Example 4. Find $x_{t}^{\prime}$ at $t=1$, if $t \ln x - x \ln t = 1$.
Solution. This equation defines $x$ as an implicit function of $t$.

Differentiate with respect to $t$:

$$
t_{t}^{\prime} \cdot \ln x + t \cdot \frac{1}{x} x_{t}^{\prime} - \left(x_{t}^{\prime} \ln t + x \cdot \frac{1}{t}\right) = 0
$$

Since $t_{i}^{\prime}=1$, we have

$$
\ln x + \frac{t}{x} x_{t}^{\prime} - x_{t}^{\prime} \ln t - \frac{x}{t} = 0, \quad x_{t}^{\prime}\left(\frac{t}{x} - \ln t\right) - \left(\frac{x}{t} - \ln x\right) = 0
$$

from which

$$
x_{i}^{\prime} = \frac{\frac{x}{t} - \ln x}{\frac{t}{x} - \ln t}
$$

To find $x_{t}^{\prime}$ at $t=1$, we need to determine the value of $x$ corresponding to this value of $t$. Substituting $t=1$ into the original equation, we get $1 \cdot \ln x - x \ln 1 = 1$, from which $\ln x = 1$, hence $x = e$.

Now substitute the values of $t$ and $x$ into the formula for the derivative. We get

$$
x_{i}^{\prime} = \frac{\frac{e}{1} - 1}{\frac{1}{e} - 0} = \frac{e - 1}{\frac{1}{e}} = e(e - 1), \quad x_{i}^{\prime} = e(e - 1)
$$

## Problems

Find the derivatives of implicit functions:

1. $x^{2} + y^{2} - 4 x y = 0$.
2. $x y^{2} + x^{2} y = 2$.
3. $x y + \sin y = 0$.
4. $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
5. $e^{y} + x = y$.
6. $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.
7. $\ln y - 2 x = 0$.
8. $y^{2} + x^{4} = x^{2}$.
9. Find $y^{\prime}$ at the point $M(1,1)$, if $\frac{y}{x} + x y = 2$.
10. Find $y^{\prime}$ when $y = 0$, if $x \cos y - \sin y + \sin 2 y = 1$.

## Answers

1. $\frac{x - 2 y}{2 x - y}$.
2. $-\frac{y(y + 2 x)}{x(x + 2 y)}$.
3. $-\frac{y}{x + \cos y}$.
4. $\frac{b^{2} x}{a^{2} y}$.
5. $\frac{1}{1 - e^{y}}$.
6. $-\left(\frac{y}{x}\right)^{\frac{1}{3}} \cdot 7.2 y$.
7. $\frac{x(1 - 2 x^{2})}{y}$.
8. 0.
9. -1.

## § 2.6. Higher-Order Derivatives

The second derivative or the second-order derivative of a function $y = f(x)$ is the derivative of its first derivative $y^{\prime} = f^{\prime}(x)$. It is denoted by:

$$
y^{\prime \prime} = \left(y^{\prime}\right)^{\prime} \text{ or } f^{\prime \prime}(x) = \left[f^{\prime}(x)\right]^{\prime} \text{ or } \frac{d^{2} y}{d x^{2}} = \frac{d y^{\prime}}{d x} \text{.}
$$

Mechanical meaning of the second derivative. If $x = f(t)$ is the law of rectilinear motion of a point, then $x^{\prime \prime} = f^{\prime \prime}(t)$ is the acceleration of this motion.

Similarly, third, fourth, and other higher-order derivatives are defined and denoted as follows:

$$
\begin{gathered}
y^{\prime \prime \prime} = f^{\prime \prime \prime}(x) = \left(y^{\prime \prime}\right)^{\prime} = \frac{d^{3} y}{d x^{3}}, y^{\mathrm{IV}} = f^{\mathrm{IV}}(x) = \left(y^{\prime \prime \prime}\right)^{\prime} = \frac{d^{4} y}{d x^{4}}, \ldots, \\
y^{(n)} = f^{(n)}(x) = \left(y^{(n-1)}\right)^{\prime} = \frac{d^{n} y}{d x^{n}}
\end{gathered}
$$

Leibniz's formula for the $n$-th order derivative of the product of two functions:

$$
\begin{aligned}
&(u v)^{(n)} = u^{(n)} v + n u^{(n-1)} v^{\prime} + \frac{n(n-1)}{2!} u^{(n-2)} v^{\prime \prime} + \cdots + \\
&+ n u^{\prime} v^{(n-1)} + u v^{(n)}
\end{aligned}
$$