Example 1. Find $\frac{d u}{d x}$, if $u=e^{z-2 y}$, where $z=\sin x, y=x^{2}$.
Solution. The function $u=e^{z-2 y}$ depends on two intermediate arguments $v_{1}=y, v_{2}=z$. According to formula (7.21), which in this case takes the form

$$
\frac{d u}{d x}=\frac{\partial u}{\partial y} \cdot \frac{d y}{d x}+\frac{\partial u}{\partial z} \cdot \frac{d z}{d x}
$$

we obtain

$$
\begin{gathered}
\frac{d u}{d x}=e^{z-2 y}(-2) 2 x+e^{z-2 y} \cos x=e^{z-2 y}(\cos x-4 x)= \\
=e^{\sin x-2 x^{2}}(\cos x-4 x)
\end{gathered}
$$

since

$$
\frac{\partial u}{\partial y}=e^{z-2 y}(-2), \frac{\partial u}{\partial z}=e^{z-2 y}, \frac{d y}{d x}=2 x, \frac{d z}{d x}=\cos x
$$

Remark. The same result can be obtained in another way, by expressing $u$ as an explicit function of $x$

$$
u=e^{\sin x-2 x^{2}}
$$

and differentiating the latter:

$$
\frac{d u}{d x}=\left(e^{\sin x-2 x^{2}}\right)^{\prime}\left(\sin x-2 x^{2}\right)^{\prime}=e^{\sin x-2 x^{2}}(\cos x-4 x)
$$