Question 12.1.1: Find dz/dt when z = f (x, y) = x² + y³ with x = t² and y = 2......

In this case f^{\prime}_{1}(x, y) = 2x, f ^{\prime}_{2}(x, y) = 3y^{2}, dx/dt = 2t, and dy/dt = 2. So formula (12.1.1) gives

{\frac{\mathrm{d}z}{\mathrm{d}t}}=f_{1}^{\prime}(x,y){\frac{\mathrm{d}x}{\mathrm{d}t}}+f_{2}^{\prime}(x,y){\frac{\mathrm{d}y}{\mathrm{d}t}}       (12.1.1)

where the last equality comes from substituting the appropriate functions of t for x and y respectively. In a simple case like this, we can verify the chain rule by substituting x = t² and y = 2t in the formula for f (x, y) and then differentiating w.r.t. t. The result is

as before.