Example 3.26. Verify that the expression

$$
y\left(e^{x y}+6\right) d x+x\left(e^{x y}+6\right) d y
$$

is a total differential of some function $f(x, y)$, and find this function.
Solution.

$$
\begin{gathered}
\text { 1) } P(x, y)=y\left(e^{x y}+6\right) ; \frac{\partial P}{\partial y}=\left(1+x y e^{x y}\right)+6 \\
Q(x, y)=x\left(e^{x y}+6\right)
\end{gathered} \begin{aligned}
\frac{\partial Q}{\partial x}=6+\left(1+x y e^{x y}\right) ; \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} ; \text { for all }(x, y) \in E^{2} \\
\text { 2) } f(x, y)=\int P d x=e^{x y}+6 x y+U(y) ; f(x, y)= \\
=\int Q d x=e^{x y}+6 x y+V(x)
\end{aligned}
$$

Since all known terms in the two found expressions for $f(x, y)$ coincide, then, consequently,

$$
f(x, y)=e^{x y}+6 x y+C, \quad(C=\text { const })
$$

## 3.4. DIFFERENTIATION OF COMPOSITE AND IMPLICIT FUNCTIONS

Let the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ be differentiable at the point $P\left(x_{1}^{0}, x_{2}^{0}, \ldots, x_{n}^{0}\right)$, and the functions $x_{i}=\varphi_{i}\left(t_{1}, t_{2}, \ldots, t_{m}\right),(i=1$, $2, \ldots, k)$ be differentiable at the point $M_{0}\left(t_{1}^{0}, t_{2}^{0}, \ldots, t_{m}^{0}\right)$ and such that $x_{i}^{0}=\varphi_{i}\left(t_{1}^{0}, t_{2}^{0}, \ldots, t_{m}^{0}\right)$. Then the composite function $y=$ $=f\left(x_{1}, x_{2}, \ldots, x_{n}\right), x_{i}=\varphi_{i}\left(t_{1}, t_{2}, \ldots, t_{m}\right), i=1,2, \ldots, k$ is differentiable at the point $M_{0}$ and its partial derivatives are determined by the formula

$$
\frac{\partial f}{\partial t_{j}}=\sum_{i=1}^{n} \frac{\partial f}{\partial x_{i}} \frac{\partial x_{i}}{\partial t_{j}}, j=1,2, \ldots, m
$$

where all partial derivatives $\frac{\partial f}{\partial x_{i}}$ are computed at the point $P_{0}$, and all partial derivatives $\frac{\partial x_{i}}{\partial t_{j}}$ - at the point $M_{0}$.

From formula (3.12), some important formulas for differentiating composite functions of several variables follow.

1) $u=f(x, y, z, t)$, where $x=x(t), y=y(t), z=z(t)$. In this case, by setting in (3.12) $n=4, m=1, x_{1}=x, x_{2}=y, x_{3}=$ $=z, x_{4}=t$, we obtain the following formula for the total (or material) derivative of the function $f(x, y, z, t)$ :

$$
\frac{d f}{d t}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x} \frac{d x}{d t}+\frac{\partial f}{\partial y} \frac{d y}{d t}+\frac{\partial f}{\partial z} \frac{d z}{d t}
$$

2) $z=f(u, v)$, where $u=u(x, y), v=v(x, y)$. The formulas for the partial derivatives $f_{x}^{\prime}, f_{y}^{\prime}$ are obtained from (3.12) by setting

$$
\begin{gathered}
x_{1}=u, x_{2}=v, t_{1}=x, t_{2}=y n=2, m=2: \\
\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \frac{\partial v}{\partial x} \\
\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \frac{\partial v}{\partial y}
\end{gathered}
$$

3) $f=f(u, v, w), u=u(x, y, z), v=v(x, y, z), w=w(x, y, z)$. The formulas for the partial derivatives $f_{x}^{\prime}, f_{y}^{\prime} f_{z}^{\prime}$ are obtained from (3.12) similarly to case 2, setting $n=3, m=3$ and adding the corresponding variables $x_{3}=w, t_{3}=z$ :

$$
\begin{aligned}
& \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \frac{\partial v}{\partial x}+\frac{\partial f}{\partial w} \frac{\partial w}{\partial x} \\
& \frac{\partial f}{\partial y}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial y}