1. **State the problem:** We need to find the differential $df$ of the function $f(x,y) = \sqrt{x^3 + y^2}$ at the point $(2,3)$. Then, use this differential to estimate $f(2.1, 3.06)$. 2. **Recall the formula for the differential:** The differential $df$ of a function $f(x,y)$ is given by: $$df = f_x(x,y) \, dx + f_y(x,y) \, dy$$ where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively. 3. **Find the partial derivatives:** Given $f(x,y) = \sqrt{x^3 + y^2} = (x^3 + y^2)^{1/2}$, - Partial derivative with respect to $x$: $$f_x = \frac{1}{2}(x^3 + y^2)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + y^2}}$$ - Partial derivative with respect to $y$: $$f_y = \frac{1}{2}(x^3 + y^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{x^3 + y^2}}$$ 4. **Evaluate the partial derivatives at the point $(2,3)$:** Calculate $x^3 + y^2$ at $(2,3)$: $$2^3 + 3^2 = 8 + 9 = 17$$ Then, $$f_x(2,3) = \frac{3 \cdot 2^2}{2 \sqrt{17}} = \frac{3 \cdot 4}{2 \sqrt{17}} = \frac{12}{2 \sqrt{17}} = \frac{6}{\sqrt{17}}$$ $$f_y(2,3) = \frac{3}{\sqrt{17}}$$ 5. **Write the differential at $(2,3)$:** $$df = f_x(2,3) \, dx + f_y(2,3) \, dy = \frac{6}{\sqrt{17}} \, dx + \frac{3}{\sqrt{17}} \, dy$$ 6. **Use the differential to estimate $f(2.1, 3.06)$:** Calculate the increments: $$dx = 2.1 - 2 = 0.1$$ $$dy = 3.06 - 3 = 0.06$$ Calculate $df$: $$df = \frac{6}{\sqrt{17}} \cdot 0.1 + \frac{3}{\sqrt{17}} \cdot 0.06 = \frac{6 \cdot 0.1 + 3 \cdot 0.06}{\sqrt{17}} = \frac{0.6 + 0.18}{\sqrt{17}} = \frac{0.78}{\sqrt{17}}$$ 7. **Calculate $f(2,3)$ exactly:** $$f(2,3) = \sqrt{17}$$ 8. **Estimate $f(2.1, 3.06)$ using the differential:** $$f(2.1, 3.06) \approx f(2,3) + df = \sqrt{17} + \frac{0.78}{\sqrt{17}}$$ **Final answers:** $$df = \frac{6}{\sqrt{17}} \, dx + \frac{3}{\sqrt{17}} \, dy$$ $$f(2.1, 3.06) \approx \sqrt{17} + \frac{0.78}{\sqrt{17}}$$