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anonymous
 4 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.
anonymous
 4 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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you take two half derivatives?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think not (don't know what it is)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you found another way, you probably did calculus without realizing it..

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No differentiation rules at all

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The same method can be applied to y=1/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, well if noone else has anything to say, I'll proceed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have an audience now. may we see the method?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does your method generalize? (past quadratic)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It might, so far only works for x^2 and 1/x

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, just rearranging to show clearly what the a,b,c values are

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@estudier Try it for 1/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will, just trying to get it first...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is no calculus, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347872777981:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0D is always under (or possibly on) the parabola

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, should have clarified

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Two tangent lines, then?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, however then they are two different values of m. The initial point D kinda goes away

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347873037871:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, @Rohangrr, what is that supposed to be?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Choosing a different parabola....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the answer is 1/2 x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }x^2=2x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What happens with ax^2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, I'll think about that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this works for any quadratic iirc.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0iirc ? (you have heard of this before? @Algebraic! )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If I recall correctly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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".

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks everyone. I appreciate the encouragement
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