6.40. Find the fourth differential of the function $y=$ $=x^{3} \ln x$.
Solution.

$$
\begin{aligned}
d y & =\left(x^{3} \ln x\right)^{\prime} d x=\left(3 x^{2} \cdot \ln x+x^{3} \cdot \frac{1}{x}\right) d x=\left(3 x^{2} \cdot \ln x+x^{2}\right) \cdot d x \\
d^{2} y & =\left((3 \ln x+1) x^{2}\right)^{\prime} \cdot(d x)^{2}=\left(\frac{3}{x} \cdot x^{2}+(3 \ln x+1) 2 x\right) \cdot(d x)^{2}= \\
& =(6 x \cdot \ln x+5 x) \cdot(d x)^{2}
\end{aligned}
$$

$$
\begin{aligned}
& d^{3} y=(6 x \ln x+5 x)^{\prime} \cdot(d x)^{3}=\left(6 \ln x+6 x \cdot \frac{1}{x}+5\right) \cdot(d x)^{3}= \\
& \quad=(6 \ln x+11) \cdot(d x)^{3} \\
& d^{4} y=(6 \ln x+11)^{\prime} \cdot(d x)^{4}=\frac{6}{x} \cdot(d x)^{4}
\end{aligned}
$$