1. **State the problem:** Find the first and second derivatives of the function $$w = 3z^{-2} - \frac{1}{z}$$. 2. **Recall the power rule for derivatives:** - If $$f(z) = z^n$$, then $$f'(z) = nz^{n-1}$$. - For negative exponents, the rule applies the same way. 3. **Rewrite the function for clarity:** $$w = 3z^{-2} - z^{-1}$$ 4. **Find the first derivative $$w'$$:** $$w' = \frac{d}{dz}(3z^{-2}) - \frac{d}{dz}(z^{-1})$$ Using the power rule: $$\frac{d}{dz}(3z^{-2}) = 3 \times (-2) z^{-3} = -6z^{-3}$$ $$\frac{d}{dz}(z^{-1}) = -1 \times z^{-2} = -z^{-2}$$ So, $$w' = -6z^{-3} - (-z^{-2}) = -6z^{-3} + z^{-2}$$ 5. **Find the second derivative $$w''$$:** $$w'' = \frac{d}{dz}(-6z^{-3} + z^{-2}) = \frac{d}{dz}(-6z^{-3}) + \frac{d}{dz}(z^{-2})$$ Using the power rule again: $$\frac{d}{dz}(-6z^{-3}) = -6 \times (-3) z^{-4} = 18z^{-4}$$ $$\frac{d}{dz}(z^{-2}) = -2 z^{-3}$$ So, $$w'' = 18z^{-4} - 2z^{-3}$$ 6. **Final answers:** $$w' = -6z^{-3} + z^{-2}$$ $$w'' = 18z^{-4} - 2z^{-3}$$ These derivatives tell us how the function changes and how the rate of change itself changes with respect to $$z$$.