1. **Problem statement:** Given the function $m(q) = 2q e^{-q}$, we need to state the differentiation rules applicable and find the derivative at $q=0$. 2. **Differentiation rules:** - Product rule: Used when differentiating a product of two functions. Here, $m(q)$ is a product of $2q$ and $e^{-q}$. - Chain rule: Used when differentiating a composite function. Here, $e^{-q}$ is a composite function where the exponent is $-q$. 3. **Applying the product rule:** If $m(q) = u(q) v(q)$, then $m'(q) = u'(q) v(q) + u(q) v'(q)$. Here, $u(q) = 2q$ and $v(q) = e^{-q}$. 4. **Find derivatives of $u(q)$ and $v(q)$:** - $u'(q) = \frac{d}{dq}(2q) = 2$ - $v'(q) = \frac{d}{dq}(e^{-q}) = e^{-q} \cdot (-1) = -e^{-q}$ (by chain rule) 5. **Calculate $m'(q)$:** $$m'(q) = u'(q) v(q) + u(q) v'(q) = 2 e^{-q} + 2q (-e^{-q}) = 2 e^{-q} - 2q e^{-q} = 2 e^{-q} (1 - q)$$ 6. **Evaluate $m'(0)$:** $$m'(0) = 2 e^{0} (1 - 0) = 2 \times 1 \times 1 = 2$$ **Final answer:** $m'(0) = 2$