Answer
$f''(x)=\dfrac{56}{(x-4)^3}$
Work Step by Step
First Derivative:
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(x^2+3x); u'(x)=(2x+3)$
$v(x)=(x-4) ; v'(x)=1$
$f'(x)=\frac{(2x+3)(x-4)-(1)(x^2+3x)}{(x-4)^2}=\dfrac{x^2-8x-12}{(x-4)^2}$
Second Derivative:
Using the quotient rule: $fâ'(x)=\frac{d}{dx}f'(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(x^2-8x-12); u'(x)=(2x-8)$
$v(x)=(x-4)^2; v'(x)=(2x-8)$
$f''(x)=\frac{(2x-8)(x-4)^2-(x^2-8x-12)(2x-8)}{((x-4)^2)^2}=\frac{56}{(x-4)^3}.$
