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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
was not differentiable at the origin. Let’s calculate the partial derivatives ${f}_{x}$ and ${f}_{y}\text{:}$
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 $\underset{\left(x,y\right)\to \left(0,0\right)}{\text{lim}}{f}_{x}\left(0,0\right)={f}_{x}\left(0,0\right)\text{:}$
Let $x=ky.$ Then
If $y>0,$ then this expression equals $1\text{/}{\left({k}^{2}+1\right)}^{3\text{/}2};$ if $y<0,$ then it equals $\text{\u2212}\left(1\text{/}{\left({k}^{2}+1\right)}^{3\text{/}2}\right).$ In either case, the value depends on $k,$ so the limit fails to exist.
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 $\text{\Delta}y=f\left(x+\text{\Delta}x\right)-f\left(x\right),$ where $\text{\Delta}x=dx.$ 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.
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 $\text{\Delta}x$ and $\text{\Delta}y$ be chosen so that $\left({x}_{0}+\text{\Delta}x,{y}_{0}+\text{\Delta}y\right)$ 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
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
Notice that the symbol $\partial $ is not used to denote the total differential; rather, $d$ appears in front of $z.$ Now, let’s define $\text{\Delta}z=f\left(x+\text{\Delta}x,y+\text{\Delta}y\right)-f\left(x,y\right).$ We use $dz$ to approximate $\text{\Delta}z,$ so
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 $\text{\Delta}x$ and $\text{\Delta}y.$ Since $\text{\Delta}z=f\left(x+\text{\Delta}x,y+\text{\Delta}y\right)-f\left(x,y\right),$ this can be used further to approximate $f\left(x+\text{\Delta}x,y+\text{\Delta}y\right)\text{:}$
See the following figure.
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.
Find the differential $dz$ of the function $f\left(x,y\right)=3{x}^{2}-2xy+{y}^{2}$ and use it to approximate $\text{\Delta}z$ at point $\left(2,\mathrm{-3}\right).$ Use $\text{\Delta}x=0.1$ and $\text{\Delta}y=\mathrm{-0.05}.$ What is the exact value of $\text{\Delta}z?$
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 ${y}_{0}=\mathrm{-3}\text{:}$
Then, we substitute these quantities into [link] :
This is the approximation to $\text{\Delta}z=f\left({x}_{0}+\text{\Delta}x,{y}_{0}+\text{\Delta}y\right)-f\left({x}_{0},{y}_{0}\right).$ The exact value of $\text{\Delta}z$ is given by
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