if ` Y= cos^(-1) "" ""sqrt((sqrt(1+x^2)+1)/(2 sqrt(1+x^2))` then ` (dy)/(dx)`=

Options

A. `(1)/(2 (1+x^2))`
B. `(1)/(1+x^2)`
C. `-(1)/(2 (1+x^2))`
D. None

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ Y = \cos^{-1} \left( \sqrt{ \frac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}} } \right), \]

we will follow these steps:

### Step 1: Simplify the expression inside the inverse cosine

Let

\[ u = \sqrt{ \frac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}} }. \]

### Step 2: Rewrite \( u \) in terms of trigonometric functions

Notice that

\[ \sqrt{1+x^2} = \sec(\theta) \quad \text{where} \quad x = \tan(\theta). \]

Thus, we can rewrite \( u \):

\[ u = \sqrt{ \frac{\sec(\theta) + 1}{2\sec(\theta)} } = \sqrt{ \frac{1 + \cos(\theta)}{2} }. \]

### Step 3: Use the half-angle identity

From the half-angle identity, we know:

\[ \sqrt{\frac{1 + \cos(\theta)}{2}} = \cos\left(\frac{\theta}{2}\right). \]

Thus, we can write:

\[ Y = \cos^{-1} \left( \cos\left(\frac{\theta}{2}\right) \right). \]

### Step 4: Simplify \( Y \)

Since \( \cos^{-1}(\cos(x)) = x \) for \( x \) in the range of \( \cos^{-1} \):

\[ Y = \frac{\theta}{2}. \]

### Step 5: Substitute back for \( \theta \)

Recall that \( \theta = \tan^{-1}(x) \):

\[ Y = \frac{1}{2} \tan^{-1}(x). \]

### Step 6: Differentiate \( Y \)

Now we differentiate \( Y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx} \left( \tan^{-1}(x) \right). \]

Using the derivative of \( \tan^{-1}(x) \):

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

we have:

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}. \]

### Final Answer

Thus, the derivative is:

\[ \frac{dy}{dx} = \frac{1}{2(1+x^2)}. \]

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