Answer
\[
4 \frac{^{\circ} C}{c m}=T_{x}(1,2)
\]
\[
8 \frac{^{\circ} C}{c m}=T_{y}(1,2)
\]
Work Step by Step
We are given that
\[
x+x^{3}+2 y^{2}=T(x, y)
\]
We we have to evaluate the rate of change of the Temperature with respect to distance $x$, while $y$ remains constant
\[
\begin{array}{l}
\frac{\partial}{\partial x} T(x, y) =T_{x}(x, y)\\
T_{x}(x, y)=\frac{\partial}{\partial x}\left(x^{3}+y^{2}+x+2\right) \\
\left.T_{x}(x, y)=\frac{d}{d x}\left(x^{3}+x\right)+2 y^{2} \frac{d}{d x}(1)\right) \\
T_{x}(x, y)=1+3 x^{2} \\
1+3(1)^{2}= T_{x}(1,2) \\
T_{x}(1,2)=4 \frac{\mathrm{c}}{\mathrm{cm}}
\end{array}
\]
Similarly, we obtain $T_{y}(x, y)$
\[
\begin{array}{l}
\frac{\partial}{\partial y} T(x, y) =T_{y}(x, y) \\
\frac{\partial}{\partial y}\left(x+x^{3}+2 y^{2}\right)= T_{y}(x, y) \\
\left.T_{y}(x, y)=2 \frac{d}{d y}\left(y^{2}\right)\right)+\frac{d}{d y}\left(x^{3}+x\right) \\
4 y=T_{y}(x, y) \\
4(2)= T_{y}(1,2) \\
8\frac{^{\circ} C}{c m}=T_{y}(1,2)
\end{array}
\]
