Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 54

Determine $\displaystyle \frac{dy}{dx}$ for $y =\sqrt{7 - 3u}$ and $u = x^2 - 9$.

We have $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ with


$\begin{equation} \begin{aligned} \frac{dy}{du} =& \frac{1}{2} (7-3u)^{\frac{-1}{2}} \cdot \frac{d}{du} (7-3u) \qquad \text{ and } &&& \frac{du}{dx} =& \frac{d}{dx} (x^2) - \frac{d}{dx} (9) \\ \\ =& \frac{1}{2} (7 - 3u)^{\frac{-1}{2}} (-3) &&& =& 2x \\ \\ =& \frac{-3}{2(7-3u)^{\frac{1}{2}}} \end{aligned} \end{equation}$



Thus,


$\begin{equation} \begin{aligned} \frac{dy}{dx} = \frac{-3}{2(7 - 3u)^{\frac{1}{2}}} \cdot 2x \\ =& \frac{-3x}{(7 - 3u)^{\frac{1}{2}}} \\ =& \frac{-3x}{[7-3(x^2 - 9)]^{\frac{1}{2}}} \qquad \text{Substitute $x^2 - 9$ for $u$} \\ =& \frac{-3x}{(7 - 3x^2 + 27)^{\frac{1}{2}}} \\ =& \frac{-3x}{(34 - 3x^2)^{\frac{1}{2}}} \end{aligned} \end{equation}$