If `f(x)=(a+sqrt(a^2-x^2)+x)/(sqrt(a^2-x^2)+a-x)` where `a>0` then `f'(0)` has the value equal to

Options

A. `sqrta`
B. a
C. `(1)/(sqrta)`
D. `(1)/(a)`

Answer

To find \( f'(0) \) for the function

\[ f(x) = \frac{a + \sqrt{a^2 - x^2} + x}{\sqrt{a^2 - x^2} + a - x} \]

where \( a > 0 \), we will follow these steps:

### Step 1: Simplify the Function
We start by simplifying the function \( f(x) \).

\[ f(x) = \frac{a + \sqrt{a^2 - x^2} + x}{\sqrt{a^2 - x^2} + a - x} \]

### Step 2: Factor the Denominator
Notice that \( a^2 - x^2 \) can be factored as \( (a - x)(a + x) \). We can rewrite \( f(x) \) using this identity.

### Step 3: Rewrite the Function
We can rewrite the function as:

\[ f(x) = \frac{(a + x) + \sqrt{a^2 - x^2}}{\sqrt{a^2 - x^2} + (a - x)} \]

### Step 4: Differentiate the Function
To find \( f'(0) \), we will differentiate \( f(x) \) using the quotient rule:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

where \( u = a + \sqrt{a^2 - x^2} + x \) and \( v = \sqrt{a^2 - x^2} + a - x \).

### Step 5: Compute \( u' \) and \( v' \)
We need to compute the derivatives \( u' \) and \( v' \):

1. \( u' = \frac{d}{dx}(a + \sqrt{a^2 - x^2} + x) = \frac{-x}{\sqrt{a^2 - x^2}} + 1 \)
2. \( v' = \frac{d}{dx}(\sqrt{a^2 - x^2} + a - x) = \frac{-x}{\sqrt{a^2 - x^2}} - 1 \)

### Step 6: Substitute \( x = 0 \)
Now we substitute \( x = 0 \) into \( u \) and \( v \):

- \( u(0) = a + \sqrt{a^2} + 0 = a + a = 2a \)
- \( v(0) = \sqrt{a^2} + a - 0 = a + a = 2a \)

Now, substitute \( x = 0 \) into \( u' \) and \( v' \):

- \( u'(0) = 1 \)
- \( v'(0) = -1 \)

### Step 7: Apply the Quotient Rule
Now we can apply the quotient rule:

\[ f'(0) = \frac{(1)(2a) - (2a)(-1)}{(2a)^2} \]

### Step 8: Simplify
This simplifies to:

\[ f'(0) = \frac{2a + 2a}{4a^2} = \frac{4a}{4a^2} = \frac{1}{a} \]

### Final Answer
Thus, the value of \( f'(0) \) is:

\[ \boxed{\frac{1}{a}} \]