Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 40

We need to find the function $f \circ g \circ h$

$\displaystyle f(x) = \tan x , \qquad \quad g(x) = \frac{x}{x-1},\qquad \quad h(x) = \sqrt[3]{x}$


$
\begin{equation}
\begin{aligned}

f \circ g \circ h =& f(g(h(x)))\\

\text{ Solving for $g \circ h$}\\

\displaystyle g(h(x))=& \frac{x}{x-1}\\

g(\sqrt[3]{x})=& \frac{x}{x-1}
&& \text{ Substitute the given function $h(x)$ to the value of $x$ of the function $g(x)$}\\

\displaystyle g \circ h =& \frac{\sqrt[3]{x}}{\sqrt[3]{x} -1}\\

\text{ Solving for $f \circ g \circ h$}\\

\displaystyle g(x)=& \frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}\\

f \circ g \circ h =& f(g(h(x)))\\

f(\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1})=& \tan x
&& \text{ Substitute the value of $x$ }\\

\end{aligned}
\end{equation}
$


$\displaystyle \boxed{ f \circ g \circ h = \tan \left( \frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1} \right)}$