Solutions

a) For fear that the handwriting parser cannot recognize the cursive letter used by Spivak, let $D$ be the operator of the Schwarzian derivative. Now, by the chain rule we have

$$(f\circ g)'=f'(g(x))g'(x),$$$$(f\circ g)''=f''(g(x))g'(x)^2+f'(g(x))g''(x),$$$$(f\circ g)'''=f'''(g(x))g'(x)^3+2f''(g(x))g'(x)g''(x)+f''(g(x))g'(x)g''(x)+f'(g(x))g'''(x).$$

Thus,

$$\begin{align*} D(f\circ g)&=\frac{(f\circ g)'''(x)}{(f\circ g)'(x)}-\frac{3}{2}\left(\frac{(f\circ g)''(x)}{(f\circ g)'(x)}\right)^2 \\ &=\frac{f'''(g(x))g'(x)^3+3f''(g(x))g'(x)g''(x)+f'(g(x))g'''(x)}{f'(g(x))g'(x)} \\ &\quad-\frac{3}{2}\left(\frac{f''(g(x))g'(x)^2+f'(g(x))g''(x)}{f'(g(x))g'(x)}\right)^2 \\ &=\frac{f'''(g(x))g'(x)^2}{f'(g(x))}+\frac{3f''(g(x))g''(x)}{f'(g(x))}+\frac{g'''(x)}{g'(x)} \\ &\quad-\frac{3}{2}\left(\frac{f''(g(x))g'(x)}{f'(g(x))}+\frac{g''(x)}{g'(x)}\right)^2 \\ &=\left(\frac{f'''\circ g}{f'\circ g}-\frac{3}{2}\left(\frac{f''\circ g}{f'\circ g}\right)^2\right)(g')^2+\frac{g'''}{g'}-\frac{3}{2}\left(\frac{g''}{g'}\right)^2 \\ &=(D(f)\circ g)(g')^2+D(g). \end{align*}$$

b) If

$$f(x)=\frac{ax+b}{cx+d},$$

then

$$f'(x)=\frac{a(cx+d)-c(ax+b)}{(cx+d)^2}=\frac{acx+ad-acx-bc}{(cx+d)^2}=\frac{ad-bc}{(cx+d)^2},$$$$f''(x)=\frac{-2c(ad-bc)}{(cx+d)^3},$$$$f'''(x)=\frac{6c^2(ad-bc)}{(cx+d)^4}.$$

Hence,

$$\begin{align*} D(f)&=\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2 \\ &=\frac{6c^2(ad-bc)}{(cx+d)^4}\cdot\frac{(cx+d)^2}{ad-bc}-\frac{3}{2}\left(\frac{-2c(ad-bc)}{(cx+d)^3}\cdot\frac{(cx+d)^2}{ad-bc}\right)^2 \\ &=\frac{6c^2}{(cx+d)^2}-\frac{3}{2}\left(\frac{-2c}{cx+d}\right)^2 \\ &=\frac{6c^2}{(cx+d)^2}-\frac{3}{2}\cdot\frac{4c^2}{(cx+d)^2} \\ &=0. \end{align*}$$

Consequently,

$$\begin{align*} D(f\circ g)&=(D(f)\circ g)(g')^2+D(g) \\ &=0\cdot(g')^2+D(g) \\ &=D(g). \end{align*}$$