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: Define the function
Let \(y = f(x) = \sqrt{x}\).

### Step 2: Choose a value of \(x\) close to 0.26
We can choose \(x = 0.25\) because it is a perfect square and close to 0.26.

### Step 3: Calculate \(y\) at \(x = 0.25\)
Now, we calculate:
\[ y = f(0.25) = \sqrt{0.25} = 0.5 \]

### Step 4: Determine \(dx\)
Since we are moving from \(0.25\) to \(0.26\), we have:
\[ dx = 0.26 - 0.25 = 0.01 \]

### Step 5: Find the derivative \(f'(x)\)
To find \(dy\), we need to compute the derivative of \(f(x)\):
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Now, we evaluate this at \(x = 0.25\):
\[ f'(0.25) = \frac{1}{2\sqrt{0.25}} = \frac{1}{2 \cdot 0.5} = 1 \]

### Step 6: Calculate \(dy\)
Now we can find \(dy\):
\[ dy = f'(0.25) \cdot dx = 1 \cdot 0.01 = 0.01 \]

### Step 7: Approximate the value of \(\sqrt{0.26}\)
Now we can approximate \(\sqrt{0.26}\) using:
\[ \sqrt{0.26} \approx y + dy = 0.5 + 0.01 = 0.51 \]

Thus, the approximate value of \(\sqrt{0.26}\) is **0.51**.

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