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apple_pi Group Title

Is there a method for finding the derivative of y=x^2 without using calculus OR LIMITS? I think I have found a way, but would first like to hear if you have heard of any.

  • one year ago
  • one year ago

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  1. Algebraic! Group Title
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    did you take two half derivatives?

    • one year ago
  2. apple_pi Group Title
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    I think not (don't know what it is)

    • one year ago
  3. Algebraic! Group Title
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    If you found another way, you probably did calculus without realizing it..

    • one year ago
  4. apple_pi Group Title
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    No, definitely not

    • one year ago
  5. ParthKohli Group Title
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    Power Rule? Well, that is Calculus too.

    • one year ago
  6. apple_pi Group Title
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    No differentiation rules at all

    • one year ago
  7. apple_pi Group Title
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    The same method can be applied to y=1/x

    • one year ago
  8. apple_pi Group Title
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    Ok, well if no-one else has anything to say, I'll proceed

    • one year ago
  9. Algebraic! Group Title
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    you have an audience now. may we see the method?

    • one year ago
  10. Algebraic! Group Title
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    :(

    • one year ago
  11. apple_pi Group Title
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    Let D(0,y1) be a point on the y-axis We want to find the equation of a line such that it passes through the point D and is a tangent to the parabola y=x^2. The equation can be written in the form: y=mx+y1 To find where the line and parabola contact, solve simultaneously. x^2=mx+y1 x^2-mx-y1=0 x= -b±√∆ ------ 2a but since it is a tangent, only one point of contact. So ∆=0 x = m/2 m=2x Tada

    • one year ago
  12. estudier Group Title
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    Does your method generalize? (past quadratic)

    • one year ago
  13. apple_pi Group Title
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    It might, so far only works for x^2 and 1/x

    • one year ago
  14. estudier Group Title
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    x^2=mx+y1 x^2-mx-y1=0 Aren't these the same?

    • one year ago
  15. apple_pi Group Title
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    yes, just rearranging to show clearly what the a,b,c values are

    • one year ago
  16. apple_pi Group Title
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    @estudier Try it for 1/x

    • one year ago
  17. estudier Group Title
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    I will, just trying to get it first...

    • one year ago
  18. apple_pi Group Title
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    This is no calculus, right?

    • one year ago
  19. estudier Group Title
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    |dw:1347872777981:dw|

    • one year ago
  20. apple_pi Group Title
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    D is always under (or possibly on) the parabola

    • one year ago
  21. apple_pi Group Title
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    sorry, should have clarified

    • one year ago
  22. estudier Group Title
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    Two tangent lines, then?

    • one year ago
  23. apple_pi Group Title
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    yes, however then they are two different values of m. The initial point D kinda goes away

    • one year ago
  24. Rohangrr Group Title
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    |dw:1347873037871:dw|

    • one year ago
  25. apple_pi Group Title
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    sorry, @Rohangrr, what is that supposed to be?

    • one year ago
  26. estudier Group Title
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    Choosing a different parabola....

    • one year ago
  27. Rohangrr Group Title
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    the answer is 1/2 x^2

    • one year ago
  28. apple_pi Group Title
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    \[\frac{ d }{ dx }x^2=2x\]

    • one year ago
  29. apple_pi Group Title
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    @mathslover

    • one year ago
  30. estudier Group Title
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    What happens with ax^2?

    • one year ago
  31. apple_pi Group Title
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    hmm, I'll think about that

    • one year ago
  32. Algebraic! Group Title
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    still works.

    • one year ago
  33. Algebraic! Group Title
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    this works for any quadratic iirc.

    • one year ago
  34. apple_pi Group Title
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    well you end up with: ax^2-mx-y1=0 x=m/2a 2ax=m So yeah

    • one year ago
  35. estudier Group Title
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    iirc ? (you have heard of this before? @Algebraic! )

    • one year ago
  36. apple_pi Group Title
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    iirc? no clue

    • one year ago
  37. estudier Group Title
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    If I recall correctly

    • one year ago
  38. apple_pi Group Title
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    oh

    • one year ago
  39. estudier Group Title
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    I think what you have done is to reinvent some of the work done leading up to the calculus. Namely, the so called "Tangent Line Problem" "In the second book of his Geometry, René Descartes said of the problem of constructing the tangent to a curve, "And I dare say that this is not only the most useful and most general problem in geometry that I know, but even that I have ever desired to know".

    • one year ago
  40. nipunmalhotra93 Group Title
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    Actually, you're able to apply this method because of the formula of roots of a quadratic equation. This can not be applied to a lot of functions including polynomials of degree 5 or higher. That's because polynomials of degree 5 or higher do not have a general formula for the roots. So, if you do not have a formula for the roots, you can't really find the derivative with this method. Good work though :)

    • one year ago
  41. apple_pi Group Title
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    Thanks everyone. I appreciate the encouragement

    • one year ago
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