Solution
= Solution
The <second derivative chain rule> is
$$
\frac{d^2z}{dx^2}=\frac{d^2z}{dy^2}(y')^2+\frac{dz}{dy}y''.
$$
Put $s=\sin x>0$ and $c_0=\cos x$. Substituting the preceding results and $y'=2ysc_0$ gives
$$
\frac{d^2z}{dx^2}
=-\frac{(2s^2+1)c_0^2}{s}
+2sc_0^2+\frac{c_0^2-s^2}{s}=-s.
$$
Therefore
$$
\boxed{\frac{d^2z}{dx^2}=-\sin x}.
$$