# 4.4 Tangent planes and linear approximations  (Page 5/11)

 Page 5 / 11

## Continuity of first partials implies differentiability

Let $z=f\left(x,y\right)$ be a function of two variables with $\left({x}_{0},{y}_{0}\right)$ in the domain of $f.$ If $f\left(x,y\right),$ ${f}_{x}\left(x,y\right),$ and ${f}_{y}\left(x,y\right)$ all exist in a neighborhood of $\left({x}_{0},{y}_{0}\right)$ and are continuous at $\left({x}_{0},{y}_{0}\right),$ then $f\left(x,y\right)$ is differentiable there.

Recall that earlier we showed that the function

$f\left(x,y\right)=\left\{\begin{array}{cc}\frac{xy}{\sqrt{{x}^{2}+{y}^{2}}}\hfill & \left(x,y\right)\ne \left(0,0\right)\hfill \\ 0\hfill & \left(x,y\right)=\left(0,0\right)\hfill \end{array}$

was not differentiable at the origin. Let’s calculate the partial derivatives ${f}_{x}$ and ${f}_{y}\text{:}$

$\frac{\partial f}{\partial x}=\frac{{y}^{3}}{{\left({x}^{2}+{y}^{2}\right)}^{3\text{/}2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=\frac{{x}^{3}}{{\left({x}^{2}+{y}^{2}\right)}^{3\text{/}2}}.$

The contrapositive of the preceding theorem states that if a function is not differentiable, then at least one of the hypotheses must be false. Let’s explore the condition that ${f}_{x}\left(0,0\right)$ must be continuous. For this to be true, it must be true that $\underset{\left(x,y\right)\to \left(0,0\right)}{\text{lim}}{f}_{x}\left(0,0\right)={f}_{x}\left(0,0\right)\text{:}$

$\underset{\left(x,y\right)\to \left(0,0\right)}{\text{lim}}{f}_{x}\left(x,y\right)=\underset{\left(x,y\right)\to \left(0,0\right)}{\text{lim}}\frac{{y}^{3}}{{\left({x}^{2}+{y}^{2}\right)}^{3\text{/}2}}.$

Let $x=ky.$ Then

$\begin{array}{cc}\hfill \underset{\left(x,y\right)\to \left(0,0\right)}{\text{lim}}\frac{{y}^{3}}{{\left({x}^{2}+{y}^{2}\right)}^{3\text{/}2}}& =\underset{y\to 0}{\text{lim}}\frac{{y}^{3}}{{\left({\left(ky\right)}^{2}+{y}^{2}\right)}^{3\text{/}2}}\hfill \\ & =\underset{y\to 0}{\text{lim}}\frac{{y}^{3}}{{\left({k}^{2}{y}^{2}+{y}^{2}\right)}^{3\text{/}2}}\hfill \\ & =\underset{y\to 0}{\text{lim}}\frac{{y}^{3}}{{|y|}^{3}{\left({k}^{2}+1\right)}^{3\text{/}2}}\hfill \\ & =\frac{1}{{\left({k}^{2}+1\right)}^{3\text{/}2}}\underset{y\to 0}{\text{lim}}\frac{|y|}{y}.\hfill \end{array}$

If $y>0,$ then this expression equals $1\text{/}{\left({k}^{2}+1\right)}^{3\text{/}2};$ if $y<0,$ then it equals $\text{−}\left(1\text{/}{\left({k}^{2}+1\right)}^{3\text{/}2}\right).$ In either case, the value depends on $k,$ so the limit fails to exist.

## Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of $y,$ written $dy,$ is defined as ${f}^{\prime }\left(x\right)dx.$ The differential is used to approximate $\text{Δ}y=f\left(x+\text{Δ}x\right)-f\left(x\right),$ where $\text{Δ}x=dx.$ Extending this idea to the linear approximation of a function of two variables at the point $\left({x}_{0},{y}_{0}\right)$ yields the formula for the total differential for a function of two variables.

## Definition

Let $z=f\left(x,y\right)$ be a function of two variables with $\left({x}_{0},{y}_{0}\right)$ in the domain of $f,$ and let $\text{Δ}x$ and $\text{Δ}y$ be chosen so that $\left({x}_{0}+\text{Δ}x,{y}_{0}+\text{Δ}y\right)$ is also in the domain of $f.$ If $f$ is differentiable at the point $\left({x}_{0},{y}_{0}\right),$ then the differentials $dx$ and $dy$ are defined as

$dx=dx\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dy=\text{Δ}y.$

The differential $dz,$ also called the total differential    of $z=f\left(x,y\right)$ at $\left({x}_{0},{y}_{0}\right),$ is defined as

$dz={f}_{x}\left({x}_{0},{y}_{0}\right)dx+{f}_{y}\left({x}_{0},{y}_{0}\right)dy.$

Notice that the symbol $\partial$ is not used to denote the total differential; rather, $d$ appears in front of $z.$ Now, let’s define $\text{Δ}z=f\left(x+\text{Δ}x,y+\text{Δ}y\right)-f\left(x,y\right).$ We use $dz$ to approximate $\text{Δ}z,$ so

$\text{Δ}z\approx dz={f}_{x}\left({x}_{0},{y}_{0}\right)dx+{f}_{y}\left({x}_{0},{y}_{0}\right)dy.$

Therefore, the differential is used to approximate the change in the function $z=f\left({x}_{0},{y}_{0}\right)$ at the point $\left({x}_{0},{y}_{0}\right)$ for given values of $\text{Δ}x$ and $\text{Δ}y.$ Since $\text{Δ}z=f\left(x+\text{Δ}x,y+\text{Δ}y\right)-f\left(x,y\right),$ this can be used further to approximate $f\left(x+\text{Δ}x,y+\text{Δ}y\right)\text{:}$

$\begin{array}{cc}\hfill f\left(x+\text{Δ}x,y+\text{Δ}y\right)& =f\left(x,y\right)+\text{Δ}z\hfill \\ & \approx f\left(x,y\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\text{Δ}x+{f}_{y}\left({x}_{0},{y}_{0}\right)\text{Δ}y.\hfill \end{array}$

See the following figure. The linear approximation is calculated via the formula f ( x + Δ x , y + Δ y ) ≈ f ( x , y ) + f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y .

One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and are off by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume of the gadget.

## Approximation by differentials

Find the differential $dz$ of the function $f\left(x,y\right)=3{x}^{2}-2xy+{y}^{2}$ and use it to approximate $\text{Δ}z$ at point $\left(2,-3\right).$ Use $\text{Δ}x=0.1$ and $\text{Δ}y=-0.05.$ What is the exact value of $\text{Δ}z?$

First, we must calculate $f\left({x}_{0},{y}_{0}\right),{f}_{x}\left({x}_{0},{y}_{0}\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}_{y}\left({x}_{0},{y}_{0}\right)$ using ${x}_{0}=2$ and ${y}_{0}=-3\text{:}$

$\begin{array}{ccc}\hfill f\left({x}_{0},{y}_{0}\right)& =\hfill & f\left(2,-3\right)=3{\left(2\right)}^{2}-2\left(2\right)\left(-3\right)+{\left(-3\right)}^{2}=12+12+9=33\hfill \\ \hfill {f}_{x}\left(x,y\right)& =\hfill & 6x-2y\hfill \\ \hfill {f}_{y}\left(x,y\right)& =\hfill & -2x+2y\hfill \\ \hfill {f}_{x}\left({x}_{0},{y}_{0}\right)& =\hfill & {f}_{x}\left(2,-3\right)=6\left(2\right)-2\left(-3\right)=12+6=18\hfill \\ \hfill {f}_{y}\left({x}_{0},{y}_{0}\right)& =\hfill & {f}_{y}\left(2,-3\right)=-2\left(2\right)+2\left(-3\right)=-4-6=-10.\hfill \end{array}$

Then, we substitute these quantities into [link] :

$\begin{array}{c}dz={f}_{x}\left({x}_{0},{y}_{0}\right)dx+{f}_{y}\left({x}_{0},{y}_{0}\right)dy\hfill \\ dz=18\left(0.1\right)-10\left(-0.05\right)=1.8+0.5=2.3.\hfill \end{array}$

This is the approximation to $\text{Δ}z=f\left({x}_{0}+\text{Δ}x,{y}_{0}+\text{Δ}y\right)-f\left({x}_{0},{y}_{0}\right).$ The exact value of $\text{Δ}z$ is given by

$\begin{array}{cc}\text{Δ}z\hfill & =f\left({x}_{0}+\text{Δ}x,{y}_{0}+\text{Δ}y\right)-f\left({x}_{0},{y}_{0}\right)\hfill \\ & =f\left(2+0.1,-3-0.05\right)-f\left(2,-3\right)\hfill \\ & =f\left(2.1,-3.05\right)-f\left(2,-3\right)\hfill \\ & =2.3425.\hfill \end{array}$

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