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

Options

A. ` (4x^(3))/(1-x^(4))`
B. `(-4x)/(1-x^(4))`
C. ` (1)/(4-x^(4))`
D. `(-4x^(3))/(1-x^(4))`

Explanation

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

### Step 1: Differentiate \( y \)
We start by differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \log\left(\frac{1 - x^2}{1 + x^2}\right) \right) \]

### Step 2: Apply the Chain Rule
Using the chain rule for logarithmic differentiation, we have:
\[ \frac{dy}{dx} = \frac{1}{\frac{1 - x^2}{1 + x^2}} \cdot \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1 + x^2}{1 - x^2} \cdot \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 3: Differentiate the Quotient
Now we need to differentiate \( \frac{1 - x^2}{1 + x^2} \) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( u = 1 - x^2 \) and \( v = 1 + x^2 \).

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = -2x, \quad \frac{dv}{dx} = 2x \]

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

### Step 4: Simplify the Derivative
Now we simplify the numerator:
\[ = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 + x^2)^2} \]
\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} \]
\[ = \frac{-4x}{(1 + x^2)^2} \]

### Step 5: Substitute Back
Now substituting back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1 + x^2}{1 - x^2} \cdot \frac{-4x}{(1 + x^2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-4x(1 + x^2)}{(1 - x^2)(1 + x^2)^2} \]
The \( 1 + x^2 \) in the numerator and one in the denominator cancels out:
\[ \frac{dy}{dx} = \frac{-4x}{(1 - x^2)(1 + x^2)} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{-4x}{1 - x^4} \]