Find y'_x for x = e^-t, y = e^t
Find $y'_{x}$ for $x = e^{-t}$, $y = e^{t}$.
Explanation of Method
A hyperbola parametrised by exponentials.
Step 1: Differentiate with respect to $t$
\[ x'_{t} = -e^{-t}, \qquad y'_{t} = e^{t}. \]
Step 2: Form the ratio
\[ y'_{x} = \frac{y'_{t}}{x'_{t}} = \frac{e^{t}}{-e^{-t}} = -e^{2t}. \]
Step 3: Express in terms of $x$ and $y$
Note that $xy = e^{-t}e^{t} = 1$, so the curve is the hyperbola $xy = 1$. Also $e^{2t} = \dfrac{y}{x} = y^{2}$, since $x = \dfrac1y$: \[y'_{x} = -e^{2t} = -\frac{y}{x} = -y^{2}.\]
Step 4: Verification
From $xy = 1$ we get $y = \dfrac1x$, whose derivative was found in Problem 13: $y' = -\frac{1}{x^{2}} = -y^{2} \;\;\checkmark$.
\[t = 0:\quad (x,y) = (1,1),\quad y'_{x} = -1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'_{x} = -e^{2t} = -\dfrac{y}{x} = -y^{2}\,}\]
Remark
Remark. As $t$ runs over $\mathbb{R}$, the point $(e^{-t},e^{t})$ traces only the branch of $xy = 1$ in the first quadrant — a parametrisation can miss part of its implicit curve. The derivative is negative throughout, matching the fact that this branch is strictly decreasing.