Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 70

Find $h'$ in terms of $f'$ and $g'$ in the equation $\displaystyle h(x) = \sqrt{\frac{f(x)}{g(x)}}$.

Taking the derivative of $h(x)$ using Chain Rule as well as Quotient Rule



$
\begin{equation}
\begin{aligned}


h'(x) =& \frac{1}{2} \left[ \frac{f9x)}{g(x)} \right] ^{\frac{-1}{2}} \cdot \left( \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2} \right]
\\
\\
h'(x) =& \frac{g(x) f'(x) - f(x) g'(x)}{\displaystyle 2 \sqrt{\frac{f(x)}{g(x)}} [g(x)]^2 }
\\
\\
h' =& \frac{gf' - fg'}{2 \sqrt{fg^3}}

\end{aligned}
\end{equation}
$