Find y''_xx for x = ln (t+ sqrt(1+t²)), y = sqrt(1+t²)

Find $y''_{xx}$ for $x = \ln\!\left(t+\sqrt{1+t^{2}}\right),\ y = \sqrt{1+t^{2}}$.

Step 1: The derivative of the logarithm

The inner derivative combines with the outer one in a way that simplifies dramatically: \[\dot x = \frac{1}{t+\sqrt{1+t^{2}}}\left(1+\frac{t}{\sqrt{1+t^{2}}}\right) = \frac{1}{t+\sqrt{1+t^{2}}}\cdot\frac{\sqrt{1+t^{2}}+t}{\sqrt{1+t^{2}}} = \frac{1}{\sqrt{1+t^{2}}}.\]

The bracket reproduces the denominator exactly, and it cancels.

Step 2: The other derivative

\[ \dot y = \frac{t}{\sqrt{1+t^{2}}}. \]

Step 3: First derivative

\[ y'_{x} = \frac{t/\sqrt{1+t^{2}}}{1/\sqrt{1+t^{2}}} = t. \]

Step 4: Second derivative

\[ \bigl(y'_{x}\bigr)'_{t} = 1, \qquad y''_{xx} = \frac{1}{1/\sqrt{1+t^{2}}} = \sqrt{1+t^{2}}. \]

Step 5: Verification by eliminating the parameter

\[x = \ln\!\left(t+\sqrt{1+t^{2}}\right) = \operatorname{arsinh}t \;\Longrightarrow\; t = \sinh x\]

\[y = \sqrt{1+\sinh^{2}x} = \cosh x, \quad\text{so } y'_{x} = \sinh x = t \;\;\checkmark \text{ and } y''_{xx} = \cosh x = y \;\;\checkmark\]

\[t = 0:\ x = 0,\ y = 1,\ y'_{x} = 0,\ y''_{xx} = 1 \;\Longrightarrow\; \text{a minimum at } (0,1) \;\;\checkmark\]