Find y'''_xxx for x = e^t, y = t³

Find $y'''_{xxx}$ for $x = e^{t},\ y = t^{3}$.

Step 1: Set up

\[ \dot x = e^{t}, \qquad \dot y = 3t^{2}. \]

Each successive derivative is obtained by differentiating the previous one in $t$ and dividing by $\dot x = e^{t}$, so every step adds one power of $e^{-t}$.

Step 2: First derivative

\[ y'_{x} = \frac{3t^{2}}{e^{t}} = 3t^{2}e^{-t}. \]

Step 3: Second derivative

\[ \bigl(y'_{x}\bigr)'_{t} = \bigl(6t-3t^{2}\bigr)e^{-t} = 3t\bigl(2-t\bigr)e^{-t}, \]

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

Step 4: Third derivative

Differentiating $3t(2-t)e^{-2t} = \bigl(6t-3t^{2}\bigr)e^{-2t}$: \[\bigl(y''_{xx}\bigr)'_{t} = \bigl(6-6t\bigr)e^{-2t}-2\bigl(6t-3t^{2}\bigr)e^{-2t} = \bigl(6-6t-12t+6t^{2}\bigr)e^{-2t} = 6\bigl(t^{2}-3t+1\bigr)e^{-2t},\]

\[ y'''_{xxx} = \frac{6\bigl(t^{2}-3t+1\bigr)e^{-2t}}{e^{t}} = \frac{6\bigl(t^{2}-3t+1\bigr)}{e^{3t}}. \]

Step 5: Verification by eliminating the parameter

Here $t = \ln x$, so the curve is $y = \bigl(\ln x\bigr)^{3}$. Writing $L = \ln x$: \[y'_{x} = \frac{3L^{2}}{x}, \qquad y''_{xx} = \frac{3L\bigl(2-L\bigr)}{x^{2}}, \qquad y'''_{xxx} = \frac{6\bigl(L^{2}-3L+1\bigr)}{x^{3}}\]

\[\text{With } L = t \text{ and } x = e^{t}, \text{ these are exactly the three results above} \;\;\checkmark\]

\[t = 0:\ y'_{x} = 0,\ y''_{xx} = 0,\ y'''_{xxx} = 6 \;\;\checkmark\]