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)

spanDEFINITIONspan 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 \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \Delta z=f(x_{0}+\Delta x, y_{0}+\Delta y)-f(x_{0}, y_{0})}} \]

Finding the Change in \(z\)

1. Find the change \(\Delta z\) in \(z=f(x,y)=x^{2}y-1\) from \((x_{0},y_{0})\) to \((x_{0}+\Delta x, y_{0}+\Delta y)\).
2. Use the answer to calculate the change in \(z\) from \((1,2)\) to \((1.1,1.9)\).

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.

spanDEFINITIONspan 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 \(D\).

Showing That a Function of Two Variables Is Differentiable

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 \((x_{0},y_{0})\).

spanDEFINITIONspan 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 \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ dz=f_{x}(x_{0},y_{0})\,dx+f_{y}(x_{0},y_{0})\,dy }} \]

Finding the Differential \(dz\) of \(z=f(x,y)\)

Find the differential \(dz\) of each function:

1. \(f(x,y) =e^{x}\cos\;y\)
2. \(f(x,y) =\dfrac{\ln\;x}{y}\)

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.

Using the Differential \(dz\) to Approximate the Change in \(z\)

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.

Using the Differential in Astronomy

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 \(10.6\%\).

Using the Differential in Error Analysis

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.

Finding the Differential of a Function of Three Variables

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.