Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 47

Find the first and second derivatives of $h(x) = \sqrt{x^2+1}$
Solving for the first derivative of the given function


$
\begin{equation}
\begin{aligned}
h'(x) & = \frac{d}{dx} \left( \sqrt{x^2+1} \right)\\
\\
h'(x) & = \frac{d}{dx} (x^2+1)^{\frac{1}{2}}\\
\\
h'(x) & = \frac{1}{2} (x^2+1)^{\frac{-1}{2}} \frac{d}{dx} ( x^2 +1 )\\
\\
h'(x) & = \frac{1}{2} (x^2+1)^{\frac{-1}{2}} \left[ \frac{d}{dx} (x^2) + \frac{d}{dx} (1) \right]\\
\\
h'(x) & = \frac{1}{2} (x^2+1)^{\frac{-1}{2}} (2x+0)\\
\\
h'(x) &= \frac{\cancel{2}x}{\cancel{2}(x^2+1)^{\frac{1}{2}}}\\
\\
h'(x) &= \frac{x}{\sqrt{x^2}+1}
\end{aligned}
\end{equation}
$


Solving for the second derivative of the given function


$
\begin{equation}
\begin{aligned}
h''(x) &= \frac{d}{dx} \left( \frac{x}{\sqrt{x+1}} \right)\\
\\
h''(x) &= \frac{d}{dx} \left[ \frac{x}{(x^2+1)^{\frac{1}{2}}} \right]\\
\\
h''(x) &= \frac{\left[ (x^2+1)^{\frac{1}{2}} \cdot \frac{d}{dx}(x) \right]- \left[ (x) \cdot \frac{d}{dx} (x^2+1)^{\frac{1}{2}}\right] }{\left[ (x^2+1)^{\frac{1}{2}}\right]^2}\\
\\
h''(x) &= \frac{\left[ (x^2+1)^{\frac{1}{2}} (1)\right] - \left[ (x) \left( \frac{1}{2}\right) (x^2+1)^{\frac{-1}{2}} \cdot \frac{d}{dx} (x^2+1) \right] }{x^2+1}\\
\\
h''(x) &= \frac{(x^2+1)^{\frac{1}{2}} - \left[ \left(\frac{x}{2} \right)(x^2+1)^{\frac{-1}{2}} (2x+0)\right]}{x^2+1}\\
\\
h''(x) &= \frac{(x^2+1)^{\frac{1}{2}} - \left[ \left( \frac{x}{\cancel{2}} \right) (x^2+1)^{\frac{-1}{2}} (\cancel{2}x)\right] }{x^2+1}\\
\\
h''(x) &= \frac{(x^2+1)^{\frac{1}{2}} - (x^2)(x^2+1)^{\frac{-1}{2}}}{x^2+1}\\
\\
h''(x) &= \frac{(x^2+1)^{\frac{1}{2}} - \frac{x^2}{(x^2+1)^{\frac{1}{2}}}}{x^2+1}\\
\\
h''(x) &= \frac{\frac{\cancel{x^2} +1 - \cancel{x^2}}{(x^2+1)^{\frac{1}{2}}}}{x^2+1}\\
\\
h''(x) &= \frac{1}{(x^2+1)(x^2+1)^{\frac{1}{2}}}\\
\\
h''(x) &= \frac{1}{(x^2+1)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$