The derivation of ` tan^(-1) "" {(1+x)/(1-x)} ` is

Options

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

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \frac{1+x}{1-x} \right) \), we will use the chain rule and the quotient rule. Here’s the step-by-step solution:

### Step 1: Identify the outer and inner functions
Let:
- \( u = \frac{1+x}{1-x} \)
- \( y = \tan^{-1}(u) \)

### Step 2: Differentiate the outer function
The derivative of \( y = \tan^{-1}(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{1+u^2} \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = \frac{1+x}{1-x} \) using the quotient rule. The quotient rule states that if \( u = \frac{f(x)}{g(x)} \), then:
\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Here, \( f(x) = 1+x \) and \( g(x) = 1-x \).

- \( f'(x) = 1 \)
- \( g'(x) = -1 \)

Applying the quotient rule:
\[ \frac{du}{dx} = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} \]
\[ = \frac{(1-x) + (1+x)}{(1-x)^2} = \frac{1 - x + 1 + x}{(1-x)^2} = \frac{2}{(1-x)^2} \]

### Step 4: Apply the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{2}{(1-x)^2} \]

### Step 5: Substitute back for \( u \)
Now substitute \( u = \frac{1+x}{1-x} \) back into the equation:
\[ \frac{dy}{dx} = \frac{1}{1+\left(\frac{1+x}{1-x}\right)^2} \cdot \frac{2}{(1-x)^2} \]

### Step 6: Simplify the expression
To simplify \( 1 + \left(\frac{1+x}{1-x}\right)^2 \):
\[ = 1 + \frac{(1+x)^2}{(1-x)^2} = \frac{(1-x)^2 + (1+x)^2}{(1-x)^2} \]
Calculating the numerator:
\[ (1-x)^2 + (1+x)^2 = (1 - 2x + x^2) + (1 + 2x + x^2) = 2 + 2x^2 \]
Thus,
\[ 1 + \left(\frac{1+x}{1-x}\right)^2 = \frac{2 + 2x^2}{(1-x)^2} \]

Now substituting this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\frac{2 + 2x^2}{(1-x)^2}} \cdot \frac{2}{(1-x)^2} = \frac{(1-x)^2}{2 + 2x^2} \cdot \frac{2}{(1-x)^2} \]
\[ = \frac{2}{2 + 2x^2} = \frac{1}{1 + x^2} \]

### Final Result
Thus, the derivative of \( y = \tan^{-1} \left( \frac{1+x}{1-x} \right) \) is:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

---