Answer
Therefore, the linear function of two or three variables can be differentiable everywhere.
Work Step by Step
According to definition 13.4.1, a function is said to be differentiable at $\left(x_{0}, y_{0}\right)$ if both partial derivatives exist and
\[
\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{\Delta f-f_{x}\left(x_{0}, y_{0}\right) \Delta x-f_{y}\left(x_{0}, y_{0}\right) \Delta y}{\sqrt{(\Delta y)^{2}+(\Delta x)^{2}}}=0
\]
Thus, we find:
\[
f_{x}(x, y)=a \\ f_{y}\left(x_{0}, y_{0}\right)=b
\]
Also,
\[
\begin{array}{l}
\Delta f=f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right) \\
\Delta f=a\left(x_{0}+\Delta x\right)+b\left(y_{0}+\Delta y\right)+c-a x_{0}+b y_{0}+c \\
b \Delta y+a \Delta x=\Delta f
\end{array}
\]
$\mathrm{Thus}$
\[
\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{\Delta f-f_{x}\left(x_{0}, y_{0}\right) \Delta x-f_{y}\left(x_{0}, y_{0}\right) \Delta y}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}=\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{a \Delta x+b \Delta y-a \Delta x-b \Delta y}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}
\]
Therefore, the linear function of two or three variables can be differentiable everywhere.
