`d/(dx)(cos^(-1)((1 - x^2)/(1 + x^2)))` =.............. If x is positive

Options

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

Answer

To find the derivative of the function \( y = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \), we can follow these steps:

### Step 1: Substitute \( x \) with \( \tan(\theta) \)
Let \( x = \tan(\theta) \). This substitution will help us simplify the expression inside the inverse cosine function.

### Step 2: Rewrite the function
Using the identity for tangent, we can rewrite:
\[ \frac{1 - x^2}{1 + x^2} = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Using the trigonometric identity for cosine, we know:
\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \]
Thus, we can rewrite \( y \) as:
\[ y = \cos^{-1}(\cos(2\theta)) \]

### Step 3: Simplify the expression
Since \( \cos^{-1}(\cos(2\theta)) = 2\theta \) (for \( \theta \) in the appropriate range), we have:
\[ y = 2\theta \]

### Step 4: Substitute back for \( \theta \)
Recall that \( \theta = \tan^{-1}(x) \), so:
\[ y = 2\tan^{-1}(x) \]

### Step 5: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Thus:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2} \]

### Final Answer
The derivative is:
\[ \frac{d}{dx}\left(\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)\right) = \frac{2}{1 + x^2} \]

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