https://r2.erweima.ai/mathbot/image/17258047775267aqc1q9l.jpg

The image you've uploaded contains three main tasks related to differentials and approximations. Let me walk you through the solution of each task:

### I. Approximation Using Differentials
This section asks for the approximation of square roots using the concept of differentials.

For each of these expressions, we use the approximation formula:
\[
f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x
\]
Where:
- \( f(x) \) is the original function (in this case, the square root).
- \( \Delta x \) is the small change from \( x \) (we want to approximate the square root of a number near \( x \)).

Let’s approximate the square roots:

#### 1. \( \sqrt{3.98} \)
Let \( f(x) = \sqrt{x} \) and take \( x = 4 \) (since 3.98 is close to 4). We have:
- \( f(x) = \sqrt{4} = 2 \)
- \( f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{4} \)
- \( \Delta x = -0.02 \)

Now, apply the approximation:
\[
f(3.98) \approx f(4) + f'(4) \cdot \Delta x = 2 + \frac{1}{4} \cdot (-0.02) = 2 - 0.005 = 1.995
\]

#### 2. \( \sqrt{24.2} \)
Take \( x = 25 \), as it’s close to 24.2. We have:
- \( f(x) = \sqrt{25} = 5 \)
- \( f'(x) = \frac{1}{2\sqrt{25}} = \frac{1}{10} \)
- \( \Delta x = -0.8 \)

Approximation:
\[
f(24.2) \approx 5 + \frac{1}{10} \cdot (-0.8) = 5 - 0.08 = 4.92
\]

#### 3. \( \sqrt{9.05} \)
Take \( x = 9 \). We have:
- \( f(x) = \sqrt{9} = 3 \)
- \( f'(x) = \frac{1}{2\sqrt{9}} = \frac{1}{6} \)
- \( \Delta x = 0.05 \)

Approximation:
\[
f(9.05) \approx 3 + \frac{1}{6} \cdot 0.05 = 3 + 0.0083 \approx 3.0083
\]

#### 4. \( \sqrt{15.86} \)
Take \( x = 16 \). We have:
- \( f(x) = \sqrt{16} = 4 \)
- \( f'(x) = \frac{1}{2\sqrt{16}} = \frac{1}{8} \)
- \( \Delta x = -0.14 \)

Approximation:
\[
f(15.86) \approx 4 + \frac{1}{8} \cdot (-0.14) = 4 - 0.0175 = 3.9825
\]

---

### II. Calculating Differentials of Functions
We are asked to find the differential of given functions.

The differential \( dy \) of a function \( y = f(x) \) is given by:
\[
dy = f'(x) \cdot dx
\]

#### 1. \( y = \sin(2x) + \tan(5x) \)
- Derivative: \( y' = 2\cos(2x) + 5\sec^2(5x) \)
- Differential: \( dy = [2\cos(2x) + 5\sec^2(5x)] dx \)

#### 2. \( y = \frac{x^2 + 2}{x^3 + 1} \)
Using the quotient rule:
\[
y' = \frac{(2x)(x^3 + 1) - (x^2 + 2)(3x^2)}{(x^3 + 1)^2}
\]
Differential:
\[
dy = \frac{(2x(x^3 + 1) - (x^2 + 2)(3x^2))}{(x^3 + 1)^2} dx
\]

#### 3. \( y = x^3 + e^x \)
- Derivative: \( y' = 3x^2 + e^x \)
- Differential: \( dy = (3x^2 + e^x) dx \)

---

### III. Application of Differentials to Approximations
This section seems to deal with applying differentials to estimate changes in functions, similar to the method used in part I. We apply differentials to estimate the values of expressions like \( (2.99)^3 \), which can be done similarly to the square root approximations above.

Would you like to go through part III in detail as well, or need further clarification on any of these steps? Let me know!

---

**Here are some follow-up questions to test your understanding:**

1. How is the differential approximation method useful for estimating values?
2. What is the relationship between differentials and derivatives?
3. Why is it important to choose a nearby value \( x \) when using the differential approximation?
4. Can you apply the same approximation technique for higher-order functions like \( x^n \)?
5. How do differentials help in error estimation in scientific calculations?

**Tip**: When approximating values using differentials, always choose a base value \( x \) that is easy to compute and close to the target value.

Solve a worksheet on differentials and approximations, covering the approximation of square roots, finding differentials of functions using the quotient rule, and applying differentials to estimate changes. Learn key mathematical concepts related to derivatives, trigonometry, and approximation methods in calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation