In the following find the approximate values, using differentials :
`sqrt(25.3)`

Answer

To find the approximate value of \(\sqrt{25.3}\) using differentials, we can follow these steps:

### Step 1: Define the function
Let \( y = \sqrt{x} \). We will use this function to find the approximate value of \(\sqrt{25.3}\).

**Hint:** Choose a value of \(x\) that is close to 25.3 and whose square root you know exactly.

### Step 2: Choose a point close to 25.3
Select \( x = 25 \) because it is a perfect square and close to 25.3.

### Step 3: Determine \(\Delta x\)
Calculate \(\Delta x\):
\[ \Delta x = 25.3 - 25 = 0.3 \]

**Hint:** \(\Delta x\) is the difference between the value you want to approximate and the value you chose.

### Step 4: Calculate \(y\) at \(x = 25\)
Now, calculate \(y\) when \(x = 25\):
\[ y = \sqrt{25} = 5 \]

### Step 5: Differentiate the function
Differentiate \(y = \sqrt{x}\):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 6: Substitute \(x = 25\) into the derivative
Now, substitute \(x = 25\) into the derivative:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{25}} = \frac{1}{10} = 0.1 \]

### Step 7: Calculate \(\Delta y\)
Using the differential, we can find \(\Delta y\):
\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x = 0.1 \cdot 0.3 = 0.03 \]

### Step 8: Find the approximate value of \(\sqrt{25.3}\)
Now, we can find the approximate value of \(\sqrt{25.3}\):
\[ \sqrt{25.3} \approx y + \Delta y = 5 + 0.03 = 5.03 \]

### Final Answer
Thus, the approximate value of \(\sqrt{25.3}\) is \(5.03\).

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