Find y''_xx for the parametric function x = ln t, y = t³-1
Find $y''_{xx}$ for the parametric function $x = \ln t,\ y = t^{3}-1$.
Explanation of Method
The rule: differentiate with respect to the parameter, then divide again.
Step 1: The rule
For a curve given parametrically, $y'_{x} = \frac{\dot y}{\dot x}, \qquad y''_{xx} = \frac{\bigl(y'_{x}\bigr)'_{t}}{\dot x}$.
The second formula is the point that is easy to get wrong: one differentiates the already computed $y'_{x}$ with respect to $t$, then divides by $\dot x$ once more.
Step 2: Derivatives with respect to $t$
\[ \dot x = \frac1t, \qquad \dot y = 3t^{2}. \]
Step 3: First derivative
\[ y'_{x} = \frac{3t^{2}}{1/t} = 3t^{3}. \]
Step 4: Second derivative
\[ \bigl(y'_{x}\bigr)'_{t} = 9t^{2}, \qquad y''_{xx} = \frac{9t^{2}}{1/t} = 9t^{3}. \]
Step 5: Verification by eliminating the parameter
\[x = \ln t \;\Longrightarrow\; t = e^{x} \;\Longrightarrow\; y = e^{3x}-1\]
\[y'_{x} = 3e^{3x} = 3t^{3} \;\;\checkmark, \qquad y''_{xx} = 9e^{3x} = 9t^{3} \;\;\checkmark\]
\[t = 1:\ x = 0,\ y = 0,\ y'_{x} = 3,\ y''_{xx} = 9 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y''_{xx} = 9t^{3}\,}\]
Remark
Remark. A frequent error is to write $y''_{xx} = \dfrac{\ddot y}{\ddot x}$, which here would give $\dfrac{6t}{-1/t^{2}} = -6t^{3}$ — wrong in both magnitude and sign. The correct formula can be remembered as "differentiate the slope along the curve, then convert back to $x$".