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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.

Mathematics
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did you take two half derivatives?
I think not (don't know what it is)
If you found another way, you probably did calculus without realizing it..

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Other answers:

No, definitely not
Power Rule? Well, that is Calculus too.
No differentiation rules at all
The same method can be applied to y=1/x
Ok, well if no-one else has anything to say, I'll proceed
you have an audience now. may we see the method?
:(
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
Does your method generalize? (past quadratic)
It might, so far only works for x^2 and 1/x
x^2=mx+y1 x^2-mx-y1=0 Aren't these the same?
yes, just rearranging to show clearly what the a,b,c values are
@estudier Try it for 1/x
I will, just trying to get it first...
This is no calculus, right?
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D is always under (or possibly on) the parabola
sorry, should have clarified
Two tangent lines, then?
yes, however then they are two different values of m. The initial point D kinda goes away
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sorry, @Rohangrr, what is that supposed to be?
Choosing a different parabola....
the answer is 1/2 x^2
\[\frac{ d }{ dx }x^2=2x\]
What happens with ax^2?
hmm, I'll think about that
still works.
this works for any quadratic iirc.
well you end up with: ax^2-mx-y1=0 x=m/2a 2ax=m So yeah
iirc ? (you have heard of this before? @Algebraic! )
iirc? no clue
If I recall correctly
oh
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".
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 :)
Thanks everyone. I appreciate the encouragement

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