In this section we argued that if \(f=f(x, y)\) is a function of two variables
and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk
containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable
at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is
differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it
means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this
exercise we explore the definition of differentiability of a function of two
variables in more detail. Throughout, let \(g\) be the function defined by \(g(x,
y)=\sqrt{|x y|}\)
a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1]
\times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0)
b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally
linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\)
at (0,0)\(?\)
c. Recall that if a function \(f=f(x)\) of a single variable is differentiable
at \(x=x_{0},\) then
$$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0}
\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$
exists. We saw in single variable calculus that the existence of
\(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear
at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization
\(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\)
for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely
approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how
good the approximation of \(L(x)\) is to \(f(x)\) with the error function
$$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$
As \(x\) approaches \(x_{0}, E(x)\) approaches
\(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\),
and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\)
gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not
locally linear at \((0,0),\) its error term
$$
E(x, y)=L(x, y)-g(x, y)
$$
at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the
linearization you found in part (b).) This shows that just because an error
term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot
conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\).
d. As the previous part illustrates, having the error term go to 0 does not
ensure that a function of two variables is locally linear. Instead, we need a
notation of a relative error. To see how this works, let us return to the
single variable case for a moment and consider \(f=f(x)\) as a function of one
variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to
\(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with
\(L\left(x_{0}+h\right)\) is
$$\frac{E\left(x_{0}+h\right)}{h}$$
Show that, for a function \(f=f(x)\) of a single variable, the limit of the
relative error is 0 as \(h\) approaches 0 .
e. Even though the error term for a function of two variables might have a
limit of 0 at a point, our example shows that the function may not be locally
linear at that point. So we use the concept of relative error to define
differentiability of a function of two variables. When we consider
differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0},
y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\)
to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)
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