Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 48

Determine $\displaystyle \frac{dy}{du}, \frac{du}{dx}$ and $\displaystyle \frac{dy}{dx}$ if $\displaystyle y = \frac{u+1}{u-1}$ and $u = 1 + \sqrt{x}$.

We first find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$.



$\begin{equation} \begin{aligned} \frac{dy}{du} =& \frac{\displaystyle (u-1) \cdot \frac{d}{du} (u+1) - (u+1) \cdot \frac{d}{du} (u-1)}{(u-1)^2} \qquad \text{ and } &&& \frac{du}{dx} =& \frac {d}{dx} (1) + \frac{d}{dx} (x)^{\frac{1}{2}} \\ \\ =& \frac{(u-1) (1) - (u + 1)(1)}{(u - 1)^2} &&& =& 0 + \frac{1}{2} (x)^{\frac{-1}{2}} \\ \\ =& \frac{u-1-u-1}{(u-1)^2} &&& =& \frac{1}{2 \sqrt{x}} \\ \\ =& \frac{-2}{(u-1)^2} \end{aligned} \end{equation}$


Then,


$\begin{equation} \begin{aligned} \frac{dy}{dx} =& \frac{dy}{du} \cdot \frac{du}{dx} \\ \\ =& \frac{-2}{(u-1)^2} \cdot \frac{1}{2 \sqrt{x}} \\ \\ =& \frac{-2}{2(u - 1)^2 \sqrt{x}} \\ \\ =& \frac{-1}{(1 + \sqrt{x} - 1)^2 \sqrt{x}} \qquad \text{Substitute $1 + \sqrt{x}$ for $u$} \\ \\ =& \frac{-1}{(\sqrt{x})^2 \sqrt{x}} \\ \\ =& \frac{-1}{x \sqrt{x}} \end{aligned} \end{equation}$