Need help with derivative

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June 12th 2006, 02:13 PM
c_323_h

Need help with derivative

Find $f'(a)$

$f(x)=3-2x+4x^2$
June 12th 2006, 03:11 PM
Jameson

First, what's f'(x)?

$f'(x)=-2+8x$ That's just using basic power rules.

How would you find f'(2)? Easy. You would just take the derivative and plug in 2. Same concept here.

$f'(a)=-2+8a$

How about f'(a+h)?
June 12th 2006, 03:34 PM
c_323_h

i haven't learned power rules yet. only difference quotient. i've figured it out though
June 12th 2006, 06:12 PM
ThePerfectHacker

Quote:

Originally Posted by c_323_h

i haven't learned power rules yet. only difference quotient. i've figured it out though

In that case.
Given, $f(x)=3-2x+4x^2$
Then,
$f'(x)=(3)'-(2x)'+(4x^2)'$ ---->Sum-Difference Rule
Thus,
$f'(x)=(3)'-2(x)'+4(x^2)'$ ----->Constant-Multiplte Rule
Thus,
$f'(x)=0-2(1)+4(x^2)'$ ----->Derivative of constant and "x",
$f'(x)=-2+4(x\cdot x)'$ ----->Definition of exponent,
$f'(x)=-2+4[x(x)'+(x)'x]$ ---->Product Rule
Thus,
$f'(x)=-2+4[x(1)+(1)x]$ ---->Derivative of 'x',
Thus,
$f'(x)=-2+4[2x]=-2+8x$