using the function f(x)=sqrt(4x^3+3) +2, find dy and Δy as x changes from 2 to 2.01

To find the differential \( dy \) and the change in \( y \) (\( \Delta y \)) for the function \( f(x) = \sqrt{4x^3 + 3} + 2 \) as \( x \) changes from \( 2 \) to \( 2.01 \), we’ll go through the following steps:

### Step 1: Find \( f'(x) \) to compute \( dy \)
The differential \( dy \) is given by:
\[
dy = f'(x) \, dx
\]
where \( dx \) is the small change in \( x \), in this case \( dx = 0.01 \).

#### Differentiate \( f(x) \)
Given:
\[
f(x) = \sqrt{4x^3 + 3} + 2
\]
Rewrite \( f(x) \) as:
\[
f(x) = (4x^3 + 3)^{1/2} + 2
\]

Using the chain rule, differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = \frac{1}{2}(4x^3 + 3)^{-1/2} \cdot (12x^2)
\]
Simplify \( f'(x) \):
\[
f'(x) = \frac{6x^2}{\sqrt{4x^3 + 3}}
\]

### Step 2: Calculate \( f'(2) \) and then \( dy \)
Evaluate \( f'(x) \) at \( x = 2 \):
\[
f'(2) = \frac{6 \cdot 2^2}{\sqrt{4 \cdot 2^3 + 3}} = \frac{6 \cdot 4}{\sqrt{32 + 3}} = \frac{24}{\sqrt{35}}
\]

Now, \( dy = f'(2) \cdot dx \):
\[
dy = \frac{24}{\sqrt{35}} \cdot 0.01
\]

### Step 3: Calculate \( \Delta y = f(2.01) - f(2) \)
Now, compute \( f(2) \) and \( f(2.01) \) to find \( \Delta y \).

1. **Calculate \( f(2) \):**
   \[
   f(2) = \sqrt{4 \cdot 2^3 + 3} + 2 = \sqrt{32 + 3} + 2 = \sqrt{35} + 2
   \]

2. **Calculate \( f(2.01) \):**
   \[
   f(2.01) = \sqrt{4 \cdot (2.01)^3 + 3} + 2
   \]
   \( 2.01^3 \approx 8.120601 \), so:
   \[
   f(2.01) = \sqrt{4 \cdot 8.120601 + 3} + 2 = \sqrt{32.482404 + 3} + 2 = \sqrt{35.482404} + 2
   \]

3. **Calculate \( \Delta y \):**
   \[
   \Delta y = f(2.01) - f(2) = \left( \sqrt{35.482404} + 2 \right) - \left( \sqrt{35} + 2 \right) = \sqrt{35.482404} - \sqrt{35}
   \]

### Summary of Results
1. \( dy \approx \frac{24}{\sqrt{35}} \cdot 0.01 \)
2. \( \Delta y \approx \sqrt{35.482404} - \sqrt{35} \)

Would you like me to proceed with exact decimal calculations for \( dy \) and \( \Delta y \) or any additional explanation?

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#### Related Questions:
1. What are differentials and how are they applied in approximations?
2. How does the chain rule assist in differentiating composite functions?
3. Why is \( \Delta y \) an approximation to the exact change in \( y \)?
4. What role does \( dx \) play in the concept of differentials?
5. How would you interpret \( dy \) geometrically on the graph of \( f(x) \)?

#### Tip:
Differentials provide an efficient way to approximate small changes in function values, which is especially useful in applications like physics and engineering.

Using the function f(x) = sqrt(4x^3 + 3) + 2, find dy and Δy as x changes from 2 to 2.01

Explore how to find the differential dy and the change in y (Δy) for the function f(x) = sqrt(4x^3 + 3) + 2 as x changes from 2 to 2.01. This detailed guide covers differentiation using the chain rule and differential approximations. Ideal for advanced high school students (Grade 11-12).

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