math questionWhat is dy/dx from first principle for y=3x^2-2x ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate `(dy)/(dx)` using the first principle, such that:

`(dy)/(dx) = lim_(Delta x -> 0) (f(x + Delta x) - f(x))/(Delta x)`

`(dy)/(dx) = lim_(Delta x -> 0) (3(x + Delta x)^2 - 2(x + Delta x) - 3x^2 + 2x)/(Delta x)`

Expanding the square yields:

`(dy)/(dx) = lim_(Delta x -> 0) (3x^2 + 6x*Delta x + 3(Delta x)^2 - 2x - 2Delta x - 3x^2 + 2x)/(Delta x)`

Reducing duplicate terms yields:

`(dy)/(dx) = lim_(Delta x -> 0) (6x*Delta x + 3(Delta x)^2 - 2Delta x )/(Delta x)`

Factoring out `Delta x` yields:

`(dy)/(dx) = lim_(Delta x -> 0) (Delta x*(6x + 3(Delta x) - 2))/(Delta x)`

Reducing duplicate factors yields:

`(dy)/(dx) = lim_(Delta x -> 0) (6x + 3(Delta x) - 2)`

Substituting 0 for `Delta x` yields:

`(dy)/(dx) = 6x + 3*0 - 2 => (dy)/(dx) = 6x - 2`

Hence, evaluating `(dy)/(dx)` using the first principle, yields `(dy)/(dx) = 6x - 2.`

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