derivative

Hello, mr.burger!

Are you in self-study?
Or are you simply browsing through a Calculus text?
Your questions span the entire spectrum of Calculus I, you see,
. . starting with this one.

If y = 2x<sup>2</sup> + 3x – 5 find dy/dx from the definition of the first derivative.
. . . . . . . . . . . . . . . . . . . . . f(x + h) - f(x)
Definition: . f '(x) . = . lim . ----------------
. . . . . . . . . . . . . . . . h->0 . . . . . h

I always make <u>four</u> steps out of it:

. . . [1] Find f(x + h) . . . replace x with x + h
. . . [2] Subtract f(x) . . . the original function
. . . [3] Divide by h . . . . there's usually some cancelling possible.
. . . [4] Take the limit . . .let h → 0


[1] .f(x + h) . = . (2(x + h)<sup>2</sup> + 3(x + h) - 5
. . . . . . . . . . .= . 2x<sup>2</sup> + 4xh + 2h<sup>2</sup> + 3x + 3h - 5


[2] .f(x + h) - f(x) . = . [2x<sup>2</sup> + 4xh + 2h<sup>2</sup> + 3x + 3h - 5] - [2x<sup>2</sup> + 3x - 5]
. . . . . . . . . . . . . . .= . 4xh + 2h<sup>2</sup> + 3h

. . . . f(x + h) - f(x) . . . . .4xh + 2h<sup>2</sup> + 3h . . . . h(4x + 2h + 3)
[3] . ----------------- . = . -------------------- . = . ------------------- . = . 4x + 2h + 3
. . . . . . . . .h . . . . . . . . . . . . . h . . . . . . . . . . . . . . . h


[4] . f '(x) . = . lim .(4x + 2h + 3) . = . 4x + 3 . . . There!
. . . . . . . . . . .h->0
 
i am simply trying to teach myself how to do these problems and i need a little help because the book isnt helping me any.
 
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