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apple_pi
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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.
 2 years ago
 2 years ago
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.
 2 years ago
 2 years ago

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Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
did you take two half derivatives?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
I think not (don't know what it is)
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
If you found another way, you probably did calculus without realizing it..
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
No, definitely not
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Power Rule? Well, that is Calculus too.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
No differentiation rules at all
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
The same method can be applied to y=1/x
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
Ok, well if noone else has anything to say, I'll proceed
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
you have an audience now. may we see the method?
 2 years ago

apple_pi Group TitleBest 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
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Does your method generalize? (past quadratic)
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
It might, so far only works for x^2 and 1/x
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
x^2=mx+y1 x^2mxy1=0 Aren't these the same?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
yes, just rearranging to show clearly what the a,b,c values are
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
@estudier Try it for 1/x
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
I will, just trying to get it first...
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
This is no calculus, right?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
dw:1347872777981:dw
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
D is always under (or possibly on) the parabola
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
sorry, should have clarified
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Two tangent lines, then?
 2 years ago

apple_pi Group TitleBest 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
 2 years ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.1
dw:1347873037871:dw
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
sorry, @Rohangrr, what is that supposed to be?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Choosing a different parabola....
 2 years ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.1
the answer is 1/2 x^2
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{ d }{ dx }x^2=2x\]
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
@mathslover
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
What happens with ax^2?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
hmm, I'll think about that
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
still works.
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
this works for any quadratic iirc.
 2 years ago

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

estudier Group TitleBest ResponseYou've already chosen the best response.0
iirc ? (you have heard of this before? @Algebraic! )
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
iirc? no clue
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
If I recall correctly
 2 years ago

estudier Group TitleBest 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".
 2 years ago

nipunmalhotra93 Group TitleBest 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 :)
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.2
Thanks everyone. I appreciate the encouragement
 2 years ago
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