ここで勉強すれば数学検定１級の壁は超えられるか。

$\begin{aligned}\dfrac {d^{2}y}{dx^{2}}=\dfrac {d}{dx}\left( \dfrac {dy}{dx}\right) =\dfrac {d}{d\theta }\left( \dfrac {y'\left( \theta \right) }{x'\left( \theta \right) }\right) \cdot \dfrac {1}{x'\left( \theta \right) }\end{aligned}$
 

上記の式を使って答えを計算する。

$\begin{aligned}\dfrac {dx}{d\theta}=\cos \theta\end{aligned}$
 
 

$\begin{aligned}\dfrac {dy}{d\theta }=\dfrac {\dfrac {1}{\cos ^{2}\theta }}{2\tan \dfrac {\theta }{2}}+\sin \theta \\ =\dfrac {1}{2\sin \dfrac {\theta }{2}\cos \dfrac {\theta }{2}}+\sin \theta \end{aligned}$

 
$\begin{aligned}\dfrac {dy}{dx}=\dfrac {1}{\cos \theta }\left( \sin \theta -\dfrac {1}{\sin \theta }\right) \\ =\dfrac {1}{\cos \theta }\left( \dfrac {\sin ^{2}\theta -1}{\sin \theta }\right) \end{aligned}$
 

$=\dfrac {-\cos ^{2}\theta }{\cos \theta \sin \theta }=-\dfrac {\cos \theta }{\sin \theta }$
 

となる。
 

$\begin{aligned}\dfrac {d^{2}y}{dx^{2}}=\dfrac {d}{d\theta }\left( \dfrac {dy}{dx}\right) \cdot \dfrac {d\theta }{dx}\end{aligned}$

上の式に当てはめて、２段階に分けて計算する。

$\begin{aligned}\dfrac {d}{d\theta }\left( \dfrac {dy}{dx}\right) =\dfrac {d}{d\theta }\left( \dfrac {-\cos \theta }{\sin \theta }\right) =\dfrac {1}{\sin ^{2}\theta }\end{aligned}$

$\begin{aligned}\dfrac {d^{2}y}{dx^{2}}=\dfrac {1}{\cos \theta \cdot \sin ^{2}\theta }\end{aligned}$

$\theta =\dfrac {\pi }{3}\\ $
 

を上の式に代入すると

$\begin{aligned}\dfrac {1}{\dfrac {1}{2}\times \left( \dfrac {\sqrt {3}}{2}\right) ^{2}}=\dfrac {8}{3}\end{aligned}$

参考事項

$\dfrac {d^{2}y}{dx^{2}}=\dfrac {d}{dx}\left( \dfrac {y'\left( \theta \right) }{x'\left( \theta \right) }\right) \dfrac {1}{x'\left( \theta \right) }$

$=\dfrac {y''\left( \theta \right) x'\left( \theta \right) -y'\left( \theta \right) x''\left( \theta \right) }{\left( x'\left( \theta \right) \right) ^{2}}\cdot \dfrac {1}{x'\left( \theta \right) }$

$=\dfrac {y''\left( \theta \right) x'\left( \theta \right) -y'\left( \theta \right) x''\left( \theta \right) }{\left( x'\left( \theta \right) \right) ^{3}}$

上の式に代入してもよい。