1. The problem asks to find the partial derivatives $f_u$, $f_v$, and $f_w$ of the function $$f(u,v,w) = e^{uv} \ln w.$$ 2. Recall the rules for partial derivatives: when differentiating with respect to one variable, treat the other variables as constants. 3. To find $f_u$, differentiate $f(u,v,w)$ with respect to $u$: $$f_u = \frac{\partial}{\partial u} \left(e^{uv} \ln w\right) = \ln w \cdot \frac{\partial}{\partial u} e^{uv} = \ln w \cdot e^{uv} \cdot \frac{\partial}{\partial u} (uv) = \ln w \cdot e^{uv} \cdot v = v e^{uv} \ln w.$$ 4. To find $f_v$, differentiate $f(u,v,w)$ with respect to $v$: $$f_v = \frac{\partial}{\partial v} \left(e^{uv} \ln w\right) = \ln w \cdot \frac{\partial}{\partial v} e^{uv} = \ln w \cdot e^{uv} \cdot \frac{\partial}{\partial v} (uv) = \ln w \cdot e^{uv} \cdot u = u e^{uv} \ln w.$$ 5. To find $f_w$, differentiate $f(u,v,w)$ with respect to $w$: $$f_w = \frac{\partial}{\partial w} \left(e^{uv} \ln w\right) = e^{uv} \cdot \frac{\partial}{\partial w} \ln w = e^{uv} \cdot \frac{1}{w} = \frac{e^{uv}}{w}.$$ 6. Now match these results with the given options: - For $f_u$, the correct answer is $v e^{uv} \ln w$, which is not listed exactly, but the closest correct form is none of the options given (A is $u e^{uv} \ln w$, B is $u e^u \ln w$, C is $e^{uv} \ln w$, D is $-e^{uv} \ln w$). Since none matches $v e^{uv} \ln w$, the correct derivative is $v e^{uv} \ln w$. - For $f_v$, the correct answer is $u e^{uv} \ln w$, which matches option B. - For $f_w$, the correct answer is $\frac{e^{uv}}{w}$, which matches options A and B. Final answers: 14. $f_u = v e^{uv} \ln w$ (not listed exactly) 15. $f_v = u e^{uv} \ln w$ (option B) 16. $f_w = \frac{e^{uv}}{w}$ (options A or B)