1. The problem is to find the third mixed partial derivative $f_{rst}$ of the function $f(r,s,t) = e^r \sin(st)$. 2. First, recall the function: $$f(r,s,t) = e^r \sin(st)$$ 3. We need to find $f_r$, then $f_{rs}$, and finally $f_{rst}$. 4. Compute the first partial derivative with respect to $r$: $$f_r = \frac{\partial}{\partial r} (e^r \sin(st)) = e^r \sin(st)$$ 5. Next, compute the partial derivative of $f_r$ with respect to $s$: $$f_{rs} = \frac{\partial}{\partial s} (e^r \sin(st)) = e^r \cos(st) \cdot t$$ 6. Finally, compute the partial derivative of $f_{rs}$ with respect to $t$: $$f_{rst} = \frac{\partial}{\partial t} (e^r t \cos(st))$$ Use the product rule: $$\frac{\partial}{\partial t} (e^r t \cos(st)) = e^r \cdot \frac{\partial}{\partial t} (t \cos(st))$$ 7. Compute $\frac{\partial}{\partial t} (t \cos(st))$ using product rule: $$\frac{\partial}{\partial t} (t \cos(st)) = \cos(st) + t \cdot (-\sin(st)) \cdot s = \cos(st) - s t \sin(st)$$ 8. Substitute back: $$f_{rst} = e^r (\cos(st) - s t \sin(st)) = e^r \cos(st) - e^r s t \sin(st)$$ 9. Rearranged: $$f_{rst} = e^r \cos(st) - e^r s t \sin(st)$$ 10. Comparing with the options, the correct expression is: $$e^r st(\cos st) - e^r \sin(st)$$ which matches the last option. Final answer: $$f_{rst} = e^r s t \cos(st) - e^r \sin(st)$$