- #1
evinda
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Hello! (Wave)
Suppose that we want to check if $f(x,y)=\frac{x}{y}+\frac{y}{x}$ is differentiable at each point of its domain and if it is $C^1$.
The domain is $D=\{ (x,y) \in \mathbb{R}^2: x \neq 0 \text{ and } y \neq 0\}$.The partial derivatives :
$$\frac{\partial{f}}{\partial{x}}=\frac{x^2-y^2}{x^2 y} \\ \frac{\partial{f}}{\partial{y}}=\frac{-x^2+y^2}{xy^2}$$
are continuous on $D$. So $f$ is $C^1$ and so it is differentiable.
Does this suffice?
Because there is the following definition:
Let $f: \mathbb{R}^2 \to \mathbb{R}$. We say that $f$ is differentiable in $(x_0, y_0)$, if $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ exist in $(x_0, y_0) $ and
$$\frac{f(x,y)-f(x_0, y_0)- \left[ \frac{\partial{f}}{\partial{x}}(x_0, y_0)\right](x-x_0)-\left[ \frac{\partial{f}}{\partial{y}}(x_0, y_0)\right](y-y_0) }{||(x,y)-(x_0, y_0)||} \to 0$$while $(x,y) \to (x_0, y_0)$.Don't we have to show the latter?
Suppose that we want to check if $f(x,y)=\frac{x}{y}+\frac{y}{x}$ is differentiable at each point of its domain and if it is $C^1$.
The domain is $D=\{ (x,y) \in \mathbb{R}^2: x \neq 0 \text{ and } y \neq 0\}$.The partial derivatives :
$$\frac{\partial{f}}{\partial{x}}=\frac{x^2-y^2}{x^2 y} \\ \frac{\partial{f}}{\partial{y}}=\frac{-x^2+y^2}{xy^2}$$
are continuous on $D$. So $f$ is $C^1$ and so it is differentiable.
Does this suffice?
Because there is the following definition:
Let $f: \mathbb{R}^2 \to \mathbb{R}$. We say that $f$ is differentiable in $(x_0, y_0)$, if $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ exist in $(x_0, y_0) $ and
$$\frac{f(x,y)-f(x_0, y_0)- \left[ \frac{\partial{f}}{\partial{x}}(x_0, y_0)\right](x-x_0)-\left[ \frac{\partial{f}}{\partial{y}}(x_0, y_0)\right](y-y_0) }{||(x,y)-(x_0, y_0)||} \to 0$$while $(x,y) \to (x_0, y_0)$.Don't we have to show the latter?