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Q: If z = x^2 y, x = t^2, y = t^3, find dz/dt at t = 1. fx = 2xy, fy = x^2. dx/dt = 2t, dy/dt = 3t^2. dz/dt = (2xy)(2t) + (x^2)(3t^2). With x = t^2, y = t^3: 2xy = 2t^5, so the first term is 2t^5 times 2t = 4t^6; x^2 = t^4, so the second term is t^4 times 3t^2 = 3t^6. Total: 7t^6. At t=1: 7. Check: z = t^7, dz/dt = 7t^6 = 7 at t=1!