1. **State the problem:** Differentiate the function $$f(z) = (6 - e^z)(6z + e^z)$$ with respect to $$z$$. 2. **Recall the product rule:** For two functions $$u(z)$$ and $$v(z)$$, the derivative is $$f'(z) = u'(z)v(z) + u(z)v'(z)$$. 3. **Identify:** Let $$u(z) = 6 - e^z$$ and $$v(z) = 6z + e^z$$. 4. **Compute derivatives:** - $$u'(z) = -e^z$$ (since derivative of $$6$$ is $$0$$ and derivative of $$-e^z$$ is $$-e^z$$). - $$v'(z) = 6 + e^z$$ (derivative of $$6z$$ is $$6$$ and derivative of $$e^z$$ is $$e^z$$). 5. **Apply product rule:** $$ f'(z) = u'(z)v(z) + u(z)v'(z) = (-e^z)(6z + e^z) + (6 - e^z)(6 + e^z) $$ 6. **Expand terms:** $$ (-e^z)(6z) + (-e^z)(e^z) + 6(6) + 6(e^z) - e^z(6) - e^z(e^z) $$ which is $$ -6ze^z - e^{2z} + 36 + 6e^z - 6e^z - e^{2z} $$ 7. **Simplify:** Notice $$6e^z - 6e^z = 0$$, so $$ -6ze^z - e^{2z} - e^{2z} + 36 = -6ze^z - 2e^{2z} + 36 $$ **Final answer:** $$ f'(z) = 36 - 6ze^z - 2e^{2z} $$