1. **Problem 8:** Given $x = r \cos \theta$ and $y = r \sin \theta$, find $\frac{\partial r}{\partial x}$. 2. **Step 1:** Express $r$ in terms of $x$ and $y$. Since $x = r \cos \theta$ and $y = r \sin \theta$, we have $$r = \sqrt{x^2 + y^2}$$ 3. **Step 2:** Differentiate $r$ with respect to $x$ treating $y$ as constant: $$\frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2 + y^2} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}}$$ 4. **Step 3:** Substitute back $r = \sqrt{x^2 + y^2}$ to get $$\frac{\partial r}{\partial x} = \frac{x}{r}$$ --- 5. **Problem 9:** Given $J = \frac{\partial(x,y)}{\partial(u,v)}$ and $J' = \frac{\partial(u,v)}{\partial(x,y)}$, find $JJ'$. 6. **Step 1:** Recall that the Jacobian matrices $J$ and $J'$ are inverses of each other, so their determinants satisfy $$JJ' = 1$$ --- 7. **Problem 10:** Evaluate $$\lim_{(x,y) \to (1,1)} \frac{x - y}{\sqrt{x} - \sqrt{y}}$$ 8. **Step 1:** Notice that direct substitution gives $\frac{0}{0}$, an indeterminate form. Use algebraic manipulation. 9. **Step 2:** Multiply numerator and denominator by the conjugate of the denominator: $$\frac{x - y}{\sqrt{x} - \sqrt{y}} \cdot \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}} = \frac{(x - y)(\sqrt{x} + \sqrt{y})}{x - y}$$ 10. **Step 3:** Simplify numerator and denominator: $$= \sqrt{x} + \sqrt{y}$$ 11. **Step 4:** Take the limit as $(x,y) \to (1,1)$: $$\lim_{(x,y) \to (1,1)} (\sqrt{x} + \sqrt{y}) = \sqrt{1} + \sqrt{1} = 2$$ **Final answers:** - $\frac{\partial r}{\partial x} = \frac{x}{r}$ - $JJ' = 1$ - $\lim_{(x,y) \to (1,1)} \frac{x - y}{\sqrt{x} - \sqrt{y}} = 2$