Find $\displaystyle f'(a)$
$\displaystyle f(x)=3-2x+4x^2$
In that case.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,
$\displaystyle f'(x)=-2+4[2x]=-2+8x$