# Need help with derivative

• Jun 12th 2006, 02:13 PM
c_323_h
Need help with derivative
Find $\displaystyle f'(a)$

$\displaystyle f(x)=3-2x+4x^2$
• Jun 12th 2006, 03:11 PM
Jameson
First, what's f'(x)?

$\displaystyle f'(x)=-2+8x$ 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.

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

• Jun 12th 2006, 03:34 PM
c_323_h
i haven't learned power rules yet. only difference quotient. i've figured it out though
• Jun 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, $\displaystyle f(x)=3-2x+4x^2$
Then,
$\displaystyle f'(x)=(3)'-(2x)'+(4x^2)'$---->Sum-Difference Rule
Thus,
$\displaystyle f'(x)=(3)'-2(x)'+4(x^2)'$----->Constant-Multiplte Rule
Thus,
$\displaystyle f'(x)=0-2(1)+4(x^2)'$----->Derivative of constant and "x",
$\displaystyle f'(x)=-2+4(x\cdot x)'$----->Definition of exponent,
$\displaystyle f'(x)=-2+4[x(x)'+(x)'x]$---->Product Rule
Thus,
$\displaystyle f'(x)=-2+4[x(1)+(1)x]$---->Derivative of 'x',
Thus,
$\displaystyle f'(x)=-2+4[2x]=-2+8x$