Example 3. Find the total differential of the function $z=z(x, y)$, given by the equation $z^{2}-2 x y=c$.
Solution. Let's take the differentials of both sides of the given equation:

$$
\begin{gathered}
d\left(z^{2}-2 x y\right)=d c, d\left(z^{2}\right)-d(2 x y)=d c \\
2 z d z-2(x d y+y d x)=0, z d z-(x d y+y d x)=0 \\
z d z=x d y+y d x, d z=\frac{y}{z} d x+\frac{x}{z} d y
\end{gathered}
$$

## Problems

Find the derivatives of implicit functions:

1. $e^{x} \sin y-e^{y} \cos x=0$.
2. $\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)=0$.

Find the partial derivatives of implicit functions:
3. $3 x^{2}+4 y^{2}+5 z^{2}-60=0$.
4. $\frac{y}{x}-\operatorname{tg} \frac{z}{c}=0$.

Find the total differentials of functions $z=z(x, y)$, given by the equations:
5. $x^{2}+y^{2}+z^{2}-2 y-1=0$.
6. $z-y e^{\frac{x}{z}}=0$.

## Answers

1. $y'=-\frac{e^{x} \sin y+e^{y} \sin x}{e^{x} \cos y-e^{y} \cos x}$. 2. $y'=-\frac{2 x^{3}+2 x y^{2}-a^{2} x}{2 x^{2} y+2 y^{3}+a^{2} y}$. 3. $\frac{\partial z}{\partial x}=-\frac{3 x}{5 z}$, $\frac{\partial z}{\partial y}=\frac{4 y}{5 z}$. 4. $\frac{\partial z}{\partial x}=-\frac{c y}{x^{2}+y^{2}}$; $\frac{\partial z}{\partial y}=\frac{c x}{x^{2}+y^{2}}$. 5. $d z=-\frac{x}{z} d x+\frac{1-y}{z} d y$.
2. $d z=\frac{z e^{\frac{x}{z}}(z d y+y d x)}{z^{2}+x y e^{\frac{x}{z}}}$.

## § 7.4. Differentiation of Composite Functions

If

$$
u=F\left(v_{1}, v_{2}, \ldots, v_{n}\right)
$$

where

$$
\begin{gathered}
v_{1}=f_{1}(x, y, z, \ldots, t), v_{2}=f_{2}(x, y, z, \ldots, t), \ldots \\
v_{n}=f_{n}(x, y, z, \ldots, t)
\end{gathered}
$$

then the function $u$ is called a composite function of the independent variables $x, y, z, \ldots, t$. The variables $v_{1}, v_{2}, \ldots, v_{n}$ are called intermediate arguments.

The partial derivative of a composite function with respect to one of the independent variables is equal to the sum of the products of its partial derivatives with respect to the intermediate arguments and the partial derivatives of these arguments with respect to the given independent variable:

$$
\begin{aligned}
& \frac{\partial u}{\partial x}=\frac{\partial u}{\partial v_{1}} \cdot \frac{\partial v_{1}}{\partial x}+\frac{\partial u}{\partial v_{2}} \cdot \frac{\partial v_{2}}{\partial x}+\cdots+\frac{\partial u}{\partial v_{n}} \cdot \frac{\partial v_{n}}{\partial x} \\
& \frac{\partial u}{\partial y}=\frac{\partial u}{\partial v_{1}} \cdot \frac{\partial v_{1}}{\partial y}+\frac{\partial u}{\partial v_{2}} \cdot \frac{\partial v_{2}}{\partial y}+\cdots+\frac{\partial u}{\partial v_{n}} \cdot \frac{\partial v_{n}}{\partial y} \\
& \frac{\partial u}{\partial t}=\frac{\partial u}{\partial v_{1}} \cdot \frac{\partial v_{1}}{\partial t}+\frac{\partial u}{\partial v_{2}} \cdot \frac{\partial v_{2}}{\partial t}+\cdots+\frac{\partial u}{\partial v_{n}} \cdot \frac{\partial v_{n}}{\partial t} \cdot
\end{aligned}
$$

In particular, if all the intermediate arguments are functions of one independent variable $x$, then the function (7.18) will be a composite function of $x$. The derivative of this function is called the total derivative and is found using the formula

$$
\frac{d u}{d x}=\frac{\partial u}{\partial v_{1}} \cdot \frac{d v_{1}}{d x}+\frac{\partial u}{\partial v_{2}} \cdot \frac{d v_{2}}{d x}+\cdots+\frac{\partial u}{\partial v_{n}} \cdot \frac{d v_{n}}{d x}
$$