 3 years ago
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    here u r

  30. YTheManifold
    • 3 years ago
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    raheen there is still no explanation of part b :D

  31. raheen
    • 3 years ago
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    been answered all

  32. raheen
    • 3 years ago
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    part b asking to find using the rule only

  33. YTheManifold
    • 3 years ago
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    ta123 do you have more questions?

  34. raheen
    • 3 years ago
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    whears the other 2 parts are to prove

  35. raheen
    • 3 years ago
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    @ta got that?

  36. YTheManifold
    • 3 years ago
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    lol raheen are you just trying to get a medal, because I can give you one if thats what you want :D

  37. ta123
    • 3 years ago
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    your answer is wrong raheen, I got -1-ke^(s)/(s+ke^(s))^2

  38. raheen
    • 3 years ago
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    no YThe, I always try to help , pls don read the things in wrong way

  39. YTheManifold
    • 3 years ago
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    Your answer is correct ta.

  40. ta123
    • 3 years ago
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    yes, thanks for the help Y and raheen

  41. raheen
    • 3 years ago
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    ta why some ppl are using an offensive words?

  42. raheen
    • 3 years ago
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    I think I stayed with u b4 more than 2 hrs, without any annyance, right?

  43. ta123
    • 3 years ago
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    yes

  44. raheen
    • 3 years ago
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    thank u.

  45. ta123
    • 3 years ago
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    no prob!

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