6.31. Find the first differential of the function $y(x)=$ $=e^{3 x} \ln \left(1+x^{2}\right)$ and calculate its value at $x=0, d x=$ $=\Delta x=0.1$.
Solution. The first method is based on the direct application of the formula $d y(x)=y^{\prime}(x) d x$.

We have

$$
y^{\prime}(x)=3 e^{3 x} \cdot \ln \left(1+x^{2}\right)+\frac{e^{3 x} \cdot 2 x}{1+x^{2}}
$$

hence

$$
d y(x)=\left(3 e^{3 x} \cdot \ln \left(1+x^{2}\right)+\frac{e^{3 x} \cdot 2 x}{1+x^{2}}\right) \cdot d x
$$

The second method is based on the application of the rules for finding the differential of a product:

$$
\begin{aligned}
& d y(x)=d\left(e^{3 x}\right) \ln \left(1+x^{2}\right)+e^{3 x} d\left(\ln \left(1+x^{2}\right)\right)= \\
& \quad=3 e^{3 x} \ln \left(1+x^{2}\right) d x+\frac{e^{3 x} 2 x}{1+x^{2}} d x \\
& d y(0)=\left(3 e^{3 \cdot 0} \ln \left(1+0^{2}\right)+\frac{e^{3 \cdot 0} 2 \cdot 0}{1+0^{2}}\right) \cdot 0.1=0
\end{aligned}
$$