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anonymous

  • 2 years ago

why does: y+dy=(x+dx)^-2 equal: =x^-2(1+dx/x)^-2

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  1. anonymous
    • 2 years ago
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    put x in evidence

  2. atlas
    • 2 years ago
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    Another way of thinking is divide the expression on R.H.S with x^-2 and then multiply it with x^-2 on the outside

  3. atlas
    • 2 years ago
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    multiplying and dividing Right hand side with x^-2 \[\frac{ x ^{-2}(x+dx)^{-2} }{ x^{-2} }\] \[x^{-2} (\frac{ x+dx }{ x })^{-2}\]

  4. atlas
    • 2 years ago
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    simplify it to x^-2 (1+dx/x)^-2 I hope it makes sense now

  5. anonymous
    • 2 years ago
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    I stiil dont get it Could someone please explain it step by step?

  6. myininaya
    • 2 years ago
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    Ok, do you agree that x+dx could be written as x(1+dx/x)?

  7. myininaya
    • 2 years ago
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    \[x+dx=x+\frac{x}{x} dx=x+x \cdot \frac{dx}{x}=x(1+\frac{dx}{x})\]

  8. anonymous
    • 2 years ago
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    how can x+dx be written as x(1+dx/x)?

  9. myininaya
    • 2 years ago
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    \[(x+dx)^{-2}=[(x)(1+\frac{dx}{x})]^{-2}=x^{-2}(1+\frac{dx}{x})^{-2} \]

  10. myininaya
    • 2 years ago
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    Did you see what I said above what you said?

  11. myininaya
    • 2 years ago
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    |dw:1387637203504:dw|

  12. myininaya
    • 2 years ago
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    |dw:1387637215111:dw|

  13. myininaya
    • 2 years ago
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    you can factor out that x

  14. anonymous
    • 2 years ago
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    but how did you get to x+x d/x? the equation is (x+dx)^-2

  15. myininaya
    • 2 years ago
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    You wanted step by step I'm looking at the inside, x+dx, right now.

  16. anonymous
    • 2 years ago
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    ok

  17. myininaya
    • 2 years ago
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    If you can understand why x+dx=x(1+dx/x) we can move on to the exponent part

  18. myininaya
    • 2 years ago
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    which i already did above

  19. myininaya
    • 2 years ago
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    but i can redo

  20. anonymous
    • 2 years ago
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    i dont understand why x+dx=x(1+dx/x)

  21. anonymous
    • 2 years ago
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    where did you get the 1?

  22. myininaya
    • 2 years ago
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    Like do you the distributive property. It says: a(b+c)=ab+ac

  23. anonymous
    • 2 years ago
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    and how did you get dx/x?

  24. myininaya
    • 2 years ago
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    x(1)=x

  25. myininaya
    • 2 years ago
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    \[x+dx=x+\frac{x}{x} dx \] look I multiply dx by x/x which is just one

  26. myininaya
    • 2 years ago
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    you should see an x in both terms now

  27. myininaya
    • 2 years ago
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    you can factor an x out

  28. anonymous
    • 2 years ago
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    how did you get x/x?

  29. myininaya
    • 2 years ago
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    \[x(1+\frac{1}{x}dx)\] \[x(1+\frac{dx}{x}) \]

  30. myininaya
    • 2 years ago
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    I just put it there so you would see better x/x is there if you write it because it is just one

  31. myininaya
    • 2 years ago
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    If you factor something out you are dividing the terms you factor from

  32. anonymous
    • 2 years ago
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    im sorry but i dont understnad

  33. anonymous
    • 2 years ago
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    *understand

  34. myininaya
    • 2 years ago
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    For example \[3x+9=3(\frac{3x}{3}+\frac{9}{3})=3(x+3)\] Do you understand this example?

  35. anonymous
    • 2 years ago
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    Can you please write down every step you do and how you get every variable?

  36. anonymous
    • 2 years ago
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    in one answer please

  37. myininaya
    • 2 years ago
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    If you don't understand it separately how do you expect to understand it together?

  38. myininaya
    • 2 years ago
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    Do you understand the example I just put?

  39. anonymous
    • 2 years ago
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    no sorry

  40. anonymous
    • 2 years ago
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    im new to calculus

  41. myininaya
    • 2 years ago
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    You might want to review factoring then. Because all I did was factor 3x+9 in that example

  42. anonymous
    • 2 years ago
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    i will

  43. myininaya
    • 2 years ago
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    Do you know the distributive property?

  44. anonymous
    • 2 years ago
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    no sorry

  45. myininaya
    • 2 years ago
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    Like a(b+c) means ab+ac

  46. anonymous
    • 2 years ago
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    yeah

  47. myininaya
    • 2 years ago
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    That is the distributive property

  48. anonymous
    • 2 years ago
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    ok

  49. myininaya
    • 2 years ago
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    You can look at it either way. a(b+c)=ab+ac ab+ac=a(b+c) Do you see on this side they say hey both of these terms have an a in common so let's factor it out

  50. anonymous
    • 2 years ago
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    ok

  51. myininaya
    • 2 years ago
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    What they are doing when they factor is pulling something out (multiplying) and then also dividing by what they pulled out. Like for example , ab+ac we could factor factor out ac but that means we also need to divide each term by ac like this: \[ac(\frac{ab}{ac}+\frac{ac}{ac})=ac(\frac{b}{c}+1)\]

  52. myininaya
    • 2 years ago
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    What do you get when you multiply out the ac(b/c+1) expression?

  53. myininaya
    • 2 years ago
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    i must go good luck

  54. anonymous
    • 2 years ago
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    no idea

  55. anonymous
    • 2 years ago
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    thankyou

  56. anonymous
    • 2 years ago
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    I don't understand what we are supposed to do, re read your post, y +dy = (x+dx)^-2 equal = x^-2(1+dx/x)^-2 can you use the draw box below to write it again ?

  57. anonymous
    • 2 years ago
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    and dy =y'? dx =x'?

  58. anonymous
    • 2 years ago
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    \[y+dy=(x+dx)^{-2} =x ^{-2}(1+\frac{ dx }{ x })^{-2}\]

  59. anonymous
    • 2 years ago
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    the aim is to differentiate y=x^2

  60. anonymous
    • 2 years ago
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    2 equal signs?

  61. anonymous
    • 2 years ago
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    the second equal sign is the result

  62. anonymous
    • 2 years ago
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    once we have powered the left hand side of the equation to -2

  63. anonymous
    • 2 years ago
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    what do you mean by "ODE?"

  64. anonymous
    • 2 years ago
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    one more question: from which course , you have this problem?

  65. anonymous
    • 2 years ago
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    its from a book called calculus made easy, ill post the page i found it on right now

  66. anonymous
    • 2 years ago
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    thanks for information,

  67. anonymous
    • 2 years ago
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    no problem

  68. anonymous
    • 2 years ago
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    ill post the page in a sec

  69. anonymous
    • 2 years ago
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  70. anonymous
    • 2 years ago
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    he adds dy or dx to expand it a bit

  71. anonymous
    • 2 years ago
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    I can understand (x +dx)^(-2) = x^(-2)(1+dx/x)^-2 but from y +dy , I cannot see how they get the right hand side. since y = x^-2 --> dy = -2x^-3 y + dy = x^-2 - 2 x^-3 = x^-2(1-2dx/x) which is not their right hand side :((

  72. zpupster
    • 2 years ago
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    this might clear it up i like this book his proofs are easier to understand than the way was taught.

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