In the following find the approximate values, using differentials :
`sqrt(0.26)`

Answer

To find the approximate value of \(\sqrt{0.26}\) using differentials, we can follow these steps:

### Step 1: Identify a point close to 0.26
We choose \(x = 0.25\) because it is a perfect square and close to 0.26.

### Step 2: Define the function
Let \(f(x) = \sqrt{x}\).

### Step 3: Calculate the differential
The differential \(dy\) can be expressed as:
\[ dy = f'(x) \cdot dx \]
where \(f'(x)\) is the derivative of \(f(x)\).

### Step 4: Find the derivative
The derivative of \(f(x) = \sqrt{x}\) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]

### Step 5: Evaluate the derivative at \(x = 0.25\)
Now, we evaluate \(f'(0.25)\):
\[ f'(0.25) = \frac{1}{2\sqrt{0.25}} = \frac{1}{2 \cdot 0.5} = \frac{1}{1} = 1 \]

### Step 6: Calculate \(dx\)
We find \(dx\) as follows:
\[ dx = 0.26 - 0.25 = 0.01 \]

### Step 7: Calculate \(dy\)
Now we can calculate \(dy\):
\[ dy = f'(0.25) \cdot dx = 1 \cdot 0.01 = 0.01 \]

### Step 8: Find the approximate value of \(\sqrt{0.26}\)
Using the formula:
\[ \sqrt{0.26} \approx \sqrt{0.25} + dy \]
we have:
\[ \sqrt{0.26} \approx 0.5 + 0.01 = 0.51 \]

### Final Answer
Thus, the approximate value of \(\sqrt{0.26}\) is:
\[ \sqrt{0.26} \approx 0.51 \]

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