Answer
$\frac{dx}{dt}=1+2t\cos (t^2+2)$
$\frac{dy}{dt}=2t\sec^2(t^2+2)$
$\frac{dy}{dx}=\frac{2t\sec^2(t^2+2)}{1+2t\cos(t^2+2)}$
Work Step by Step
Find $\frac{dx}{dt}$:
$\frac{dx}{dt}=\frac{d}{dt}(t+\sin(t^2+2))=1+2t\cos (t^2+2)$
Find $\frac{dy}{dt}$:
$\frac{dy}{dt}=\frac{d}{dt}(\tan(t^2+2))=2t\sec^2(t^2+2)$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t\sec^2(t^2+2)}{1+2t\cos(t^2+2)}$
