In the following find the approximate values, using differentials :
`sqrt(0.82)`

Answer

To find the approximate value of \( \sqrt{0.82} \) using differentials, we can follow these steps:

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

### Step 2: Differentiate the function
We need to find the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 3: Choose a nearby value
We know that \( \sqrt{0.81} = 0.9 \). So, we will use \( x = 0.81 \) as our nearby value.

### Step 4: Calculate \( \Delta x \)
To find \( \sqrt{0.82} \), we can express it as:
\[ \Delta x = 0.82 - 0.81 = 0.01 \]

### Step 5: Calculate \( dy \)
Using the derivative we found, we can calculate \( dy \):
\[ dy = \frac{dy}{dx} \cdot \Delta x \]
Substituting the values:
\[ dy = \frac{1}{2\sqrt{0.81}} \cdot 0.01 \]
Calculating \( \sqrt{0.81} \):
\[ \sqrt{0.81} = 0.9 \]
Now substituting:
\[ dy = \frac{1}{2 \cdot 0.9} \cdot 0.01 = \frac{0.01}{1.8} = \frac{1}{180} \]

### Step 6: Approximate \( \sqrt{0.82} \)
Now we can approximate \( \sqrt{0.82} \):
\[ \sqrt{0.82} \approx \sqrt{0.81} + dy = 0.9 + \frac{1}{180} \]
Calculating \( \frac{1}{180} \):
\[ \frac{1}{180} \approx 0.00555 \]
Thus,
\[ \sqrt{0.82} \approx 0.9 + 0.00555 = 0.90555 \]

### Final Result
The approximate value of \( \sqrt{0.82} \) is \( \approx 0.90555 \).

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