If `f(x)=sqrt(x-4sqrt(x-4))+tan^(-1)((1-2x)/(2+x)), AA 4 lt x lt 8`,
then the value of `f'(5)` is equal to

Options

A. `-(7)/(13)`
B. 0
C. `(5)/(13)`
D. `-(8)/(13)`

Answer

To find the value of \( f'(5) \) for the function

\[ f(x) = \sqrt{x - 4\sqrt{x - 4}} + \tan^{-1}\left(\frac{1 - 2x}{2 + x}\right) \]

for \( 4 < x < 8 \), we will first simplify the function and then differentiate it.

### Step 1: Simplify the function

We start with the first part of the function:

\[ \sqrt{x - 4\sqrt{x - 4}} \]

Let \( y = \sqrt{x - 4} \). Then, we have:

\[ x = y^2 + 4 \]

Substituting this into our expression gives:

\[ \sqrt{(y^2 + 4) - 4y} = \sqrt{y^2 - 4y + 4} = \sqrt{(y - 2)^2} = |y - 2| \]

Since \( y = \sqrt{x - 4} \) and \( x \) is in the range \( (4, 8) \), we have \( y = \sqrt{x - 4} \) which lies between \( 0 \) and \( 2 \). Therefore, \( y - 2 \) is negative in this range, and we can write:

\[ |y - 2| = 2 - \sqrt{x - 4} \]

Now, substituting back, we have:

\[ \sqrt{x - 4\sqrt{x - 4}} = 2 - \sqrt{x - 4} \]

Now we can rewrite \( f(x) \):

\[ f(x) = 2 - \sqrt{x - 4} + \tan^{-1}\left(\frac{1 - 2x}{2 + x}\right) \]

### Step 2: Differentiate the function

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2 - \sqrt{x - 4}) + \frac{d}{dx}\left(\tan^{-1}\left(\frac{1 - 2x}{2 + x}\right)\right) \]

The derivative of \( 2 \) is \( 0 \), and the derivative of \( -\sqrt{x - 4} \) is:

\[ -\frac{1}{2\sqrt{x - 4}} \cdot \frac{d}{dx}(x - 4) = -\frac{1}{2\sqrt{x - 4}} \]

Next, we differentiate the second part using the chain rule. Let \( u = \frac{1 - 2x}{2 + x} \):

\[ \frac{du}{dx} = \frac{(2 + x)(-2) - (1 - 2x)(1)}{(2 + x)^2} = \frac{-2(2 + x) - (1 - 2x)}{(2 + x)^2} = \frac{-4 - 2x - 1 + 2x}{(2 + x)^2} = \frac{-5}{(2 + x)^2} \]

Thus, the derivative of \( \tan^{-1}(u) \) is:

\[ \frac{1}{1 + u^2} \cdot \frac{du}{dx} = \frac{1}{1 + \left(\frac{1 - 2x}{2 + x}\right)^2} \cdot \left(\frac{-5}{(2 + x)^2}\right) \]

### Step 3: Evaluate \( f'(5) \)

Now we evaluate \( f'(5) \):

1. Calculate \( \sqrt{5 - 4} = \sqrt{1} = 1 \):

\[ f'(5) = -\frac{1}{2 \cdot 1} + \text{(the derivative of the second part evaluated at } x = 5) \]

2. For \( u \) at \( x = 5 \):

\[ u = \frac{1 - 10}{7} = -\frac{9}{7} \]

Thus,

\[ 1 + u^2 = 1 + \left(-\frac{9}{7}\right)^2 = 1 + \frac{81}{49} = \frac{49 + 81}{49} = \frac{130}{49} \]

So,

\[ \frac{1}{1 + u^2} = \frac{49}{130} \]

Thus,

\[ f'(5) = -\frac{1}{2} + \frac{49}{130} \cdot \frac{-5}{(7)^2} = -\frac{1}{2} - \frac{245}{130} = -\frac{65}{130} - \frac{245}{130} = -\frac{310}{130} = -\frac{31}{13} \]

### Final Answer

Therefore, the value of \( f'(5) \) is:

\[ \boxed{-\frac{31}{13}} \]