Use differential to approximate `sqrt(51)`.

Answer

To approximate \(\sqrt{51}\) using differentials, we can follow these steps:

### Step 1: Identify the function and the point of approximation
We want to approximate \(\sqrt{51}\). We can express this as \(\sqrt{49 + 2}\) because \(49\) is a perfect square.

Let \(f(x) = \sqrt{x}\). We will use \(x = 49\) (since \(\sqrt{49} = 7\)) and \(h = 2\) (the increment to reach \(51\)).

### Step 2: Write the differential approximation formula
The differential approximation formula is given by:
\[ f(x + h) \approx f(x) + f'(x) \cdot h \]
where \(f'(x)\) is the derivative of \(f(x)\).

### Step 3: Calculate \(f(49)\)
Now, we calculate:
\[ f(49) = \sqrt{49} = 7 \]

### Step 4: Calculate the derivative \(f'(x)\)
The derivative of \(f(x) = \sqrt{x}\) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Now, we find \(f'(49)\):
\[ f'(49) = \frac{1}{2\sqrt{49}} = \frac{1}{2 \cdot 7} = \frac{1}{14} \]

### Step 5: Apply the differential approximation
Now, we can apply the values into the approximation formula:
\[ f(51) \approx f(49) + f'(49) \cdot h \]
Substituting the known values:
\[ f(51) \approx 7 + \left(\frac{1}{14}\right) \cdot 2 \]
Calculating the product:
\[ \frac{1}{14} \cdot 2 = \frac{2}{14} = \frac{1}{7} \]
Thus, we have:
\[ f(51) \approx 7 + \frac{1}{7} \]

### Step 6: Combine the fractions
To combine \(7\) and \(\frac{1}{7}\), we convert \(7\) into a fraction:
\[ 7 = \frac{49}{7} \]
So,
\[ f(51) \approx \frac{49}{7} + \frac{1}{7} = \frac{50}{7} \]

### Conclusion
The approximate value of \(\sqrt{51}\) using differentials is:
\[ \sqrt{51} \approx \frac{50}{7} \]

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