Chapter 4 - Calculating the Derivative - 4.1 Techniques for Finding Derivatives - 4.1 Exercises - Page 207: 29

$$f'(-2) =\frac{-25}{3}$$

Since $$f(x) = \frac{x^4}{6}-3x$$ Then by using rule $ \dfrac{d}{dx}x^n=nx^{n-1}$ \begin{align*} f'(x) &=\frac{d}{dx}\left( \frac{x^4}{6}-3x\right)\\ &=\frac{2}{3}x^3-3 \end{align*} Hence \begin{align*} f'(-2) &=\frac{2}{3}(-2)^3-3\\ &=\frac{-16}{3}-3\\ &=\frac{-25}{3} \end{align*}