Consider the function `f(x)={{:(x^(2)-5,xle3),(sqrt(x+13),xgt3):}`
What is the differential coefficient of f(x) at = 12 ?

Options

A. `5//2`
B. 5
C. `1//5`
D. `1//10`

Explanation

To find the differential coefficient of the function \( f(x) \) at \( x = 12 \), we first need to determine which part of the piecewise function applies when \( x = 12 \).

The function is defined as:
\[ f(x) = \begin{cases} x^2 - 5 & \text{if } x \leq 3 \\ \sqrt{x + 13} & \text{if } x > 3 \end{cases} \]

Since \( 12 > 3 \), we will use the second part of the function:
\[ f(x) = \sqrt{x + 13} \]

### Step 1: Differentiate the function
To find the differential coefficient (derivative) of \( f(x) \) at \( x = 12 \), we first differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sqrt{x + 13}) \]

Using the chain rule, we have:
\[ f'(x) = \frac{1}{2\sqrt{x + 13}} \cdot \frac{d}{dx}(x + 13) = \frac{1}{2\sqrt{x + 13}} \cdot 1 = \frac{1}{2\sqrt{x + 13}} \]

### Step 2: Evaluate the derivative at \( x = 12 \)
Now, we evaluate \( f'(x) \) at \( x = 12 \):
\[ f'(12) = \frac{1}{2\sqrt{12 + 13}} = \frac{1}{2\sqrt{25}} = \frac{1}{2 \cdot 5} = \frac{1}{10} \]

### Final Answer
Thus, the differential coefficient of \( f(x) \) at \( x = 12 \) is:
\[ \boxed{\frac{1}{10}} \]