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

Options

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

Answer

To find the derivative of the function \( y = \log\left(\frac{1 - x^2}{1 + x^2}\right) \), we will use the chain rule and the quotient rule. Let's go through the solution step by step.

### Step-by-Step Solution

1. **Identify the function**:
\[ y = \log\left(\frac{1 - x^2}{1 + x^2}\right) \]

2. **Apply the chain rule**:
Using the derivative of the logarithm, we have:
\[ \frac{dy}{dx} = \frac{1}{\frac{1 - x^2}{1 + x^2}} \cdot \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) \]

3. **Simplify the logarithmic derivative**:
The expression simplifies to:
\[ \frac{dy}{dx} = \frac{1 + x^2}{1 - x^2} \cdot \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) \]

4. **Use the quotient rule**:
Let \( u = 1 - x^2 \) and \( v = 1 + x^2 \). Then, using the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = -2x \) and \( \frac{dv}{dx} = 2x \).

5. **Substituting into the quotient rule**:
\[ \frac{d}{dx}\left(\frac{1 - x^2}{1 + x^2}\right) = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

6. **Simplifying the numerator**:
Expanding the numerator:
\[ = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 + x^2)^2} \]
\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} \]
\[ = \frac{-4x}{(1 + x^2)^2} \]

7. **Combine the results**:
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{1 + x^2}{1 - x^2} \cdot \frac{-4x}{(1 + x^2)^2} \]
\[ = \frac{-4x(1 + x^2)}{(1 - x^2)(1 + x^2)^2} \]

8. **Final simplification**:
The \( (1 + x^2) \) cancels one power in the denominator:
\[ = \frac{-4x}{(1 - x^2)(1 + x^2)} \]

Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{-4x}{(1 - x^2)(1 + x^2)} \]