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.

1. Algebraic!

did you take two half derivatives?

2. apple_pi

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

3. Algebraic!

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

4. apple_pi

No, definitely not

5. ParthKohli

Power Rule? Well, that is Calculus too.

6. apple_pi

No differentiation rules at all

7. apple_pi

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

8. apple_pi

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

9. Algebraic!

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

10. Algebraic!

:(

11. apple_pi

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

13. apple_pi

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

14. estudier

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

15. apple_pi

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

16. apple_pi

@estudier Try it for 1/x

17. estudier

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

18. apple_pi

This is no calculus, right?

19. estudier

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20. apple_pi

D is always under (or possibly on) the parabola

21. apple_pi

sorry, should have clarified

22. estudier

Two tangent lines, then?

23. apple_pi

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

24. Rohangrr

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25. apple_pi

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

26. estudier

Choosing a different parabola....

27. Rohangrr

the answer is 1/2 x^2

28. apple_pi

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

29. apple_pi

@mathslover

30. estudier

What happens with ax^2?

31. apple_pi

hmm, I'll think about that

32. Algebraic!

still works.

33. Algebraic!

this works for any quadratic iirc.

34. apple_pi

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

35. estudier

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

36. apple_pi

iirc? no clue

37. estudier

If I recall correctly

38. apple_pi

oh

39. estudier

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. apple_pi

Thanks everyone. I appreciate the encouragement