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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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Algebraic!Best ResponseYou've already chosen the best response.0
did you take two half derivatives?
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
I think not (don't know what it is)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
If you found another way, you probably did calculus without realizing it..
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Power Rule? Well, that is Calculus too.
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
No differentiation rules at all
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
The same method can be applied to y=1/x
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
Ok, well if noone else has anything to say, I'll proceed
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
you have an audience now. may we see the method?
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
Let D(0,y1) be a point on the yaxis 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^2mxy1=0 x= b±√∆  2a but since it is a tangent, only one point of contact. So ∆=0 x = m/2 m=2x Tada
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Does your method generalize? (past quadratic)
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
It might, so far only works for x^2 and 1/x
 one year ago

estudierBest ResponseYou've already chosen the best response.0
x^2=mx+y1 x^2mxy1=0 Aren't these the same?
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
yes, just rearranging to show clearly what the a,b,c values are
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
@estudier Try it for 1/x
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I will, just trying to get it first...
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
This is no calculus, right?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
dw:1347872777981:dw
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
D is always under (or possibly on) the parabola
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
sorry, should have clarified
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Two tangent lines, then?
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
yes, however then they are two different values of m. The initial point D kinda goes away
 one year ago

RohangrrBest ResponseYou've already chosen the best response.1
dw:1347873037871:dw
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
sorry, @Rohangrr, what is that supposed to be?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Choosing a different parabola....
 one year ago

RohangrrBest ResponseYou've already chosen the best response.1
the answer is 1/2 x^2
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
\[\frac{ d }{ dx }x^2=2x\]
 one year ago

estudierBest ResponseYou've already chosen the best response.0
What happens with ax^2?
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
hmm, I'll think about that
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
this works for any quadratic iirc.
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
well you end up with: ax^2mxy1=0 x=m/2a 2ax=m So yeah
 one year ago

estudierBest ResponseYou've already chosen the best response.0
iirc ? (you have heard of this before? @Algebraic! )
 one year ago

estudierBest ResponseYou've already chosen the best response.0
If I recall correctly
 one year ago

estudierBest ResponseYou've already chosen the best response.0
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".
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.0
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 :)
 one year ago

apple_piBest ResponseYou've already chosen the best response.2
Thanks everyone. I appreciate the encouragement
 one year ago
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