Using differentials, find the approximate values of the following :
(i) `sqrt(37)`
(ii) `sqrt(401)`.

Answer

To approximate the values of \(\sqrt{37}\) and \(\sqrt{401}\) using differentials, we can follow these steps:

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

### Step 2: Find the derivative
To find the differential \( dy \), we first need to compute the derivative of \( f(x) \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 3: Express \( dy \)
The differential \( dy \) can be expressed as:
\[ dy = \frac{dy}{dx} \cdot dx = \frac{1}{2\sqrt{x}} \cdot dx \]

### Step 4: Approximate \(\sqrt{37}\)
For \(\sqrt{37}\):
- Choose \( x = 36 \) (since \( 36 \) is a perfect square).
- Thus, \( dx = 37 - 36 = 1 \).

Now, calculate \( y \) and \( dy \):
\[ y = \sqrt{36} = 6 \]
\[ dy = \frac{1}{2\sqrt{36}} \cdot 1 = \frac{1}{2 \cdot 6} = \frac{1}{12} \approx 0.0833 \]

Now, approximate \(\sqrt{37}\):
\[ \sqrt{37} \approx y + dy = 6 + 0.0833 \approx 6.0833 \]

### Step 5: Approximate \(\sqrt{401}\)
For \(\sqrt{401}\):
- Choose \( x = 400 \) (since \( 400 \) is a perfect square).
- Thus, \( dx = 401 - 400 = 1 \).

Now, calculate \( y \) and \( dy \):
\[ y = \sqrt{400} = 20 \]
\[ dy = \frac{1}{2\sqrt{400}} \cdot 1 = \frac{1}{2 \cdot 20} = \frac{1}{40} \approx 0.025 \]

Now, approximate \(\sqrt{401}\):
\[ \sqrt{401} \approx y + dy = 20 + 0.025 \approx 20.025 \]

### Final Answers
Thus, the approximate values are:
- \(\sqrt{37} \approx 6.0833\)
- \(\sqrt{401} \approx 20.025\)