Answer
$a.$
The temperature is decreasing at a rate of $(\displaystyle \frac{20}{3})^{o}C $in the $x$-direction at point (2,1).
$b.$
The temperature is decreasing at a rate of $(\displaystyle \frac{10}{3})^{o}C $in the $y$-direction at point (2,1).
Work Step by Step
$a.$
$T(x,y)=60(1+x^{2}+y^{2})^{-1}$
For $T_{x}$, treat y as constant. We need the chain rule
$T_{x}(x,y)=-60(1+x^{2}+y^{2})^{-2}\displaystyle \cdot(2x)=\frac{-120x}{(1+x^{2}+y^{2})^{2}}$,
At $(2, 1)$,
$T_{x}(2, 1)=\displaystyle \frac{-240}{(1+4+1)^{2}}=-\frac{240}{36}=-\frac{20}{3}$.
The temperature is decreasing at a rate of $(\displaystyle \frac{20}{3})^{o}C $in the $x$-direction at point (2,1).
$b.$
$T_{x}(x,y)=-60(1+x^{2}+y^{2})^{-2}\displaystyle \cdot(2y)=\frac{-120y}{(1+x^{2}+y^{2})^{2}}$,
$T_{y}(2, 1)=\displaystyle \frac{-120}{36}=-\frac{10}{3}$.
The temperature is decreasing at a rate of $(\displaystyle \frac{10}{3})^{o}C $in the $y$-direction at point (2,1).
