If ` y = cos^(-1) ((2x)/(1 + x^(2))), - 1 lt x lt 1 " then " (dy)/(dx)` is equal to

Options

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

Explanation

To solve the problem, we need to find \(\frac{dy}{dx}\) for the function given by:

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

where \(-1 < x < 1\).

### Step 1: Substitute \(x = \tan(\theta)\)

We start by substituting \(x\) with \(\tan(\theta)\). Thus, we have:

\[ y = \cos^{-1}\left(\frac{2\tan(\theta)}{1 + \tan^2(\theta)}\right) \]

### Step 2: Simplify the expression

Using the identity for the double angle of tangent, we know:

\[ \frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \sin(2\theta) \]

Thus, we can rewrite \(y\) as:

\[ y = \cos^{-1}(\sin(2\theta)) \]

### Step 3: Rewrite using cosine

Using the identity \(\cos^{-1}(\sin(2\theta))\), we can express this as:

\[ y = \cos^{-1}(\sin(2\theta)) = \frac{\pi}{2} - 2\theta \]

### Step 4: Substitute back for \(\theta\)

Since \(\theta = \tan^{-1}(x)\), we can substitute back:

\[ y = \frac{\pi}{2} - 2\tan^{-1}(x) \]

### Step 5: Differentiate with respect to \(x\)

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

\[ \frac{dy}{dx} = 0 - 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]

Using the derivative of \(\tan^{-1}(x)\), which is \(\frac{1}{1+x^2}\):

\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1+x^2} \]

### Final Result

Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = -\frac{2}{1+x^2} \]

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