Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=2x^2-3x+6`?

Answer 1

`(df(x))/(dx) = 2x+3`
(see below for method of derivation)

Explanation:

Given a function `f(x)`, the derivative of `f(x)` is defined as:
`color(white)("XXX")(df(x))/dx=lim_(hrarr0) (f(x+h)-f(x))/h`

If `f(x)=2x^2-3x+6`
then `f(x+h)= 2(x+h)^2-3(x+h)+6`
`color(white)("XXXXXXXX")=2x^2+2xh+h^2-3x-3h+6`
So
`f(x+h)-f(x) = color(white)("XX")2x^2+2xh+h^2-3x-3h+6`
`color(white)("XXXXXXXXXXX")-(underline(2x^2color(white)("XXXXXXX")-3xcolor(white)("XXX")+6))`
`color(white)("XXXXXXXXX")=color(white)("XXXXXX")2xh+h^2color(white)("XXX")+3h`

and
`color(white)("XXX")(f(x+h)-f(x))/h = 2x+h+3`

Therefore
`color(white)("XXX")(df(x))/(dx)=lim_(hrarr0) 2x+h+3 = 2x+3`

Answer 2

Evaluate the limit of the expression `lim_(h->0) (f(x + h) -f(x))/(h)`

Explanation:

So `lim_(h->0) ((2(x+h)^2 - 3(x+h) + 6) -f(3x^2 - 3x +6))/(h)`