\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{description}
\item[(i)]
Show that if $f'(x)=(x-a)(x-b)(x-c)$ with $a,\ b,\ c$ all distinct
then the Schwarzian derivative $Sf(x)$ is negative everywhere
(where defined).

\item[(ii)]
Show that if $Sf(x)<0$ and $S_g(x)<0$ for all $x$ then $S(f\circ
g)(x)<0$.
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$\ds\frac{f"(x)}{f'(x)}=\ds\frac{1}{(x-a)}+\ds\frac{1}{(x-b)}+\frac{1}{(x-c)}$

$\ds\frac{f'''(x)}{f'(x)}=\ds\frac{2}{(x-b)(x-c)}+\ds\frac{2}{(x-c)(x-a)}+\frac{2}{(x-a)(x-b)}$

so
$\ds\frac{f'''(x)}{f'(x)}-\left(\ds\frac{f"(x)}{f'(x)}\right)^2=-\left(\ds\frac{1}{(x-a)^2}+\ds\frac{1}{(x-b)^2}+\frac{1}{(x-c)^2}\right)<0$

and so certainly $Sf(x)<0$.

\item[(ii)]

$$\begin{array} {rcl} {\rm Write}\ h & = & f(g)\ {\rm to\ denote}\
h(x)=f(g(x0):\ {\rm then}\\ h' & = & f'(g)(g')^2+f'(g)g''\\ h'' &
= & f''(g)(g')^2+f'(g)g''\\ h''' & = &
f'''(g)(g')^3+3f''(g)g'g''+f'(g)g''' \end{array}$$

and we find

$2h'h'''-3(h'')^2=[2f'(g)f'''(g)-3(f''(g))^2](g')^4+2(f'(g))^2[2g'g'''-3(g'')^2]$

which gives the result since both square brackets $[\ ]\ $ are
$<0$.
\end{description}

\end{document}