Show that $f(x,y)=7x{e}^{xy}$ is differentiable at $(1,0)$ and find its linearization there. Then use it to approximate $f(1\mathrm{.}1,-0\mathrm{.}1).$

Solution

The partial derivatives are as follows.

$f_x(x,y)=$

$f_y(x,y)=$

$f_x(1,0)=7$

$f_y(1,0)=7$

Both $f_x$ and $f_y$ are continuous functions, so $f$ is differentiable. The linearization is

The corresponding linear approximation is $7x{e}^{xy}=,$ so

$f(1\mathrm{.}1,-0\mathrm{.}1)=L(1\mathrm{.}1,-0\mathrm{.}1)=$

$$

Compare this with the actual value. $($Round your answer to five decimal places.$)$

$f(1\mathrm{.}1,-0\mathrm{.}1)=7\mathrm{.}7e^{-0\mathrm{.}11}=7\mathrm{.}0,x$