Using differentials, the approximate valueof `sqrt(25.5)` .

Answer

To approximate the value of \( \sqrt{25.5} \) using differentials, we can follow these steps:

### Step 1: Define the function
Let \( y = \sqrt{x} \).

### Step 2: Choose a point close to 25.5
We choose \( x = 25 \) because it is a perfect square, and we know \( \sqrt{25} = 5 \).

### Step 3: Calculate \( \Delta x \)
Since we are approximating \( \sqrt{25.5} \), we have:
\[ \Delta x = 25.5 - 25 = 0.5 \]

### Step 4: Find the derivative \( \frac{dy}{dx} \)
The derivative of \( y = \sqrt{x} \) is:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]
At \( x = 25 \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{25}} = \frac{1}{2 \cdot 5} = \frac{1}{10} \]

### Step 5: Calculate \( \Delta y \)
Using the formula for differentials:
\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]
Substituting the values we have:
\[ \Delta y \approx \frac{1}{10} \cdot 0.5 = \frac{0.5}{10} = 0.05 \]

### Step 6: Calculate the approximate value of \( \sqrt{25.5} \)
Now, we can find the approximate value of \( \sqrt{25.5} \):
\[ \sqrt{25.5} \approx \sqrt{25} + \Delta y = 5 + 0.05 = 5.05 \]

### Conclusion
Thus, the approximate value of \( \sqrt{25.5} \) is \( 5.05 \).

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