Let $f\left(x\right)=\sin x,g\left(x\right)=x^2\text{ and }h\left(x\right)=\ln x\text{. if }\varphi \left(x\right)=h\left(f\right(g\left(x\right)\left)\right)\text{ then }{\varphi }^{\prime \prime }\left(x\right)=$

$$\begin{gathered} \varphi \left(x\right)=h\left(f\right(g\left(x\right)\left)\right) =\ln \left(\sin x^2\right) \\ \Rightarrow {\varphi }^{\text{'}}\left(x\right)=\frac{1}{\sin x^2}\left(\cos x^2\right).2x=cot x^2.2x \end{gathered}$$
$$\begin{gathered} {\varphi }^{"}\left(x\right)=cot{x}^2.2+2x.\left(-\cos e{c}^2\left(x^2\right).2x\right) \\ =2co{t}^{ }{x}^2-4x^2\cos e{c}^2\left(x^2\right) \end{gathered}$$