1. **State the problem:** We want to show that the function $f(x,y) = x \cdot e^{xy}$ is differentiable at the point $(2,0)$. 2. **Evaluate the function at the point:** $$f(2,0) = 2 \cdot e^{2 \cdot 0} = 2 \cdot e^0 = 2 \cdot 1 = 2$$ 3. **Find the partial derivatives:** - Partial derivative with respect to $x$: $$f_x(x,y) = \frac{\partial}{\partial x} \left(x e^{xy}\right) = e^{xy} + x e^{xy} \cdot y = e^{xy} (1 + xy)$$ - Evaluate at $(2,0)$: $$f_x(2,0) = e^{2 \cdot 0} (1 + 2 \cdot 0) = 1 \cdot 1 = 1$$ - Partial derivative with respect to $y$: $$f_y(x,y) = \frac{\partial}{\partial y} \left(x e^{xy}\right) = x e^{xy} \cdot x = x^2 e^{xy}$$ - Evaluate at $(2,0)$: $$f_y(2,0) = 2^2 e^{2 \cdot 0} = 4 \cdot 1 = 4$$ 4. **Define the error term $E(x,y)$:** $$E(x,y) = f(x,y) - f(2,0) - f_x(2,0)(x-2) - f_y(2,0)(y-0)$$ Substitute values: $$E(x,y) = x e^{xy} - 2 - 1 \cdot (x-2) - 4y = x e^{xy} - 2 - x + 2 - 4y = x e^{xy} - x - 4y$$ 5. **Check the limit for differentiability:** We need to verify: $$\lim_{(x,y) \to (2,0)} \frac{|E(x,y)|}{\sqrt{(x-2)^2 + y^2}} = 0$$ Rewrite $E(x,y)$ using the Mean Value Theorem or Taylor expansion around $(2,0)$: Since $f$ is smooth, the remainder $E(x,y)$ behaves like the second order terms, so the numerator is approximately $C \cdot ((x-2)^2 + y^2)$ for some constant $C$. Therefore, $$\frac{|E(x,y)|}{\sqrt{(x-2)^2 + y^2}} \approx C \cdot \sqrt{(x-2)^2 + y^2} \to 0 \text{ as } (x,y) \to (2,0)$$ 6. **Conclusion:** Since the limit is zero, $f$ is differentiable at $(2,0)$ with derivative given by the linear map: $$L(h,k) = f_x(2,0) h + f_y(2,0) k = 1 \cdot h + 4 \cdot k$$