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 use the properties of logarithms and differentiation. Let's go through the solution step by step.

### Step 1: Rewrite the logarithm
Using the property of logarithms that states \( \log\left(\frac{A}{B}\right) = \log(A) - \log(B) \), we can rewrite \( y \) as:
\[ y = \log(1 - x^2) - \log(1 + x^2) \]

### Step 2: Differentiate \( y \)
Now, we will 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 derivative of \( \log(u) \), which is \( \frac{1}{u} \frac{du}{dx} \), we can differentiate each term:
- For \( \log(1 - x^2) \):
\[ \frac{d}{dx}[\log(1 - x^2)] = \frac{1}{1 - x^2} \cdot (-2x) = -\frac{2x}{1 - x^2} \]
- For \( \log(1 + x^2) \):
\[ \frac{d}{dx}[\log(1 + x^2)] = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2} \]

### Step 3: Combine the derivatives
Now, substituting these derivatives back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{2x}{1 - x^2} - \frac{2x}{1 + x^2} \]

### Step 4: Simplify the expression
To combine these fractions, we need a common denominator:
\[ \frac{dy}{dx} = -\frac{2x(1 + x^2) + 2x(1 - x^2)}{(1 - x^2)(1 + x^2)} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{2x(1 + x^2 + 1 - x^2)}{(1 - x^2)(1 + x^2)} = -\frac{2x(2)}{(1 - x^2)(1 + x^2)} = -\frac{4x}{(1 - x^2)(1 + x^2)} \]

### Step 5: Final simplification
Notice that \( (1 - x^2)(1 + x^2) = 1 - x^4 \) (using the difference of squares):
\[ \frac{dy}{dx} = -\frac{4x}{1 - x^4} \]

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