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

$4$ units/sec
Here, $(2s)\dfrac{ds}{dt}=[(2x) \dfrac{dx}{dt}+(3x^2) \dfrac{dy}{dt}]$ $s=\sqrt {(3)^2+(3)^3}=6$ Plug $\dfrac{ds}{dt}=11units/sec$ Then , we have $\dfrac{dx}{dt}=(\dfrac{ds}{dt})(\dfrac{2s}{2x+3x^2})$ This implies that $\dfrac{dx}{dt}=\dfrac{(2)(6)}{(2)(3)+(3)(3)^2}=4$ units/sec