Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 50

Find the first and second derivatives of $\displaystyle y = \frac{4x}{\sqrt{x+1}}$
Solving for the first derivative of the given function

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left(\frac{4x}{\sqrt{x+1}} \right)\\
\\
y' &= \frac{d}{dx} \left[\frac{4x}{(x+1)^{\frac{1}{2}}} \right]\\
\\
y' &= \frac{\left[ (x+1)^{\frac{1}{2}} \cdot \frac{d}{dx} (4x) - (4x) \cdot \frac{d}{dx} (x+1)^{\frac{1}{2}}\right]}{\left[ (x+1)^{\frac{1}{2}}\right]^2}\\
y\ &= \frac{(x+1)^{\frac{1}{2}} (4) - (4x) \left( \frac{1}{2} \right) (x+1)^{\frac{-1}{2}} \cdot \frac{d}{dx} (x+1) }{x+1}\\
\\
y' &= \frac{(x+1)^{\frac{1}{2}}(4) - (2x)(x+1)^{\frac{-1}{2}}(1+0)}{x+1}\\
\\
y' &= \frac{(x+1)^{\frac{1}{2}} (4) - \frac{2x}{(x+1)^{\frac{1}{2}}}}{x+1}\\
\\
y' &= \frac{(4)(x+1)-2x}{(x+1)(x+1)^{\frac{1}{2}}}\\
\\
y' &= \frac{4x+4-2x}{(x+1)^{\frac{3}{2}}}\\
\\
y' &= \frac{2x+4}{(x+1)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$

Solving for the second derivative of the given function


$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} \left[ \frac{2x+4}{(x+1)^{\frac{3}{2}}} \right]\\
\\
y'' &= \frac{\left[ (x+1)^{\frac{3}{2}} \cdot \frac{d}{dx} (2x+4) - (2x+4) \cdot \frac{d}{dx}(x+1)^{\frac{3}{2}} \right]}{\left[ (x+1)^{\frac{3}{2}}\right]^2}
\\
y'' &= \frac{\left[ (x+1)^{\frac{3}{2}} (2+0) - (2x+4) \left( \frac{3}{2} \right) (x+1)^{\frac{1}{2}} \cdot \frac{d}{dx} (x+1)\right]}{(x+1)^3}\\
\\
y'' &= \frac{(2)(x+1)^{\frac{3}{2}} - \left( \frac{3}{2} \right)(2x+4)(x+1)^{\frac{1}{2}}(1+0)}{(x+1)^3}\\
\\
y'' &= \frac{(2) (x+1)^{\frac{3}{2}} - \left( \frac{3}{\cancel{2}}\right)(\cancel{2})(x+2) (x+1)^{\frac{1}{2}} }{(x+1)^3}\\
\\
y'' &= \frac{2(x+1)^{\frac{3}{2}} - 3 (x+2)(x+1)^{\frac{1}{2}}}{(x+1)^3}\\
\\
y'' &= \frac{2(x+1)^{\frac{3}{2}}-(3x+6)(x+1)^{\frac{1}{2}}}{(x+1)^3}
\end{aligned}
\end{equation}
$