If $f$ is three times differentiable and $f'(x)\neq 0$, the Schwarz derivative of $f$ at $x$ is defined by $$\mathscr{D}f(x)=\dfrac{f'''(x)}{f'(x)}-\dfrac32\left( \dfrac{f''(x)}{f'(x)}\right)^2.$$ (a) Demonstrate that $$\mathscr{D}(f\circ g)=[\mathscr{D}f\circ g]\cdot g'^2+\mathscr{D}g.$$
My solution: using the chain rule and the product rule and, we obtain
$$(f\circ g)'(x)=f'(g(x))\cdot g'(x)$$ $$(f\circ g)''(x)=f''(g(x))\cdot (g'(x))^2+f'(g(x))\cdot g''(x)$$ $$(f\circ g)'''(x)=f'''(g(x))\cdot (g'(x))^3+2f''(g(x))\cdot g'(x)g''(x)+f''(g(x))\cdot g''(x)g'(x)+f'(g(x))\cdot g'''(x)=f'''(g(x))\cdot (g'(x))^3+3f''(g(x))\cdot g'(x)g''(x)+f'(g(x))\cdot g'''(x)$$
\begin{align*} \mathscr{D}(f\circ g) &= \dfrac{(f\circ g)'''(x)}{(f\circ g)'(x)}-\dfrac32\left( \dfrac{(f\circ g)''(x)}{(f\circ g)'(x)}\right)^2=\\ &= \dfrac{f'''(g(x))\cdot (g'(x))^3+3f''(g(x))\cdot g'(x)g''(x)+f'(g(x))\cdot g'''(x)}{f'(g(x))\cdot g'(x)}-\dfrac32 \left( \dfrac{f''(g(x))\cdot (g'(x))^2+f'(g(x))\cdot g''(x)}{f'(g(x))\cdot g'(x)}\right)^2 =\\ &= \dfrac{f'''(g(x))}{f'(g(x))}(g'(x))^2+3\dfrac{f''(g(x))}{f'(g(x))}g''(x)+\dfrac{g'''(x)}{g'(x)}-\dfrac32 \left( \left( \dfrac{f''(g(x))}{f'(g(x))}\right)^2(g'(x))^2+2\dfrac{f''(g(x))}{f'(g(x))}g''(x)+\left( \dfrac{g''(x)}{g'(x)}\right)^2\right)=\\ &= \left( \dfrac{f'''(g(x))}{f'(g(x))}-\dfrac32 \left( \dfrac{f''(g(x))}{f'(g(x))}\right) \right) (g'(x))^2+\left( \dfrac{g'''(x)}{g'(x)}-\dfrac32 \left( \dfrac{g''(x)}{g'(x)}\right)^2 \right) =[\mathscr{D}f\circ g]\cdot (g')^2+\mathscr{D}g \end{align*}
(b) Show that if $f(x)=\dfrac{ax+b}{cx+d}$, with $ad-bc\neq 0$, then $\mathscr{D}f=0$. Therefore, $\mathscr{D}(f\circ g)=\mathscr{D}g$.
My solution: $f(x)=\dfrac{ax+b}{cx+d}$, where $ad-bc\neq 0$.
We have $f(x)=\dfrac{a}{c}-\dfrac{ad-bc}{c(cx+d)}$.
We can find $f'$, $f''$, and $f'''$:
$$f'(x)=\dfrac{ad-bc}{(cx+d)^2}$$ $$f''(x)=-2c\cdot \dfrac{ad-bc}{(cx+d)^3}$$ $$f'''(x)=6c^2\cdot \dfrac{ad-bc}{(cx+d)^4}$$
Then we have $\dfrac{f'''(x)}{f'(x)}=\dfrac{6c^2}{(cx+d)^2}$ and $\dfrac{f''(x)}{f'(x)}=-\dfrac{2c}{cx+d}$.
$$\mathscr{D}f(x)=\dfrac{f'''(x)}{f'(x)}-\dfrac32\left( \dfrac{f''(x)}{f'(x)}\right)^2 =\dfrac{6c^2}{(cx+d)^2}-\dfrac32\dfrac{4c^2}{(cx+d)^2}=0$$
Therefore, $\mathscr{D}(f\circ g)=[\mathscr{D}f\circ g]\cdot (g')^2+\mathscr{D}g=0\cdot (g')^2+\mathscr{D}g=\mathscr{D}g$.
Is my solution correct? Anything to improve or comment on?