Bivariate Differentiability

The differentiability of bivariate functions has slightly different criteria compared to its univariate case. It involves the partial derivatives and a tangent plane.

Definition §

Differentiability

Let $f(x,y)$ be a real bivariate function. Then $f$ is differentiable at the point $(a,b)$ iff:

• The partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist at $(a,b)$ and
• The linear approximation of the tangent plane is sufficiently ‘close’. That is:

$$\underset{(x,y)\to (a,b)}{\lim }\left[\frac{f(x,y)-(f(a,b)+\frac{\partial f}{\partial x}{|}_{(a,b)}\cdot (x-a)+\frac{\partial f}{\partial y}{|}_{(a,b)}\cdot (y-b))}{\sqrt{(x-a{)}^2+(y-b{)}^2}}\right]=0$$

equivalently (using gradient vector )

$$\underset{(x,y)\to (a,b)}{\lim }\left[\frac{f(x,y)-(f(a,b)+\nabla f(a,b)\cdot (x-a,y-b))}{\text{dist}((x,y),(a,b))}\right]=0$$

A way to interpret the limit is that the numerator must always be smaller than the denominator for the limit to be 0. That means the linear approximation via the tangent plant must always be closer than the distance between the input points themselves.

Theorems §

T1: Continuous partial derivates implies differentiability

Let $f(x,y)$ be a real, bivariate function. Then:

$$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\text{ exist and are continuous}⟹f(x,y)\text{ is differentiable}$$

Examples §

1: Differentiability of polynomial function

Where is the function $f(x,y)=x^2+y^2$ differentiable?

Solution

First, check if the partial derivatives exist, and find where they are continuous:

• $\frac{\partial f}{\partial x}=2x$ is continuous everywhere (as it is a polynomial) over $x\in \mathbb{R}$
• $\frac{\partial f}{\partial y}=2y$ is also continuous everywhere for $y\in \mathbb{R}$

Hence, by Theorem 1, $f$ is differentiable for all $(x,y)\in {\mathbb{R}}^2$

Visuals

2: Differentiability of a piecewise bivariate function

Where is $f$ differentiable?

f(x,y) = \begin{cases} \dfrac{x^2}{\sqrt{ x^2 + y^2 }}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0)

\end{cases}

>[!info]- Solution >Because this function is piecewise, we will need to consider each case separately: >1. $(x,y) \neq (0,0)$: >We first need to check if the partial derivatives exist, and if they are [continuous](Continuity%20(Calculus).md). >$$ >\begin{align} > f_{x} &= \frac{ \partial f }{ \partial x } = \dfrac{2x}{\sqrt{ x^2 + y^2 }} + 2x \cdot -\dfrac{1}{2} \cdot \dfrac{x^2}{(x^2 + y^2)^{3/2}} \\ > &= \dfrac{2x}{\sqrt{ x^2 + y^2 }} - \dfrac{x^3}{(x^2 + y^2)^{3/2}} >\end{align} >$$ >This function is well-defined for all points except $(0,0)$. Also, we can use [continuity theorems](Continuity%20(Calculus).md#Theorems) to show that $f_{x}$ is the composition of continuous functions. Hence, it is continuous. >$$ >f_{y} = \frac{ \partial f }{ \partial y } = \dfrac{x^2}{( x^2 + y^2 )^{3/2}} \cdot -\dfrac{1}{2} \cdot 2y = \dfrac{-x^2\cdot y}{(x^2 + y^2)^{3/2}} >$$ >A similar argument can be used to show that $f_{y}$ exists and is continuous. >Since all partial derivatives exist and are continuous, by [theorem 1](#^t1), $f$ is differentiable for all points except $(0,0)$ >2. $(x,y) = (0,0)$ >This is a bit trickier to prove. We need to check if the partial derivatives actually exist, and we can't just do $\frac{ \partial f }{ \partial x }$, because we are only given a single point. That means we must do the (dreaded) [differentiation by first principles](Differentiation%20By%20First%20Principles.md): >$$ >\begin{align} >\left.\frac{ \partial f }{ \partial x }\right\vert_{(x,y) = (0,0)} &= \lim_{ h \to 0 } \left( \dfrac{f(h,0) - f(0,0)}{h} \right) \tag{Fix $y$ since we are differentiating w.r.t $x$} \\ > &= \lim_{ h \to 0 }\left( \dfrac{\frac{h^2}{\sqrt{ h^2 }} - 0}{h} \right) = \lim_{ h \to 0 }\left( \dfrac{\frac{h^{\cancel{2}}}{\lvert h \rvert } - 0}{\cancel{ h }} \right) = \lim_{ h \to 0 } \dfrac{h}{\lvert h \rvert } \tag{$\sqrt{ h^2 } = \lvert h \rvert $} >\end{align} >$$ >The [absolute value function](Absolute%20Value%20Function.md) is *not* [differentiable](Differentiability.md) at $0$, because $\lim_{ h \to 0^+ } \lvert h \rvert = 1$ but $\lim_{ h \to 0^- } \lvert h \rvert = -1$, and since the limits from both sides are different, the limit does not exist. Hence, $\frac{ \partial f }{ \partial x }$ does not exist at $(0,0)$. >Since our requirement for bivariate differentiability requires *both* partial derivatives to exist, we can stop here. $f$ is not differentiable at $(0,0)$. > >Hence $f$ is differentiable for all $(x,y) \in \mathbb{R}^2\setminus \{ (0,0) \}$ >[!info]- Visuals >