Answer
$\frac{{dy}}{{dx}} = 2tx + {t^3}{\text{ and }}\frac{{dy}}{{dt}} = {x^2} + 3{t^2}x$
Work Step by Step
$$\eqalign{
& y = t{x^2} + {t^3}x \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& {\text{Take the derivative of both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {t{x^2}} \right] + \frac{d}{{dx}}\left[ {{t^3}x} \right] \cr
& {\text{Consider }}t{\text{ as a constant}} \cr
& \frac{{dy}}{{dx}} = t\frac{d}{{dx}}\left[ {{x^2}} \right] + {t^3}\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = t\left( {2x} \right) + {t^3}\left( 1 \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = 2tx + {t^3} \cr
& \cr
& {\text{Find }}\frac{{dy}}{{dt}} \cr
& {\text{Differentiate the function with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {t{x^2}} \right] + \frac{d}{{dt}}\left[ {{t^3}x} \right] \cr
& {\text{Consider }}x{\text{ as a constant}} \cr
& \frac{{dy}}{{dt}} = {x^2}\frac{d}{{dt}}\left[ t \right] + x\frac{d}{{dt}}\left[ {{t^3}} \right] \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dt}} = {x^2}\left( 1 \right) + x\left( {3{t^2}} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dt}} = {x^2} + 3{t^2}x \cr} $$
