## 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

1. Algebraic! Group Title

did you take two half derivatives?

2. apple_pi Group Title

I think not (don't know what it is)

3. Algebraic! Group Title

If you found another way, you probably did calculus without realizing it..

4. apple_pi Group Title

No, definitely not

5. ParthKohli Group Title

Power Rule? Well, that is Calculus too.

6. apple_pi Group Title

No differentiation rules at all

7. apple_pi Group Title

The same method can be applied to y=1/x

8. apple_pi Group Title

Ok, well if no-one else has anything to say, I'll proceed

9. Algebraic! Group Title

you have an audience now. may we see the method?

10. Algebraic! Group Title

:(

11. apple_pi Group Title

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 Group Title

13. apple_pi Group Title

It might, so far only works for x^2 and 1/x

14. estudier Group Title

x^2=mx+y1 x^2-mx-y1=0 Aren't these the same?

15. apple_pi Group Title

yes, just rearranging to show clearly what the a,b,c values are

16. apple_pi Group Title

@estudier Try it for 1/x

17. estudier Group Title

I will, just trying to get it first...

18. apple_pi Group Title

This is no calculus, right?

19. estudier Group Title

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20. apple_pi Group Title

D is always under (or possibly on) the parabola

21. apple_pi Group Title

sorry, should have clarified

22. estudier Group Title

Two tangent lines, then?

23. apple_pi Group Title

yes, however then they are two different values of m. The initial point D kinda goes away

24. Rohangrr Group Title

|dw:1347873037871:dw|

25. apple_pi Group Title

sorry, @Rohangrr, what is that supposed to be?

26. estudier Group Title

Choosing a different parabola....

27. Rohangrr Group Title

28. apple_pi Group Title

$\frac{ d }{ dx }x^2=2x$

29. apple_pi Group Title

@mathslover

30. estudier Group Title

What happens with ax^2?

31. apple_pi Group Title

32. Algebraic! Group Title

still works.

33. Algebraic! Group Title

this works for any quadratic iirc.

34. apple_pi Group Title

well you end up with: ax^2-mx-y1=0 x=m/2a 2ax=m So yeah

35. estudier Group Title

iirc ? (you have heard of this before? @Algebraic! )

36. apple_pi Group Title

iirc? no clue

37. estudier Group Title

If I recall correctly

38. apple_pi Group Title

oh

39. estudier Group Title

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 Group Title

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 Group Title

Thanks everyone. I appreciate the encouragement