## Calculus (3rd Edition)

$$y'= 6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$
Recall that $(x^n)'=nx^{n-1}$ Since $y=(3t^2+20t^{-3})^6$, then by the chain rule the derivative $y'$ is given by $$y'=6 (3t^2+20t^{-3})^5(6t-60t^{-4})=6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$