# Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 34

$$y'= 6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$

#### Work Step by Step

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 $$y'=6 (3t^2+20t^{-3})^5(6t-60t^{-4})=6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$

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