Find (d³y)/(dx³) for x = e^-at, y = e^at

Find $\dfrac{d^{3}y}{dx^{3}}$ for $x = e^{-at},\ y = e^{at}$.

Step 1: Derivatives with respect to $t$

\[ \dot x = -ae^{-at}, \qquad \dot y = ae^{at}. \]

Step 2: First derivative

The constant $a$ cancels and the exponents add: $y'_{x} = \frac{ae^{at}}{-ae^{-at}} = -e^{2at}$.

Step 3: Second derivative

\[ \bigl(y'_{x}\bigr)'_{t} = -2ae^{2at}, \qquad y''_{xx} = \frac{-2ae^{2at}}{-ae^{-at}} = 2e^{3at}. \]

Step 4: Third derivative

\[ \bigl(y''_{xx}\bigr)'_{t} = 6ae^{3at}, \qquad y'''_{xxx} = \frac{6ae^{3at}}{-ae^{-at}} = -6e^{4at}. \]

Step 5: Verification by eliminating the parameter

The product $xy = e^{-at}e^{at} = 1$, so the curve is simply the hyperbola $y = \dfrac1x$: \[y = x^{-1},\quad y' = -x^{-2},\quad y'' = 2x^{-3},\quad y''' = -6x^{-4}\]

\[\text{Since } x = e^{-at},\ x^{-4} = e^{4at}: \quad y''' = -6e^{4at} \;\;\checkmark\]

\[\text{Likewise } y'_{x} = -x^{-2} = -e^{2at} \;\;\checkmark \text{ and } y''_{xx} = 2x^{-3} = 2e^{3at} \;\;\checkmark\]