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Math Help - Need help with derivative

  1. #1
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    Need help with derivative

    Find f'(a)

    f(x)=3-2x+4x^2
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  2. #2
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    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)?
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  3. #3
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    i haven't learned power rules yet. only difference quotient. i've figured it out though
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  4. #4
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    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
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