# Use the table of values of f(x, y) to estimate the values of fx(3, 2), fx(3, 2.2), and fxy(3, 2). \begin{array}{|c|c|}\hline y & 1.8 & 2.0 & 2.2 \\ \hline x & & & \\ \hline 2.5 & 12.5 & 10.2 & 9.3 \\ \hline 3.0 & 18.1 & 17.5 & 15.9 \\ \hline 3.5 & 20.0 & 22.4 & 26.1 \\ \hline \end{array}

Use the table of values of $$f(x, y)$$ to estimate the values of $$fx(3, 2)$$, $$fx(3, 2.2)$$, and $$fxy(3, 2)$$.
$$\begin{array}{|c|c|}\hline y & 1.8 & 2.0 & 2.2 \\ \hline x & & & \\ \hline 2.5 & 12.5 & 10.2 & 9.3 \\ \hline 3.0 & 18.1 & 17.5 & 15.9 \\ \hline 3.5 & 20.0 & 22.4 & 26.1 \\ \hline \end{array}$$

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Step 1
$$fx(3.2)=\frac{(f(3.5,\ 2)-f(2.5,\ 5))}{(3.5-2.5)}$$
$$=\frac{(22.4-10.2)}{1.0}$$
$$=\frac{12.2}{1.0}$$
$$=12.2$$
$$fx(3,\ 2.2)=\frac{(f(3.5,\ 2.2)-f(2.5,\ 2.2))}{(3.5-2.5)}$$
$$=\frac{26.1-9.3)}{1}=16.8$$
For $$fxy(3, 2)$$, we have choices. If we don't care about centering the y, we can use the results already gathered here.
Then, $$fxy(3.2)=\frac{(fx(3,\ 2.2)-fx(3.2))}{(2.2-2)}$$
$$=\frac{(16.8-12.2)}{2}=23$$
If we had wanted to center it, we could have used
$$\frac{(f(3.5,\ 2.2)-f(3.5,\ 1.8)-f(2.5,\ 2.2)+f(2.5,\ 1.8))}{((2.2-1.8)\cdot(3.5-2.5)}$$
$$=\frac{(26.1-20-9.3+12.5)}{(4\cdot1)}=23.25$$

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We need to estimate the values of $$f_x(3, 2)$$$$f_x(3, 2.2)$$, and $$f_{xy}(3, 2)$$

We have f(3, 2) = 17.5 and f(3, 2.2) = 15.9.

For $$f_x(3, 2)$$:

$$f_x(3, 2) = \frac{22.4 - 17.5}{3.5 - 3} = 9.8.$$

For $$f_x(3, 2.2)$$

$$f_x(3, 2.2) = \frac{26 - 15.9}{3.5 - 3} = 20.2$$

Again,

For $$f_{xy}(3, 2)$$

We know that $$f_{xy}(3, 2) = (f_x(3, 2))_y = \frac{f_x(3, 2.2) - f_x(3, 2)}{2.2 - 2}$$

$$= \frac{20.2 - 9.8}{0.2}$$

= 52