OBJECTIVES

When you finish this section, you should be able to:

1. Find the change in $z \,=\, f( x, y)$ (p. 841)
2. Show that a function of two variables is differentiable (p. 841)
3. Use the differential $dz$ to approximate a change in $z$ (p. 844)
4. Find the differential of a function of three or more variables (p. 846)

DEFINITION Change in $z=f(x,y)$

Let $z=f(x,y)$ be a function of two variables with domain $D$. Let $(x_{0},y_{0})$ be an interior point of $D$, and let $\Delta x$ and $\Delta y$ represent changes in $x$ and in $y$, respectively. If $\Delta x$ and $\Delta y$ are chosen so that the point $(x_{0}+\Delta x, y_{0}+\Delta y)$ is also in $D$, then the change in $z$, denoted by $\Delta z$, is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \Delta z=f(x_{0}+\Delta x, y_{0}+\Delta y)-f(x_{0}, y_{0})}}$$

Solution (a) $$\begin{eqnarray*} \Delta z&=&f(x_{0}+\Delta x, y_{0}+\Delta\;y)-f(x_{0},y_{0})\nonumber\\ &=& [ (x_{0}+\Delta x)^{2}(y_{0}+\Delta y)-1]-[ x_{0}^{2}\,y_{0}-1] \end{eqnarray*}$$

(b) Let $(x_{0},y_{0}) =(1,2)$ and $(x_{0}+\Delta x,y_{0}+\Delta y) =( 1.1,1.9)$. Then $$\Delta z=[ ( 1.1) ^{2}( 1.9) -1] -[(1) ^{2}( 2) -1] =0.299$$

NOW WORK

Problem 7.

NEED TO REVIEW?

The differentials $dx$ and $dy$ of a function $y=f (x)$ of one variable are discussed in Section 3.4, pp. 230-232.

NOTE

$\eta$ is the Greek letter eta.

DEFINITION Differentiability of $z=f(x,y)$ at $(x_{0},y_{0})$

Let $z=f(x,y)$ be a function of two variables whose domain is $D$. Let $(x_{0},y_{0})$ be an interior point of $D$ and let $\Delta x$ and $\Delta y$ be chosen so that the point $(x_{0}+\Delta x, y_{0}+\Delta y)$ is also in $D$. If the change $\Delta z$ from $(x_{0},y_{0})$ to $(x_{0}+\Delta x,y_{0}+\Delta y)$ can be expressed in the form $$\begin{equation*}{\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \Delta z=f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0},y_{0})\Delta y+\eta _{1}\Delta x+\eta _{2}\Delta y }}} \tag{2} \end{equation*}$$

where both $\eta _{1}$ and $\eta _{2}$ are functions of $\Delta x$ and $\Delta y$ for which $$\begin{equation*} \lim\limits_{(\Delta x, \Delta y)\rightarrow (0,\, 0)}\eta _{1}=0\qquad \hbox{and}\qquad \lim\limits_{(\Delta x, \Delta y)\rightarrow (0,\, 0)}\eta _{2}=0 \tag{3} \end{equation*}$$

then $z=f(x,y)$ is said to be differentiable at the point $(x_{0},y_{0})$.

If the domain $D$ of a function $z=f(x,y)$ is an open set and if $z=f(x,y)$ is differentiable at every point of $D$, then we say $f$ is differentiable on $D$.

Show that $z=f(x, y)=x^{2}y-1$ is differentiable.

Solution The domain of $f$ is the $xy$-plane and the partial derivatives of $f$ are $$f_{x}(x,y) =2xy\qquad \hbox{and}\qquad f_{y}(x,y) =x^{2}$$

We find the change $\Delta z$ in $z$ and express it in the form of (2). $$\begin{eqnarray*} \Delta z &=& [ (x+\Delta x)^{2}(y+\Delta y)-1 ] - [ x^{2}y-1 ]\\ &=& [ x^{2}+2x\Delta x+ ( \Delta x ) ^{2} ] ( y+\Delta y ) - x^{2}y \\ &=&x^{2}\Delta y+2xy\Delta x+2x\Delta x\Delta y+y(\Delta x)^{2}+(\Delta x)^{2}\Delta y \\ &=&\underset{\color{#0066A7}{\hbox{\(f_{x}(x,y)\)}}}{\underbrace{ ( 2xy ) }}\Delta x+\! \underset{\color{#0066A7}{\hbox{\(f_{y}(x,y)\)}}}{\underbrace{ ( x^{2} ) }} \Delta y+\,\underset{\color{#0066A7}{\hbox{\(\eta _{1}\)}}}{\underbrace{ [ y\Delta x ] }}\Delta x+\,\underset{\color{#0066A7}{\hbox{\(\eta _{2}\)}}}{\underbrace{ [ 2x\Delta x+(\Delta x)^{2} ] }}\Delta y\\ &=&f_{x}(x,y) \Delta x+f_{y}(x,y) \Delta y+\eta _{1}\Delta x+\eta _{2}\Delta y \end{eqnarray*}$$

Equation (2) is satisfied. It remains to show that equation (3) is satisfied. $$\lim_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{1}=\lim_{\Delta x\rightarrow 0}( y\Delta x) =0\quad\!\! \hbox{and}\!\!\quad \lim_{( \Delta x,\Delta y) \rightarrow (0,0) }\eta _{2}=\lim_{\Delta x\rightarrow 0}[ 2x\Delta x+(\Delta x)^{2}] =0$$

So, $z=f(x,y)$ is differentiable on its domain.

NOTE

There are other ways to choose $\eta _{1}$ and $\eta _{2}$ so that equations (2) and (3) are satisfied.

NOW WORK

Problem 29.

THEOREM Continuous Partial Derivatives Are Sufficient for Differentiability

Let $z=f(x,y)$ be a function of two variables whose domain is $D$. Let $(x_{0},y_{0})$ be an interior point of $D$. If the partial derivatives $f_{x}$ and $f_{y}$ exist at each point of some disk centered at $(x_{0},y_{0})$, and if $f_{x}$ and $f_{y}$ are each continuous at $(x_{0},y_{0})$, then $f$ is differentiable at $(x_{0},y_{0})$.

DEFINITION Differential $dz$ of $z=f(x,y)$

Let $z=f(x,y)$ be a function of two variables whose domain is $D$. Let $(x_{0},y_{0})$ be an interior point of $D$ and let $\Delta x$ and $\Delta y$ be chosen so that the point $(x_{0}+\Delta x,y_{0}+\Delta y)$ is also in $D$. If $f$ is differentiable at the point $(x_{0},y_{0})$, then the differentials $dx$ and $dy$ are defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ dx=\Delta x\qquad \hbox{and}\qquad dy=\Delta y }}$$

The differential $dz$, also called the total differential of $z=f(x,y)$ at $(x_{0},y_{0})$, is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ dz=f_{x}(x_{0},y_{0})\,dx+f_{y}(x_{0},y_{0})\,dy }}$$

Find the differential $dz$ of each function:

Solution (a) $f$ is defined everywhere in the $xy$-plane. The partial derivatives of $f$ are $$\begin{equation*} f_{x}(x,y) =e^{x}\cos\;y\qquad f_{y}(x,y) =-e^{x}\;\sin\;y \end{equation*}$$

Since $f_{x}$ and $f_{y}$ are continuous everywhere in the $xy$-plane, the function $z=f(x,y)$ is differentiable. The differential $dz$ is $$dz=e^{x}\cos\;y\,dx-e^{x}\sin\;y\,dy$$

(b) The domain of $f$ is $\left\{(x,y) \,|\,x>0,\,y\neq 0\right\} .$ The partial derivatives of $f$ are $$f_{x}(x,y) =\dfrac{1}{xy}\qquad f_{y}(x,y) =-\dfrac{\ln\;x}{y^{2}}$$

Since both partial derivatives exist and are continuous at every point $(x_{0},y_{0})$ in the domain of $f$, the function $z=f(x,y)$ is differentiable at every point $(x_{0},y_{0})$ in its domain. The differential $dz$ is $$dz=\dfrac{1}{xy}\, dx-\dfrac{\ln\;x}{y^{2}}\,dy$$

NOW WORK

Problem 11.

For the function $z=f(x, y)=x^{2}y-1,$ use the differential $dz$ to approximate the change in $z$ from $(1,2)$ to $(1.1,1.9)$.

Solution Example 2 shows $f$ is differentiable and $f_{x}(x,y) =2xy$ and $f_{y}(x,y) =x^{2}.$

Let $( x_{0},y_{0}) =( 1,2)$ and $( x_{0}+\Delta x,y_{0}+\Delta y) =( 1.1,1.9) .$ Then $$dx=\Delta x=1.1-1=0.1, \hbox{ and } dy=\Delta y=1.9-2= -0.1.$$

Using (4), an approximation to the change in $z$ is $$\Delta z\approx dz=f_{x}(1,2)\,dx+f_{y}(1,2)\,dy=2(1) (2) ( 0.1) +(1) ( -0.1) =0.4-0.1=0.3$$

NOW WORK

Problem 33.

The luminosity $L$ (total power output in watts, W) of a star is given by the formula $$L=L( R,T) =4\pi R^{2}\sigma T^{4}$$

where $R$ is the radius of the star (in meters), $T$ is its effective surface temperature (in kelvin, K), and $\sigma$ is the Stefan–Boltzmann constant. For the sun, $L_{s}=(3.90\times\;10^{26})$ $\rm{W}$, $R_{s}=(6.94\times\;10^{8})\;\rm{m}$ , and $T_{s}=4800\;\rm{K}$. Suppose in a billion years, the changes in the Sun will be $\Delta R_{s}=(0.08\;\times\;10^{8})\;\rm{m}$ and $\Delta T_{s}=100\;\rm{K}$. What will be the resulting percent increase in luminosity?

Solution We begin with $L=4\pi\;R^{2}\sigma\;T^{4}.$ Then $$dL=4\pi \sigma (2R) T^{4}dR+4\pi \sigma R^{2}( 4T^{3}) dT=8\pi \sigma RT^{3}( T dR+2R\ dT)$$

The relative error in luminosity is $$\begin{eqnarray*} \frac{\Delta L}{L}&\approx& \frac{dL}{L}=\dfrac{8\pi \sigma RT^{3}}{4\pi R^{2}\sigma T^{4}}\left( TdR+2RdT\right) =2\left( \frac{dR}{R}+2\frac{dT}{T} \right) =2\frac{\Delta R}{R}+4\frac{\Delta T}{T}\\[4pt] &=&\frac{2(0.08\times 10^{8})}{6.94\times 10^{8}}+\frac{4(100)}{4800}\approx 0.106 \end{eqnarray*}$$

The percent increase in luminosity will be approximately $10.6\%$.

A cola company requires a can in the shape of a right circular cylinder of height $10~\rm{cm}$ and radius $3~\rm{cm}$. If the manufacturer of the cans claims a percentage error of no more than $0.2\%$ in the height and no more than $0.1\%$ in the radius, what is the approximate maximum variation in the volume of the can?

Solution The volume $V$ of a right circular cylinder of height $h~\rm{cm}$ and radius $R\;\rm{cm}$ is $V=\pi\;R^{2}h\;\rm{cm}^{3}$. We find the differential $dV.$ $$dV=\frac{\partial V}{\partial R}dR+\frac{\partial V}{\partial h}dh=2\pi\;Rh\,dR+\pi\;R^{2}dh$$

The relative error in the radius $R$ is $\dfrac{|\Delta R|}{R}=\dfrac{|dR|}{R}=0.001$, and the relative error in the height $h$ is $\dfrac{\left\vert \Delta h\right\vert }{h}=\dfrac{|dh|}{h}=0.002$. The relative error in the volume $V$ is $$\begin{eqnarray*} \frac{|\Delta V|}{V} &\approx &\frac{|dV|}{V}=\frac{\left\vert 2\pi Rh\, dR+\pi R^{2} dh\right\vert }{\pi R^{2}h}=\left\vert 2\dfrac{dR}{R}+\dfrac{dh}{h}\right\vert\\[4pt] &=&\left\vert 2\dfrac{\Delta R}{R}+\dfrac{\Delta h}{h} \right\vert \leq 2\frac{\left\vert \Delta R\right\vert }{R}+\frac{\left\vert \Delta h\right\vert }{h} \\ &=&2(0.001)+0.002=0.004 \end{eqnarray*}$$

The maximum variation in the volume is approximately $0.4\%$, so the actual volume of the container varies as follows: $$\begin{equation*} V=\pi R^{2}h\pm 0.004( \pi R^{2}h) =\pi R^2 h ( 1\pm 0.004) =90\pi ( 1\pm 0.004) \rm{cm}^{3} \end{equation*}$$

The volume $V$ is between $89.64\pi \approx 281.612\;\rm{cm}^{3}$ and $90.36\pi \approx 283.874\rm{cm}^{3}$.

NOW WORK

Problem 39.

THEOREM Differentiability Is Sufficient for Continuity

Let $z=f(x,y)$ be a function of two variables whose domain is $D$. Let $(x_{0},y_{0})$ be an interior point of $D$. If $f$ is differentiable at $(x_{0},y_{0})$, then $f$ is continuous at $(x_{0},y_{0})$.

Proof

The function $z=f(x,y)$ is continuous at $(x_{0},y_{0})$ if $\lim\limits_{(x,y) \rightarrow (x_{0},y_{0}) }f(x,y) =f(x_{0},y_{0}) .$ This is equivalent to the statement $\lim\limits_{(x,y) \rightarrow (x_{0},y_{0}) }\Delta z=0.$ Since $z=f(x,y)$ is differentiable at $(x_{0},y_{0})$, then $\Delta z$ can be expressed as $$\begin{equation*} \Delta z=f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0},y_{0})\Delta y+\eta _{1}\Delta x+\eta _{2}\Delta y \end{equation*}$$

where $\lim\limits_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{1}=0$ and $\lim\limits_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{2}=0$. Then $$\begin{equation*} \Delta z=\left[ f_{x}(x_{0}, y_{0})+\eta _{1}\right] \,\Delta x+\left[ f_{y}(x_{0}, y_{0})+\eta _{2}\right] \,\Delta y \end{equation*}$$

Now let $\Delta x=x-x_{0}$ and $\Delta y=y-y_{0}$. Then $(\Delta x,\Delta y)\rightarrow (0, 0)$ is equivalent to $(x,y)\rightarrow (x_{0}, y_{0})$ and $$\begin{eqnarray*} \lim_{(x, y)\rightarrow (x_{0}, y_{0})}\Delta z&=&\lim_{(x, y)\rightarrow (x_{0}, y_{0})}\{ [ f_{x}(x_{0}, y_{0})+\eta _{1} ] (x-x_{0} )\\[4pt] &&+\, [ f_{y}(x_{0}, y_{0})+\eta _{2} ](y-y_{0})\} =0 \end{eqnarray*}$$

That is, $f$ is continuous at $(x_{0},y_{0})$.

COROLLARY Continuity of a Function of Two Variables

Let $z=f(x,y)$ be a function of two variables whose domain is $D$. Let $(x_{0},y_{0})$ be an interior point of $D$. If the partial derivatives $f_{x}$ and $f_{y}$ exist at each point of some disk centered at $(x_{0},y_{0})$, and if $f_{x}$ and $f_{y}$ are each continuous at $(x_{0},y_{0})$, then $f$ is continuous at $(x_{0},y_{0})$.

If $z=f(x,y)$,

• The continuity of the partial derivatives $f_{x}$ and $f_{y}$ implies the differentiability of $f$.
• The differentiability of $f$ implies the continuity of $f$.
• The continuity of the partial derivatives $f_{x}$ and $f_{y}$ implies the continuity of $f$.
• The existence of the partial derivatives $f_{x}$ and $f_{y}$ does not necessarily mean $f$ is differentiable.
• The existence of the partial derivatives $f_{x}$ and $f_{y}$ does not necessarily mean $f$ is continuous.

Find the differential $dw$ of the function $w=f(x,y,z)=3x^{2}\sin ^{2}y\cos z$.

Solution The function $f$ is defined everywhere in space. We begin by finding the partial derivatives of $f.$ $$\begin{eqnarray*} f_{x}(x,y,z) &=&6x\;\sin^{2}\;\!y\;\cos\;z\qquad f_{y}(x,y,z) =6x^{2}\;\sin\;y\;\cos\;y\cos\;z\\ f_{z}(x,y,z) &=&-3x^{2}\sin^{2}\;\!y\;\sin\;z \end{eqnarray*}$$

Since the partial derivatives are continuous everywhere, we have $$\begin{eqnarray*} dw&=&f_{x}(x,y,z)\ dx+f_{y}(x,y,z)\ dy+f_{z}(x,y,z)\ dz\\ &=&6x\;\sin^{2}\;\!y\;\cos\;z\,dx+6x^{2}\;\sin\;y\;\cos\;y\;\cos\;z\,dy-3x^{2}\;\sin^{2}\;\!y\;\sin\;z\,dz \end{eqnarray*}$$

NOW WORK

Problem 23.