anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
did you take two half derivatives?
anonymous
  • anonymous
I think not (don't know what it is)
anonymous
  • anonymous
If you found another way, you probably did calculus without realizing it..

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anonymous
  • anonymous
No, definitely not
ParthKohli
  • ParthKohli
Power Rule? Well, that is Calculus too.
anonymous
  • anonymous
No differentiation rules at all
anonymous
  • anonymous
The same method can be applied to y=1/x
anonymous
  • anonymous
Ok, well if no-one else has anything to say, I'll proceed
anonymous
  • anonymous
you have an audience now. may we see the method?
anonymous
  • anonymous
:(
anonymous
  • anonymous
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
anonymous
  • anonymous
Does your method generalize? (past quadratic)
anonymous
  • anonymous
It might, so far only works for x^2 and 1/x
anonymous
  • anonymous
x^2=mx+y1 x^2-mx-y1=0 Aren't these the same?
anonymous
  • anonymous
yes, just rearranging to show clearly what the a,b,c values are
anonymous
  • anonymous
@estudier Try it for 1/x
anonymous
  • anonymous
I will, just trying to get it first...
anonymous
  • anonymous
This is no calculus, right?
anonymous
  • anonymous
|dw:1347872777981:dw|
anonymous
  • anonymous
D is always under (or possibly on) the parabola
anonymous
  • anonymous
sorry, should have clarified
anonymous
  • anonymous
Two tangent lines, then?
anonymous
  • anonymous
yes, however then they are two different values of m. The initial point D kinda goes away
anonymous
  • anonymous
|dw:1347873037871:dw|
anonymous
  • anonymous
sorry, @Rohangrr, what is that supposed to be?
anonymous
  • anonymous
Choosing a different parabola....
anonymous
  • anonymous
the answer is 1/2 x^2
anonymous
  • anonymous
\[\frac{ d }{ dx }x^2=2x\]
anonymous
  • anonymous
@mathslover
anonymous
  • anonymous
What happens with ax^2?
anonymous
  • anonymous
hmm, I'll think about that
anonymous
  • anonymous
still works.
anonymous
  • anonymous
this works for any quadratic iirc.
anonymous
  • anonymous
well you end up with: ax^2-mx-y1=0 x=m/2a 2ax=m So yeah
anonymous
  • anonymous
iirc ? (you have heard of this before? @Algebraic! )
anonymous
  • anonymous
iirc? no clue
anonymous
  • anonymous
If I recall correctly
anonymous
  • anonymous
oh
anonymous
  • anonymous
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".
nipunmalhotra93
  • 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 :)
anonymous
  • anonymous
Thanks everyone. I appreciate the encouragement

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