Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 90

Differentiate $\displaystyle g(x) = \sqrt{\frac{x^2 - 4x}{2x +1}}$

By using Quotient Rule and Chain Rule, we get

$
\begin{equation}
\begin{aligned}
g'(x) &= \frac{d}{dx} \left[ \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}} \right]\\
\\
&= \frac{1}{2} \cdot \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}- 1} \cdot \frac{d}{dx} \left( \frac{x^2 - 4x}{2x + 1} \right)\\
\\
&= \frac{1}{2} \left( \frac{x^2 - 4x}{2x + 1} \right)^{-\frac{1}{2}}
\left[ \frac{(2x +1) \cdot \frac{d}{dx}(x^2 - 4x) - (x^2 - 4x) \cdot \frac{d}{dx} (2x + 1) }{(2x + 1)^2} \right]\\
\\
&= \frac{1}{2 \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}}} \left[ \frac{(2x + 1)(2x - 4) - (x^2 - 4x)(2)}{(2x + 1)^2} \right]\\
\\
&= \frac{1}{2 \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}}} \left[ \frac{4x^2 - 8x + 2x - 4 - 2x^2 + 8x}{(2x + 1)^2} \right]\\
\\
&= \frac{1}{2 \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}}} \left[ \frac{2x^2 + 2x- 4}{(2x + 1)^2} \right]\\
\\
&= \frac{1}{2 \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}}} \left[ \frac{2(x^2 + x - 2)}{(2x + 1)^2} \right]\\
\\
&= \frac{x^2 + x - 2}{(2x + 1)^2 \left( \frac{x^2 - 4x}{2x + 1} \right)^{\frac{1}{2}}}\\
\\
&= \frac{x^2 + x - 2}{(2x +1)^{\frac{3}{2}}(x^2 - 4x)^{\frac{1}{2}} }
\end{aligned}
\end{equation}
$