Find the second derivative of y = sqrt(1+x²)
Find the second derivative of $y = \sqrt{1+x^{2}}$.
Explanation of Method
A quotient that collapses to a single reciprocal power.
Step 1: First derivative
\[ y = \bigl(1+x^{2}\bigr)^{1/2} \;\Longrightarrow\; y' = \frac12\bigl(1+x^{2}\bigr)^{-1/2}\cdot2x = \frac{x}{\sqrt{1+x^{2}}}. \]
Step 2: Second derivative
Apply the product rule to $y' = x\bigl(1+x^{2}\bigr)^{-1/2}$: \[y'' = \bigl(1+x^{2}\bigr)^{-1/2} + x\cdot\left(-\frac12\right)\bigl(1+x^{2}\bigr)^{-3/2}\cdot2x = \frac{1}{\sqrt{1+x^{2}}}-\frac{x^{2}}{\bigl(1+x^{2}\bigr)^{3/2}}.\]
Step 3: Combine over one denominator
The numerator collapses completely: \[y'' = \frac{\bigl(1+x^{2}\bigr)-x^{2}}{\bigl(1+x^{2}\bigr)^{3/2}} = \frac{1}{\bigl(1+x^{2}\bigr)^{3/2}}.\]
Step 4: Verification
\[y''>0 \text{ everywhere, so the curve is convex on the whole line} \;\;\checkmark\]
\[x = 0:\ y' = 0,\ y'' = 1 \;\Longrightarrow\; \text{a minimum at } (0,1) \;\;\checkmark\]
\[x\to\pm\infty:\ y'' \to 0, \text{ matching the asymptotes } y = \pm x \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = \dfrac{1}{\bigl(1+x^{2}\bigr)^{3/2}}\,}\]
Remark
Remark. The curve is the upper branch of the hyperbola $y^{2}-x^{2} = 1$. Feeding $y'$ and $y''$ into the curvature formula gives a strikingly simple result: since $1+y'^{2} = \dfrac{1+2x^{2}}{1+x^{2}}$,
\[ k = \frac{\bigl|y''\bigr|}{\bigl(1+y'^{2}\bigr)^{3/2}} = \frac{1}{\bigl(1+x^{2}\bigr)^{3/2}}\cdot\left(\frac{1+x^{2}}{1+2x^{2}}\right)^{3/2} = \frac{1}{\bigl(1+2x^{2}\bigr)^{3/2}}, \]
largest at the vertex, where $k = 1$, and decaying to zero along the asymptotes.