Use differential to approximate `sqrt(36. 6)`

Explanation

To approximate \( \sqrt{36.6} \) using differentials, we can follow these steps:

### Step 1: Define the function
Let \( y = \sqrt{x} \). We want to approximate \( y \) when \( x = 36.6 \).

### Step 2: Choose a point close to \( 36.6 \)
We can choose \( x = 36 \) because it is a perfect square and close to \( 36.6 \).

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