`(d)/(dx)[cos^(-1)sqrt(((1+x))/(2))]`

Options

A. `(-1)/(sqrt(1-x^(2)))`
B. `(-1)/(2sqrt(1-x^(2)))`
C. `(1)/(sqrt(1-x^(2)))`
D. `(1)/(2sqrt(1-x^(2)))`

Answer

To differentiate the function \( y = \cos^{-1} \left( \sqrt{\frac{1+x}{2}} \right) \), we will follow these steps:

### Step 1: Identify the outer and inner functions
We have an outer function \( y = \cos^{-1}(u) \) where \( u = \sqrt{\frac{1+x}{2}} \).

### Step 2: Differentiate the outer function
Using the derivative of the inverse cosine function, we have:
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate the inner function
Now, we need to differentiate \( u = \sqrt{\frac{1+x}{2}} \). We can rewrite \( u \) as:
\[ u = \left( \frac{1+x}{2} \right)^{1/2} \]
Using the chain rule:
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{1+x}{2} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{1+x}{2} \right) \]
Calculating \( \frac{d}{dx} \left( \frac{1+x}{2} \right) = \frac{1}{2} \), we get:
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{1+x}{2} \right)^{-1/2} \cdot \frac{1}{2} = \frac{1}{4} \left( \frac{1+x}{2} \right)^{-1/2} \]

### Step 4: Combine the derivatives using the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{1}{4} \left( \frac{1+x}{2} \right)^{-1/2} \]

### Step 5: Substitute \( u \) back into the equation
Next, we need to substitute \( u \) back into the equation:
\[ u^2 = \frac{1+x}{2} \implies 1 - u^2 = 1 - \frac{1+x}{2} = \frac{1 - x}{2} \]
Thus, \( \sqrt{1 - u^2} = \sqrt{\frac{1-x}{2}} \).

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now substituting \( \sqrt{1 - u^2} \) into our derivative:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{\frac{1-x}{2}}} \cdot \frac{1}{4} \left( \frac{1+x}{2} \right)^{-1/2} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{4} \cdot \frac{1}{\sqrt{\frac{1-x}{2}}} \cdot \frac{1}{\sqrt{\frac{1+x}{2}}} \]
Combining the square roots:
\[ \frac{dy}{dx} = -\frac{1}{4} \cdot \frac{1}{\sqrt{\frac{(1-x)(1+x)}{4}}} = -\frac{1}{4} \cdot \frac{2}{\sqrt{1-x^2}} = -\frac{1}{2\sqrt{1-x^2}} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}} \]

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