Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 68

Determine the answer by using Chain Rule and check your answer by finding $f(g(x))$, taking the derivative and substituting.

$\displaystyle f(u) = \frac{u + 1}{u - 1}, g(x) = u = \sqrt{x}$

Find $(f \circ g)'(4)$.


First, we find


$
\begin{equation}
\begin{aligned}

f'(u) = \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 } &&& g'(x) = \frac{du}{dx} =& \frac{d}{dx} (x)^{\frac{1}{2}}
\\
\\
=& \frac{(u-1)(1) - (u+1)(1)}{(u-1)^2}
&&& =& \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{-2}{(u-1)^2} \cdot \frac{1}{2 \sqrt{x}}
\\
\\
=& \frac{-1}{(u-1)^2 \sqrt{x}}
\\
\\
=& \frac{-1}{(\sqrt{x} - 1)^2 \sqrt{x}}
\\
\\
(f \circ g)'(4) =& \frac{-1}{(\sqrt{4} - 1)^2 \sqrt{4}}
\\
\\
=& \frac{-1}{(2-1)^2 (2)}
\\
\\
=& \frac{-1}{2}


\end{aligned}
\end{equation}
$



To check, we find first $f(g(x))$ and take the derivative.


$
\begin{equation}
\begin{aligned}

f(g(x)) = f( \sqrt{x}) =& \frac{u+1}{u-1}
\\
\\
=& \frac{\sqrt{x} + 1}{\sqrt{x} - 1}
\\
\\
f'(g(x)) =& \frac{\displaystyle (\sqrt{x} - 1) \cdot \frac{d}{dx} (\sqrt{x} + 1) - (\sqrt{x} + 1) \cdot \frac{d}{dx} (\sqrt{x} - 1) }{(\sqrt{x} - 1)^2}
\\
\\
=& \frac{\displaystyle (\sqrt{x} - 1) \left( \frac{1}{2 \sqrt{x}} \right) - (\sqrt{x} + 1) \left( \frac{1}{2 \sqrt{x}} \right) }{(\sqrt{x} - 1)^2}
\\
\\
=& \frac{\sqrt{x} - 1 - \sqrt{x} - 1}{2 \sqrt{x} (\sqrt{x} - 1)^2}
\\
\\
=& \frac{-2}{2 \sqrt{x} (\sqrt{x} - 1)^2}
\\
\\
=& \frac{-1}{\sqrt{x} (\sqrt{x} - 1)^2}
\\
\\
(f \circ g)'(4) =& \frac{-1}{\sqrt{4} (\sqrt{4} - 1)^2}
\\
\\
=& \frac{-1}{2 (2-1)^2}
\\
\\
=& \frac{-1}{2}

\end{aligned}
\end{equation}
$