Let `f(x) = sqrt(x - 1)+sqrt(x+ 24 - 10sqrt(x-1)), 1 lt x lt 26` be a real valued function 1, then `f'(x)` for `1 lt x lt 26` is

Options

A. 0
B. `1/(sqrt(x-1))`
C. `2sqrt(x-1)`
D. None of these

Answer

To solve for \( f'(x) \) where \( f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}} \) in the range \( 1 < x < 26 \), we can follow these steps:

### Step 1: Simplify the function
We start with the function:
\[ f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}} \]
We can rewrite the second term. Let \( u = \sqrt{x - 1} \), then \( u^2 = x - 1 \) or \( x = u^2 + 1 \). Substituting this into the second term gives:
\[ f(x) = u + \sqrt{(u^2 + 1) + 24 - 10u} \]
This simplifies to:
\[ f(x) = u + \sqrt{u^2 - 10u + 25} \]
Notice that \( u^2 - 10u + 25 = (u - 5)^2 \), so:
\[ f(x) = u + |u - 5| \]

### Step 2: Determine the expression for \( f(x) \)
Since \( u = \sqrt{x - 1} \) and in the range \( 1 < x < 26 \), we have \( 0 < u < 5 \). Thus, \( u - 5 < 0 \) and \( |u - 5| = 5 - u \). Therefore:
\[ f(x) = u + (5 - u) = 5 \]

### Step 3: Differentiate \( f(x) \)
Since \( f(x) = 5 \) is a constant function, we differentiate:
\[ f'(x) = 0 \]

### Conclusion
Thus, the derivative \( f'(x) \) for \( 1 < x < 26 \) is:
\[ \boxed{0} \]