1. **Problem Statement:** Find the first partial derivatives of the functions: 8. $f(u,v) = e^{uv}$ 9. $g(x,y) = e^{-xy}$ 10. $f(s,t) = \ln(s^2 - t^2)$ 2. **Formulas and Rules:** - The partial derivative with respect to a variable treats other variables as constants. - For exponential functions: $\frac{\partial}{\partial x} e^{h(x)} = e^{h(x)} \cdot h'(x)$. - For logarithmic functions: $\frac{\partial}{\partial x} \ln(h(x)) = \frac{h'(x)}{h(x)}$. 3. **Calculations:** **8.** $f(u,v) = e^{uv}$ - Partial derivative with respect to $u$: $$\frac{\partial f}{\partial u} = e^{uv} \cdot \frac{\partial}{\partial u}(uv) = e^{uv} \cdot v$$ - Partial derivative with respect to $v$: $$\frac{\partial f}{\partial v} = e^{uv} \cdot \frac{\partial}{\partial v}(uv) = e^{uv} \cdot u$$ **9.** $g(x,y) = e^{-xy}$ - Partial derivative with respect to $x$: $$\frac{\partial g}{\partial x} = e^{-xy} \cdot \frac{\partial}{\partial x}(-xy) = e^{-xy} \cdot (-y) = -y e^{-xy}$$ - Partial derivative with respect to $y$: $$\frac{\partial g}{\partial y} = e^{-xy} \cdot \frac{\partial}{\partial y}(-xy) = e^{-xy} \cdot (-x) = -x e^{-xy}$$ **10.** $f(s,t) = \ln(s^2 - t^2)$ - Partial derivative with respect to $s$: $$\frac{\partial f}{\partial s} = \frac{\frac{\partial}{\partial s}(s^2 - t^2)}{s^2 - t^2} = \frac{2s}{s^2 - t^2}$$ - Partial derivative with respect to $t$: $$\frac{\partial f}{\partial t} = \frac{\frac{\partial}{\partial t}(s^2 - t^2)}{s^2 - t^2} = \frac{-2t}{s^2 - t^2}$$ 4. **Summary of results:** - $\frac{\partial f}{\partial u} = v e^{uv}$ - $\frac{\partial f}{\partial v} = u e^{uv}$ - $\frac{\partial g}{\partial x} = -y e^{-xy}$ - $\frac{\partial g}{\partial y} = -x e^{-xy}$ - $\frac{\partial f}{\partial s} = \frac{2s}{s^2 - t^2}$ - $\frac{\partial f}{\partial t} = \frac{-2t}{s^2 - t^2}$