Showing that a multivariable function is differentiable

Question

Let $D :={(x, y) ∈ \mathbb R^2 : y \not= 0} $

Let $f : D → R$ be defined by $f(x, y) := \frac{x}{y}$ if $(x, y) ∈ D$.

Answer the following questions:

1. Let $\epsilon > 0$. Show that there exists $µ > 0$ such that

$|\frac{1}{y}- (\frac{1}{y_0}-\frac{1}{y_0^2}(y-y_0))|$ ≤$\epsilon |y − y_0|$ if $|y − y_0| ≤ µ$.

2. Let $(x_0, y_0)$ ∈ D. Show by using the definition of the derivative and the previous question that $f$ has a derivative at ($x_0, y_0$) and that the derivative is given by

$Df_{(x_0,y_0)}(h, k) = \frac{1}{y_0} h - \frac{x_0}{y_0^2} k$

My attempts:

1. $|\frac{1}{y}- (\frac{1}{y_0}-\frac{1}{y_0^2}(y-y_0))|$ = $|\frac{1}{y}- \frac{1}{y_0}+\frac{1}{y_0^2}(y-y_0)|$ ≤ $|\frac{1}{y}- \frac{1}{y_0}| + |\frac{1}{y_0^2}(y-y_0)|$ = $|\frac{y_0 - y}{yy_0}| + |\frac{1}{y_0^2}(y-y_0)|$ = $|\frac{y- y_0}{yy_0}| + |\frac{1}{y_0^2}(y-y_0)|$ = $|y - y_0|× |\frac{1}{yy_0^3}| ≤ \epsilon |y-y_0|$ where $\epsilon = |\frac{1}{yy_0^3}|$

Is my answer here correct?

2. I don't know how to proceed to solve it. I only know the following definition:

$f$ is differentiable at $x_0$ if

Lim$_{h->0} \frac{||f(x_0 + h) - f(x_0) - J(h)||}{||h||} = 0$

Please could you show me how to solve this part? Thanks

Answer 1

The directional derivative reads, for any $(x,y) \in D$:

$$ \mathcal D_\theta f(x, y) = \lim_{\substack{v \rightarrow \theta \\ \varepsilon \rightarrow 0}} \frac{ \frac{x + \varepsilon v_1}{y + \varepsilon v_2} - \frac x y}{ \varepsilon } = \\ \lim_{\substack{v \rightarrow \theta \\ \varepsilon \rightarrow 0}} \frac{ xy + \varepsilon v_1 y - xy - \varepsilon v_2 x } {\varepsilon y (y + \varepsilon v_2)} = \\ \frac{\theta_1 y - \theta_2 x}{y^2} $$

The limit exists for any direction $\theta$ and point in $D$, so the function is differentiable.