Chapter 2 - Derivatives - Review - Exercises - Page 198: 70

$$h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{2g^{3/2}(x)\sqrt{f(x)}}$$

Given $$h(x)=\sqrt{\frac{f(x)}{g(x)}}$$ Then \begin{align*} h'(x)&=\frac{1}{2}\left(\frac{f(x)}{g(x)}\right)^{-1/2}\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}\\ &= \frac{g(x)f'(x)-f(x)g'(x)}{2g^{3/2}(x)\sqrt{f(x)}} \end{align*}