`d/(dx)tan^-1((1-x)/(1+x))=`

Options

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

Answer

To solve the problem \( \frac{d}{dx} \tan^{-1}\left(\frac{1-x}{1+x}\right) \), we will follow these steps:

### Step 1: Let \( y = \tan^{-1}\left(\frac{1-x}{1+x}\right) \)
We start by defining \( y \) as the inverse tangent of the expression \( \frac{1-x}{1+x} \).

### Step 2: Use the identity for tangent
We can use the identity for tangent to simplify the expression. Recall that:
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \]
By letting \( x = \tan(\theta) \), we can rewrite \( y \):
\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \]
Thus, we have:
\[ y = \frac{\pi}{4} - \theta \]

### Step 3: Express \( \theta \) in terms of \( x \)
Since \( x = \tan(\theta) \), we can express \( \theta \) as:
\[ \theta = \tan^{-1}(x) \]
Substituting this back into our expression for \( y \):
\[ y = \frac{\pi}{4} - \tan^{-1}(x) \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - \frac{d}{dx} \tan^{-1}(x) \]
Using the derivative of the inverse tangent function, we know:
\[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \]
Thus:
\[ \frac{dy}{dx} = -\frac{1}{1+x^2} \]

### Step 5: Final result
The derivative of \( \tan^{-1}\left(\frac{1-x}{1+x}\right) \) is:
\[ \frac{d}{dx} \tan^{-1}\left(\frac{1-x}{1+x}\right) = -\frac{1}{1+x^2} \]