Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 182: 36

$\frac{{dy}}{{dx}} = - \frac{{2t}}{{{x^3}}} + \frac{1}{t}{\text{ and }}\frac{{dy}}{{dt}} = \frac{1}{{{x^2}}} + \frac{x}{{{t^2}}}$

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