Find $\displaystyle f'(a)$

$\displaystyle f(x)=3-2x+4x^2$

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- Jun 12th 2006, 02:13 PMc_323_hNeed help with derivative
Find $\displaystyle f'(a)$

$\displaystyle f(x)=3-2x+4x^2$ - Jun 12th 2006, 03:11 PMJameson
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.

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

How about f'(a+h)? - Jun 12th 2006, 03:34 PMc_323_h
i haven't learned power rules yet. only difference quotient. i've figured it out though

- Jun 12th 2006, 06:12 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

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,

$\displaystyle f'(x)=-2+4[2x]=-2+8x$