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 \( \frac{dy}{dx} \) for the function \( y = \log\left(\frac{1 - x^2}{1 + x^2}\right) \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \log\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 2: Apply the logarithmic property
Using the property of logarithms, we can rewrite this as:
\[ y = \log(1 - x^2) - \log(1 + x^2) \]

### Step 3: Differentiate using the chain rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}[\log(1 - x^2)] - \frac{d}{dx}[\log(1 + x^2)] \]

Using the chain rule, we have:
\[ \frac{d}{dx}[\log(u)] = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u \) is the function inside the logarithm.

### Step 4: Differentiate each term
1. For \( \log(1 - x^2) \):
- Let \( u = 1 - x^2 \), then \( \frac{du}{dx} = -2x \).
- Thus, \( \frac{d}{dx}[\log(1 - x^2)] = \frac{1}{1 - x^2} \cdot (-2x) = \frac{-2x}{1 - x^2} \).

2. For \( \log(1 + x^2) \):
- Let \( v = 1 + x^2 \), then \( \frac{dv}{dx} = 2x \).
- Thus, \( \frac{d}{dx}[\log(1 + x^2)] = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2} \).

### Step 5: Combine the derivatives
Now we can combine the results:
\[ \frac{dy}{dx} = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2} \]

### Step 6: Find a common denominator
The common denominator for the two fractions is \( (1 - x^2)(1 + x^2) \):
\[ \frac{dy}{dx} = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 - x^2)(1 + x^2)} \]

### Step 7: Simplify the numerator
Now simplify the numerator:
\[ -2x(1 + x^2) - 2x(1 - x^2) = -2x - 2x^3 - 2x + 2x^3 = -4x \]

### Step 8: Final expression for the derivative
Thus, we have:
\[ \frac{dy}{dx} = \frac{-4x}{(1 - x^2)(1 + x^2)} = \frac{-4x}{1 - x^4} \]

### Conclusion
The final answer is:
\[ \frac{dy}{dx} = \frac{-4x}{1 - x^4} \]