Given the curve x=t^3+3t-1, y=t^2+4t-4, derive the result dy/dx=2(t+2)/3(t^2+1).
for x=t^3+3t-1, y=t^2+4t-4
dx/dt = 3t^2 + 3
dy/dt = 2t + 4
dy/dx = (dy/dt) / (dx/dt)
= (2(t+2)) / 3(t^2 + 1)
dx/dt = 3t^2 + 3
dy/dt = 2t + 4
dy/dx = (dy/dt) / (dx/dt)
= (2(t+2)) / 3(t^2 + 1)