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apple_pi

  • 2 years ago

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.

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  1. Algebraic!
    • 2 years ago
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    did you take two half derivatives?

  2. apple_pi
    • 2 years ago
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    I think not (don't know what it is)

  3. Algebraic!
    • 2 years ago
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    If you found another way, you probably did calculus without realizing it..

  4. apple_pi
    • 2 years ago
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    No, definitely not

  5. ParthKohli
    • 2 years ago
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    Power Rule? Well, that is Calculus too.

  6. apple_pi
    • 2 years ago
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    No differentiation rules at all

  7. apple_pi
    • 2 years ago
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    The same method can be applied to y=1/x

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

  9. Algebraic!
    • 2 years ago
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    you have an audience now. may we see the method?

  10. Algebraic!
    • 2 years ago
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    :(

  11. apple_pi
    • 2 years ago
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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

  12. estudier
    • 2 years ago
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    Does your method generalize? (past quadratic)

  13. apple_pi
    • 2 years ago
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    It might, so far only works for x^2 and 1/x

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

  15. apple_pi
    • 2 years ago
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    yes, just rearranging to show clearly what the a,b,c values are

  16. apple_pi
    • 2 years ago
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    @estudier Try it for 1/x

  17. estudier
    • 2 years ago
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    I will, just trying to get it first...

  18. apple_pi
    • 2 years ago
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    This is no calculus, right?

  19. estudier
    • 2 years ago
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    |dw:1347872777981:dw|

  20. apple_pi
    • 2 years ago
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    D is always under (or possibly on) the parabola

  21. apple_pi
    • 2 years ago
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    sorry, should have clarified

  22. estudier
    • 2 years ago
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    Two tangent lines, then?

  23. apple_pi
    • 2 years ago
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    yes, however then they are two different values of m. The initial point D kinda goes away

  24. Rohangrr
    • 2 years ago
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    |dw:1347873037871:dw|

  25. apple_pi
    • 2 years ago
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    sorry, @Rohangrr, what is that supposed to be?

  26. estudier
    • 2 years ago
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    Choosing a different parabola....

  27. Rohangrr
    • 2 years ago
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    the answer is 1/2 x^2

  28. apple_pi
    • 2 years ago
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    \[\frac{ d }{ dx }x^2=2x\]

  29. apple_pi
    • 2 years ago
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    @mathslover

  30. estudier
    • 2 years ago
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    What happens with ax^2?

  31. apple_pi
    • 2 years ago
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    hmm, I'll think about that

  32. Algebraic!
    • 2 years ago
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    still works.

  33. Algebraic!
    • 2 years ago
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    this works for any quadratic iirc.

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

  35. estudier
    • 2 years ago
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    iirc ? (you have heard of this before? @Algebraic! )

  36. apple_pi
    • 2 years ago
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    iirc? no clue

  37. estudier
    • 2 years ago
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    If I recall correctly

  38. apple_pi
    • 2 years ago
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    oh

  39. estudier
    • 2 years ago
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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".

  40. nipunmalhotra93
    • 2 years ago
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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 :)

  41. apple_pi
    • 2 years ago
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    Thanks everyone. I appreciate the encouragement

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