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farukk Group Title

???

  • 2 years ago
  • 2 years ago

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  1. farukk Group Title
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    which is correct? i.e a+b\a-b (after rationallization)a+b\a-b * a+b\a+b or a+b\a-b * a-b\a-b ?

    • 2 years ago
  2. UnkleRhaukus Group Title
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    |dw:1354620267974:dw|

    • 2 years ago
  3. UnkleRhaukus Group Title
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    |dw:1354620294927:dw|

    • 2 years ago
  4. farukk Group Title
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    can it be a-b rather than a+b?

    • 2 years ago
  5. UnkleRhaukus Group Title
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    it would be equal but it wouldn't helpful in rationalisation

    • 2 years ago
  6. farukk Group Title
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    all right thanks

    • 2 years ago
  7. UnkleRhaukus Group Title
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    |dw:1354620486122:dw|

    • 2 years ago
  8. farukk Group Title
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    one more question ?

    • 2 years ago
  9. farukk Group Title
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    may i ask

    • 2 years ago
  10. UnkleRhaukus Group Title
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    yes!

    • 2 years ago
  11. farukk Group Title
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    |dw:1354681791055:dw|

    • 2 years ago
  12. farukk Group Title
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    u got?

    • 2 years ago
  13. farukk Group Title
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    how will we use power rule in this question?

    • 2 years ago
  14. UnkleRhaukus Group Title
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    so the first step would be to remember that the the derivative of a sum , is the sum of derivatives

    • 2 years ago
  15. farukk Group Title
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    yes yes

    • 2 years ago
  16. UnkleRhaukus Group Title
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    \[\frac{\text d}{\text d x}(1+\sqrt{1-x^2})=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\]

    • 2 years ago
  17. farukk Group Title
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    derivative of a constant will be 0

    • 2 years ago
  18. UnkleRhaukus Group Title
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    right

    • 2 years ago
  19. UnkleRhaukus Group Title
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    so the problem is much simpler now

    • 2 years ago
  20. UnkleRhaukus Group Title
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    can you remember the derivative of a function of a function?

    • 2 years ago
  21. farukk Group Title
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    to me it's start right now:)

    • 2 years ago
  22. farukk Group Title
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    i'm sorry

    • 2 years ago
  23. farukk Group Title
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    |dw:1354682465787:dw|

    • 2 years ago
  24. UnkleRhaukus Group Title
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    \[ (f\circ g)'(x)=f(g(x))'=f'(g(x))g'(x)\]

    • 2 years ago
  25. UnkleRhaukus Group Title
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    |dw:1354621505691:dw|

    • 2 years ago
  26. UnkleRhaukus Group Title
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    you have to multiply by the derivative of the the inside function

    • 2 years ago
  27. UnkleRhaukus Group Title
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    |dw:1354621612577:dw|

    • 2 years ago
  28. UnkleRhaukus Group Title
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    following?

    • 2 years ago
  29. farukk Group Title
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    |dw:1354682900778:dw|

    • 2 years ago
  30. UnkleRhaukus Group Title
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    yeah

    • 2 years ago
  31. farukk Group Title
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    but how?

    • 2 years ago
  32. UnkleRhaukus Group Title
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    \[\frac{\text d}{\text d x}(1+\sqrt{1-x^2})\]\[=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\]\[=\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\] \[\qquad\frac{\text d}{\text dx}f(g(x))=(f\circ g)'(x)=g'(x)f'(g(x))\]\[\qquad f(x)= x^{1/2}\qquad f'(x)=\frac12x^{-1/2}\]\[\qquad g(x)=1-x^2\qquad g'(x)=-2x\] \[=-2x\times\frac12(1-x^2)^{-1/2}\]

    • 2 years ago
  33. UnkleRhaukus Group Title
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    i know this stuff is tricky, can you tell me where about you are having trouble

    • 2 years ago
  34. farukk Group Title
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    haha

    • 2 years ago
  35. farukk Group Title
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    it is but u have solved it thank u so much!!

    • 2 years ago
  36. UnkleRhaukus Group Title
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    yeah but do understand the chain rule/,

    • 2 years ago
  37. farukk Group Title
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    no will be tomorow in college

    • 2 years ago
  38. farukk Group Title
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    hey did you use chain rule?

    • 2 years ago
  39. UnkleRhaukus Group Title
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    the chain rule for the derivative of a function of a function \[(f\circ g)'(x)=g'(x)f'(g(x))\]

    • 2 years ago
  40. farukk Group Title
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    is there any other way for this question except chain rule?

    • 2 years ago
  41. farukk Group Title
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    u didt use power rule? did u?

    • 2 years ago
  42. UnkleRhaukus Group Title
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    i used the power fule to find f '(x) and g'((x)

    • 2 years ago
  43. farukk Group Title
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    confusing me

    • 2 years ago
  44. UnkleRhaukus Group Title
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    watch this video , skip the first 5 minutes http://www.khanacademy.org/math/calculus/differential-calculus/e/chain_rule_1

    • 2 years ago
  45. farukk Group Title
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    actually i have solved all my questions except this one in quite a simple way

    • 2 years ago
  46. UnkleRhaukus Group Title
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    did you get the right answers?

    • 2 years ago
  47. farukk Group Title
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    i'm solving questions using theorems

    • 2 years ago
  48. farukk Group Title
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    there is no video

    • 2 years ago
  49. farukk Group Title
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    yeah answer is right

    • 2 years ago
  50. UnkleRhaukus Group Title
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    well you will have to learn the chain rule

    • 2 years ago
  51. UnkleRhaukus Group Title
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    this is the link i ment http://www.khanacademy.org/math/calculus/differential-calculus/v/the-chain-rule?exid=chain_rule_1

    • 2 years ago
  52. farukk Group Title
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    |dw:1354747831157:dw|

    • 2 years ago
  53. farukk Group Title
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    mistake* -2x\ 2 sqrt 1 - x square

    • 2 years ago
  54. farukk Group Title
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    |dw:1354748818882:dw|

    • 2 years ago
  55. UnkleRhaukus Group Title
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    you can cancel the two

    • 2 years ago
  56. farukk Group Title
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    yeah rightly said!

    • 2 years ago
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