`sqrt((1-x^(2))/(1+x^(2)))`

Answer

To find the derivative of the function \( y = \sqrt{\frac{1 - x^2}{1 + x^2}} \), we will use the chain rule and quotient rule of differentiation. Here is the step-by-step solution:

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

### Step 2: Differentiate using the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right) \]

### Step 3: Differentiate the inner function using the quotient rule
Let \( u = 1 - x^2 \) and \( v = 1 + x^2 \). Then, using the quotient rule:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = -2x, \quad \frac{dv}{dx} = 2x \]
Substituting these into the quotient rule gives:
\[ \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right) = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

### Step 4: Simplify the derivative
Now simplifying 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 into the derivative
Now substituting back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^{-1/2} \cdot \frac{-4x}{(1 + x^2)^2} \]

### Step 6: Final simplification
Now we can simplify:
\[ = \frac{-2x}{(1 + x^2)^2 \sqrt{\frac{1 - x^2}{1 + x^2}}} \]
\[ = \frac{-2x \sqrt{1 + x^2}}{(1 + x^2)^2 \sqrt{1 - x^2}} \]
\[ = \frac{-2x}{(1 + x^2)^{3/2} \sqrt{1 - x^2}} \]

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{-2x}{(1 + x^2)^{3/2} \sqrt{1 - x^2}} \]