Example. Approximate the value of the function $y=\sqrt{x^{2}+5}$ at the point $x=1.97$ using the differential.
## Solution.

1. The point closest to 1.97 where it is easy to calculate the values of $f(a)$ and $f'(a)$ is the point $a=2$.
2. We calculate:

$$
\begin{gathered}
\Delta x=x-a=1.97-2=-0.03 \\
f(a)=f(2)=3, \quad f'(x)=\frac{x}{\sqrt{x^{2}+5}}, \quad f'(a)=y'(2)=\frac{2}{3}
\end{gathered}
$$

3. According to formula (1), we have

$$
f(1.97) \approx 3+\frac{2}{3}(-0.03)=2.98
$$

Answer. $f(1.97) \approx 2.98$.

Conditions of the Problems. Approximate the value of the function $y=f(x)$ at the point $x=a$ using the differential.

1. $y=x^{5}, \quad x=2.001$.
2. $y=\sqrt{4 x-3}, \quad x=0.98$.
3. $y=\sqrt{x^{3}}, \quad x=1.02$.
4. $y=x^{3}, \quad x=2.999$.
5. $y=\sqrt[3]{x}, \quad x=1.03$.
6. $y=\sqrt{x}, \quad x=3.996$.
7. $y=\sqrt{1+\sin x}, \quad x=0.02$.
8. $y=\sqrt{2 x+\cos x}, \quad x=0.01$.
9. $y=\sqrt[3]{2 x-\sin \frac{\pi x}{2}}, \quad x=1.03$.
10. $y=\sqrt{4 x+1}, \quad x=1.97$.

Answers. 1. 32.08. 2. 0.96. 3. 1.03. 4. 26.073. 5. 1.01. 6. 1.999. 7. 1.01. 8. 1.01. 9. 1.02. 10. 2.98.

## 4.5. Logarithmic Differentiation

Problem Statement. Find the derivative of a function of the form

$$
y=u_{1}(x)^{v_{1}(x)} \ldots u_{n}(x)^{v_{n}(x)} w_{1}(x) \ldots w_{m}(x)
$$

## PLAN OF SOLUTION.

1. The logarithm of the given function is

$\ln y=v_{1}(x) \ln u_{1}(x)+\ldots+v_{n}(x) \ln u_{n}(x)+\ln w_{1}(x)+\ldots+\ln w_{m}(x)$.

2. Differentiating both sides of this equation, we get

$$
\frac{y'}{y}=\left(v_{1}' \ln u_{1}+\frac{v_{1}}{u_{1}} u_{1}'\right)+\ldots+\left(v_{n}' \ln u_{n}+\frac{v_{n}}{u_{n}} u_{n}'\right)+\frac{w_{1}'}{w_{1}}+\ldots+\frac{w_{n}'}{w_{n}}
$$

Therefore,

$$
y'=y\left[\left(v_{1}' \ln u_{1}+\frac{v_{1}}{u_{1}} u_{1}'\right)+\ldots+\left(v_{n}' \ln u_{n}+\frac{v_{n}}{u_{n}} u_{n}'\right)+\frac{w_{1}'}{w_{1}}+\ldots+\frac{w_{n}'}{w_{n}}\right]
$$

3. Substituting the expression for $y$ into the last equation, we get the answer.