6.35. Find the second derivative of the function $y(x)=$ $=x^{2} \ln (1+\sin x)$.
Solution.

$$
\begin{gathered}
y^{\prime}(x)=2 x \ln (1+\sin x)+x^{2} \cdot \frac{\cos x}{1+\sin x} \\
y^{\prime \prime}(x)=2 \ln (1+\sin x)+\frac{2 x \cdot \cos x}{1+\sin x}+\frac{2 x \cos x}{1+\sin x}+ \\
+x^{2} \cdot \frac{-\sin x(1+\sin x)-\cos ^{2} x}{(1+\sin x)^{2}}=2 \ln (1+\sin x)+ \\
+\frac{4 x \cdot \cos x}{1+\sin x}-x^{2} \frac{(1+\sin x)}{(1+\sin x)^{2}}=2 \ln (1+\sin x)+\frac{4 x \cdot \cos x-x^{2}}{1+\sin x}
\end{gathered}
$$