## University Calculus: Early Transcendentals (3rd Edition)

We have: $\dfrac{d}{dx}[(\dfrac{x+3}{x-2})^3+C]$ $\dfrac{d}{dx}[(\dfrac{x+3}{x-2})^3+C] =(3) (\dfrac{x+3}{x-2})^2 (\dfrac{5}{(x-2)^2})=-15\dfrac{(x+3)^2}{(x-2)^4}$ Hence, the statement is true.