Use differential to approximate `sqrt(10)`.

Answer

To approximate \(\sqrt{10}\) using differentials, we can follow these steps:

### Step 1: Choose a function and a point
We start by defining the function:
\[ y = \sqrt{x} \]
Next, we choose a point \(x\) that is close to 10, for which we know the square root. A suitable choice is:
\[ x = 9 \quad (\text{since } \sqrt{9} = 3) \]

### Step 2: Calculate \(\Delta x\)
We need to find \(\Delta x\), which is the change in \(x\) from our chosen point to the point we want to approximate:
\[ \Delta x = 10 - 9 = 1 \]

### Step 3: Differentiate the function
Now, we differentiate the function \(y = \sqrt{x}\):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 4: Evaluate the derivative at the chosen point
Next, we evaluate the derivative at \(x = 9\):
\[ \frac{dy}{dx} \bigg|_{x=9} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \cdot 3} = \frac{1}{6} \]

### Step 5: Calculate \(\Delta y\)
Now we can find \(\Delta y\) using the formula:
\[ \Delta y = \frac{dy}{dx} \cdot \Delta x \]
Substituting the values we have:
\[ \Delta y = \frac{1}{6} \cdot 1 = \frac{1}{6} \approx 0.167 \]

### Step 6: Approximate \(\sqrt{10}\)
Finally, we can approximate \(\sqrt{10}\) using the value of \(y\) at \(x = 9\) and adding \(\Delta y\):
\[ \sqrt{10} \approx \sqrt{9} + \Delta y = 3 + 0.167 = 3.167 \]

### Final Answer
Thus, the approximate value of \(\sqrt{10}\) using differentials is:
\[ \sqrt{10} \approx 3.167 \]

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