Problem on chain rule with functions of several variables
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Suppose $T(t,x,y) = e^{-t} (\sin x + \cos y)$. Consider a trajectory in $xy$ such that $x(0) = 0, y(0)=0, \frac{dx}{dt}(0) = 1, \frac{dy}{dt}(0) = 1.$
Use the chain rule to compute $\frac{d}{dt}T(t,x(t), y(t))$ at $t=0$.
Solution
Recall the chain rule
Chain rule with functions of several variables
$\frac{d}{dt} T(t, x(t), y(t)) = \partial_t T + \partial_x T \cdot \frac{dx}{dt} + \partial_y T \cdot \frac{dy}{dt}$
In order to apply this formula, we need to compute all the partial derivatives of $T$.
Partial derivatives
Computing, \begin{align}
\partial_t T(t,x,y) &= - e^{-t} (\sin x + \cos y)\\
\partial_x T(t,x,y) &= e^{-t} \cos x\\
\partial_y T(t,x,y) &= -e^{-t} \sin y
\end{align}
At $t=0$, $x=0$ and $y=0$. Hence, \begin{align}
\partial_t T(0,x(0),y(0)) &= - 1\\
\partial_x T(0,x(0),y(0)) &= 1\\
\partial_y T(0,x(0),y(0)) &= 0
\end{align}
Thus,\begin{align}\frac{d}{dt} T(t, x(t), y(t)) &= -1 + 1 \cdot \frac{dx}{dt}(0) + 0 \cdot \frac{dy}{dt}(0)\\
&= -1 + 1 + 0 \\
&= 0
\end{align}
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