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

$\csc^2 (\dfrac{x-1}{3})$
Simplify: $\dfrac{d(-3 \cot(\dfrac{x-1}{3})+C)}{dx}$ Thus, we have $\dfrac{d(-3 \cot(\dfrac{x-1}{3})+C)}{dx}=(3)[\dfrac{1}{3 \sin^2 (\dfrac{x-1}{3})}]=\csc^2 (\dfrac{x-1}{3})$