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lazyaditya
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We say that a slope can be calculated as the ratio of the perpendicular to the base , Then what does differentiation of the equation shows "dy/dx"?
Differential calculus involves the evaluation of that Δy / Δx but taken to its extreme.We say that a slope can be calculated as the ratio of the perpendicular to the base of a line tangential, Then what does differentiation of the equation shows "dy/dx"?
lazyaditya said:Ya that's what i m saying suppose y =3x^2 then on differentiation i would get (dy/dx=6x) then at point x=3 the slope would be 18. Then isn't slope also calculated as (y2-y1)/(x2-x1) then at point x = 3 isn't it should be 9 ?
Yes. So if you take two points (x1,y1) and (x2,y2) on that curve, where x1 and x2 are very close to 3, then that quotient will be very close to 18.lazyaditya said:Then isn't slope also calculated as (y2-y1)/(x2-x1)
lazyaditya said:Then what is the significance of one over the other ? I mean in which situation which one should i prefer ?
Differentiation gives the exact value of the slope at any point you choose. Using the "pair of points" method gives an approximation, you are in effect approximating the curve with a straight line.lazyaditya said:Then what is the significance of one over the other ? I mean in which situation which one should i prefer ?
NascentOxygen said:Differentiation gives the exact value of the slope at any point you choose. Using the "pair of points" method gives an approximation, you are in effect approximating the curve with a straight line.
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