Answer
\[
0.1009=\Delta f
\]
\[
0.1=d f
\]
Work Step by Step
We are given a two variable function $f(x, y)$ and points $P \equiv(1,2)$ and $Q \equiv(1.01,2.04)$
\[
-4 x+x^{2}+2 x y=f(x, y)
\]
Thus, obtaining partial derivatives to approximate the function by the Local Approximation Theorem
\[
L(x, y) \approx f(x+\Delta x, y+\Delta y)=f\left(x_{0}, y_{0}\right)+\int_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)
\]
So $\Delta x=-x_{0}+x$ and $\Delta y=-y_{0}+y$ $ x_{0}=1.01,x=1, y=2,$ and $y_{0}=2.04$
\[
\begin{array}{l}
f_{x}(x, y)=\frac{\partial}{\partial x}\left(x^{2}+2 x y-4 x\right) \\
f_{x}(x, y)=2 x+2 y-4
\end{array}
\]
Similarly, $f_{y}(x, y)$
\[
\begin{aligned}
f_{y}(x, y) &=\frac{\partial}{\partial y}\left(x^{2}+2 x y-4 x\right) \\
f_{y}(x, y) &=2 x \\
d f &=f(x+\Delta x, y+\Delta y)-f\left(x_{0}, y_{0}\right) \\
d f &=f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right) \\
d f &=(2(1)+2(2)-4)(1.01-1)+2(1)(2.04-2) \\
d f &=0.1
\end{aligned}
\]
We also find:
\[
\begin{aligned}
f(1.01,2.04) &=(1.01)^{2}+2(1.01)(2.04)-4(1.01) \\
f(1.01,2.04) &=1.1009 \\
f(1,2) &=(1)^{2}+2(1)(2)-4(1) \\
f(1,2) &=1
\end{aligned}
\]
Thus, $\Delta f$ is
\[
\begin{array}{c}
\Delta f=f(1.01,2.04)-f(1,2)=1.1009-1 \\
\Delta f=0.1009
\end{array}
\]
