## Calculus 10th Edition

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