## anonymous 3 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.

1. anonymous

did you take two half derivatives?

2. anonymous

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

3. anonymous

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

4. anonymous

No, definitely not

5. ParthKohli

Power Rule? Well, that is Calculus too.

6. anonymous

No differentiation rules at all

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

:(

11. anonymous

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. anonymous

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

@estudier Try it for 1/x

17. anonymous

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

18. anonymous

This is no calculus, right?

19. anonymous

|dw:1347872777981:dw|

20. anonymous

D is always under (or possibly on) the parabola

21. anonymous

sorry, should have clarified

22. anonymous

Two tangent lines, then?

23. anonymous

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

24. anonymous

|dw:1347873037871:dw|

25. anonymous

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

26. anonymous

Choosing a different parabola....

27. anonymous

the answer is 1/2 x^2

28. anonymous

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

29. anonymous

@mathslover

30. anonymous

What happens with ax^2?

31. anonymous

hmm, I'll think about that

32. anonymous

still works.

33. anonymous

this works for any quadratic iirc.

34. anonymous

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

35. anonymous

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

36. anonymous

iirc? no clue

37. anonymous

If I recall correctly

38. anonymous

oh

39. anonymous

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

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. anonymous

Thanks everyone. I appreciate the encouragement