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Total Differential and Tangent Plane

A function $z = f(x,y)$ is defined on the region $D$. Let $\Delta x, \Delta y$ be the increment of $x$ and $y$. Then $f(x+\Delta x, y + \Delta y) - f(x,y)$ is called an increment of $z$ and denoted by $\Delta z$.

If $f(x,y)$ is partially differentiable, then

$\displaystyle \lim_{\Delta x \to 0}\frac{f(x+\Delta x, y + \Delta y) - f(x,y+\Delta y)}{\Delta x} = f_{x}(x,y).$

$\displaystyle \lim_{\Delta y \to 0}\frac{f(x,y+\Delta y) - f(x,y)}{\Delta y} = f_{y}(x,y)$

Thus

$\displaystyle \Delta z = f_{x}(x,y)\Delta x + f_{y}(x,y)\Delta y + \varepsilon(x,y).$

Now if $\lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\varepsilon(x,y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$, then we say a function $z = f(x,y)$ is totally diferentiable at $(x,y)$

NOTE Note that we can express $\Delta z$ as

$\displaystyle \Delta z = f(x+\Delta x, y + \Delta y) - f(x,y+\Delta y) + f(x,y+\Delta y) - f(x,y)$

Note that $f(x,y+\Delta y) - f(x,y)$ is an increment of $f$ as $y$ moved to $y + \Delta y$. Then we can write

$\displaystyle f(x,y+\Delta y) - f(x,y) = \frac{f(x,y+\Delta y) - f(x,y)}{\Delta y} \Delta y$

Similarly, $f(x+\Delta x,y) - f(x,y)$ is an increment of $f$ as $x$ moved to $x + \Delta x$. Then we can write

$\displaystyle f(x+\Delta x,y) - f(x,y) = \frac{f(x+\Delta x, y + \Delta y) - f(x,y+\Delta y)}{\Delta x} \Delta x$

Putting together,

$\displaystyle \Delta z = \frac{f(x+\Delta x, y + \Delta y) - f(x,y+\Delta y)}{\Delta x} \Delta x + \frac{f(x,y+\Delta y) - f(x,y)}{\Delta y} \Delta y.$

The differential of $x$ is denoted by $dx$ . Then for $z = f(x,y)$ is totally differentiable at $(x,y)$, Let $\Delta x \to 0, \Delta y \to 0$. Then

$\displaystyle{\Delta z = f_{x}(x,y)\Delta x + f_{y}\Delta y + \varepsilon(x,y)}$, where $\Delta x , \Delta y, \Delta z$ can be expressed as $dx, dy, dz$. Thus we write

$\displaystyle dz = f_{x}(x,y)dx + f_{y}(x,y)dy$

$dz$ is called taltal differential of $z$.

Example 4..9   Find the total differential of the following functions.
1. $z = x^2 e^{xy}$2. $f(x,y) = \log(x^2 + y^2)$3. $z = \sin{xy}$

SOLUTION 1. $\displaystyle{dz = z_x dx + z_y dy = (2x e^{xy} + x^2 y e^{xy})dx + x^2\cdot x e^{xy}dy }$

$\displaystyle{ = 2xe^{xy}(1+x)dx + x^3 e^{xy}dy\ensuremath{ \blacksquare}}$

2. $\displaystyle{df = f_{x}dx + f_{y}dy = \frac{2x}{x^2+y^2} dx + \frac{2y}{x^2+y^2} dy}$ $ \blacksquare$

3. $\displaystyle{dz = z_xdx + z_y dy = y\cos{xy}dx + x\cos{xy}dy}\ensuremath{ \blacksquare}$

Exercise 4..9   Approximate the following number by using total differential.

$\displaystyle \sqrt{27}\sqrt[3]{1021}$

SOLUTION Consider the function $f(x,y) = \sqrt{x} \sqrt[3]{y} = x^{\frac{1}{2}}y^{\frac{1}{3}}$. Let $x = 25, y = 1000$. Then $f(25,1000) = 50$. Now let $\Delta x = 2, \Delta y = 21$. Then $f(27,1021) = f(25 + \Delta x, 1000 + \Delta y)$. Note that $\Delta f = f(25 + \Delta x, 1000 + \Delta y) - f(25,1000) \approx df$. Thus

$\displaystyle f(25 + \Delta x, 1000 + \Delta y) \approx df + f(25,1000) $

Here, $df = f_{x}\Delta x + f_{y} \Delta y = \frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{3}}\Delta x + \frac{1}{3}x^{\frac{1}{2}}y^{-\frac{2}{3}}\Delta y.$ Thus for $x = 25, y =1000, \Delta x = 2, \Delta y = 21$, we have $df = (\frac{1}{2}\cdot \frac{1}{5} \cdot 10)2 + (\frac{1}{3} \cdot 5 \cdot \frac{1}{100} )21 = 2.35 $. From this, $\displaystyle{\sqrt{27}\sqrt[3]{1021} \approx \sqrt{25}\sqrt[3]{1000} + 2.35 = 52.35 \ensuremath{ \blacksquare}}$

Theorem 4..2   If $f(x,y)$ is totally differentiable at $(x_{0},y_{0})$, then $f(x,y)$ is continuous at $(x_{0},y_{0})$.

NOTE Suppose that $f(x,y)$ is totally differentiable at $(x_{0},y_{0})$, then

$\displaystyle f(x_{0} + \Delta x, y_{0} + \Delta y) - f(x_{0},y_{0})$ $\displaystyle =$ $\displaystyle f_{x}(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta y$  
  $\displaystyle +$ $\displaystyle \varepsilon(x,y)$  

where $\lim_{(\Delta x, \Delta y) \rightarrow (0,0)}\frac{\varepsilon(\Delta x, \Delta y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$. Now we show that $f(x,y)$ is continuous at $(x_0, y_0)$. In other words, we show

$\displaystyle \lim_{(x,y) \to (x_0,y_0)}f(x,y) = f(x_0,y_0)$

Let $x - x_0 = \Delta x, y - y_0 = \Delta y$. Then $(x,y) \to (x_0,y_0)$ is equivalent to $(\Delta x, \Delta y) \to (0,0)$. Thus,

$\displaystyle \lim_{(\Delta x, \Delta y) \rightarrow (0,0)}[f(x_{0} + \Delta x, y_{0} + \Delta y) - f(x_{0},y_{0})] = 0$

and $f(x,y)$ is continuous at $(x_{0},y_{0})$.

Theorem 4..3   Suppose that $f(x,y)$ is partially differentiable with respect to $x,y$ at $(x_0, y_0)$ and $f_x, f_y$ are continuous at $(x_0, y_0)$. Then $f(x,y)$ is totally differentiable at $(x_0, y_0)$.

Example 4..10   Determine the following function is totally differentiable or not.

$\displaystyle f(x,y) = \left\{\begin{array}{cl} \frac{xy}{x^2 + y^2} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0) \end{array} \right.$

SOLUTION $\displaystyle{\Delta f = f(0 + \Delta x, 0 + \Delta y) - f(0,0) = \frac{\Delta x \Delta y}{(\Delta x)^2 + (\Delta y)^2}.}$ Then $\displaystyle{\Delta f = f_{x}(0,0)\Delta x + f_{y}(0,0)\Delta y + \varepsilon(x,y)= \varepsilon(x,y).}$ Then

$\displaystyle \varepsilon(x,y) = \frac{\Delta x \Delta y}{(\Delta x)^2 + (\Delta y)^2}.$

$f(x,y)$ is totally differentiable at $(0,0)$ only if

$\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\varepsilon(x,y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$

Let $y = m(\Delta x)^2$. Then
$\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\varepsilon(x,y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$ $\displaystyle =$ $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\Delta x \Delta y}{\big((\Delta x)^2 + (\Delta y)^2 \big)^{3/2}}$  
  $\displaystyle =$ $\displaystyle \lim_{\Delta x \to 0}\frac{m}{(1+ m^2(\Delta x)^2)^{3/2}} = m.$  

Then this limit depends on the value of $m$. Thus it is not totally differentiable $ \blacksquare$ Alernative Solution By Example4.8, $f(x,y)$ is partially differentiable at $(0,0)$. But by Example4.4, $f(x,y)$ is not continuous at $(0,0)$. Thus by Theorem4.2, $f(x,y)$ is not totally diffrentiable at $(0,0)$ $ \blacksquare$

Exercise 4..10   Determine whether the following function is totally diffrentiable at $(0,0)$.

$\displaystyle f(x,y) = \log(1 + xy + y^2)$

SOLUTION $\displaystyle{f_{x}(x,y) = \frac{y}{1 + xy + y^2}}$ and $\displaystyle{f_{y}(x,y) = \frac{x + 2y}{1+xy + y^2}}$. Then $f_{x}(x,y), f_{y}(x,y)$ is continuous at $(0,0)$. Thus by Theorem4.3, $f(x,y)$ is totally differentiable at $(0,0)$.

Altenative Solution $\Delta f = f(0 + \Delta x, 0 + \Delta y) - f(0,0) = \log(1 + \Delta x \Delta y + (\Delta y)^2)$ and $\displaystyle{f_{x}(x,y) = \frac{y}{1 + xy + y^2}}$, $\displaystyle{f_{y}(x,y) = \frac{x + 2y}{1+xy + y^2}}$. Then

$\displaystyle \Delta f = f_{x}(0,0)\Delta x + f_{y}(0,0)\Delta y + \varepsilon(x,y)= \varepsilon(x,y)$

$\displaystyle{\varepsilon(x,y) = \log(1 + \Delta x \Delta y + (\Delta y)^2).}$ Thus,
$\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\varepsilon}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$ $\displaystyle =$ $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)}\frac{\log(1 + \Delta x \Delta y + (\Delta y)^2)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$  

Now let $\Delta x = r\cos{\theta}, \Delta y = r\sin{\theta}$. Then
$\displaystyle \frac{\log(1 + \Delta x \Delta y + (\Delta y)^2)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$ $\displaystyle =$ $\displaystyle \frac{\log(1 + r^2\cos{\theta}\sin{\theta} + r^2\sin^{2}{\theta})}{\sqrt{(r\cos{\theta})^2 + (r\sin{\theta})^2}}$  
  $\displaystyle =$ $\displaystyle \frac{\log\left(1 + r^2(\cos{\theta}\sin{\theta} + \sin^{2}{\theta})\right)}{r}$  

Let $\cos{\theta}\sin{\theta} + \sin^{2}{\theta} = k(\theta)$. Then
$\displaystyle \lim_{r \to 0}\frac{\log(1+ kr^2)}{r}$ $\displaystyle \stackrel{*}{=}$ $\displaystyle \lim_{r \to 0}\frac{2kr}{1+kr^2} = 0$  

Thus $f(x,y)$ is totally differentiable at $(0,0)$ $ \blacksquare$


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