Answer
$f(x)=\dfrac{6-4x}{(x^2-3x)^3},f'(4)=-\dfrac{5}{32}.$
Work Step by Step
$f(x)=(x^2-3x)^{-2}$
$u=x^2-3x$; $\dfrac{du}{dx}=2x-3$
$f(u)=u^{-2};\dfrac{d}{du}f(u)=-\dfrac{2}{u^3}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{6-4x}{(x^2-3x)^3}.$
$f'(4)=\dfrac{6-4(4)}{(4^2-3(4))^3}=-\dfrac{5}{32}.$
A graphing utility was used to verify this result.
