Differentiate the following w.r.t. x :
`sin^(-1)((1-x^(2))/(1+x^(2))),0ltxlt1`

Answer

To differentiate the function \( y = \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Identify the function
Let \( u = \frac{1 - x^2}{1 + x^2} \). Then, we can express \( y \) as:
\[ y = \sin^{-1}(u) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
From the derivative of the inverse sine function, we know:
\[ \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we need to differentiate \( u \):
\[ u = \frac{1 - x^2}{1 + x^2} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

### Step 4: Simplify \( \frac{du}{dx} \)
Calculating the numerator:
\[ (1 + x^2)(-2x) - (1 - x^2)(2x) = -2x - 2x^3 - 2x + 2x^3 = -4x \]
Thus,
\[ \frac{du}{dx} = \frac{-4x}{(1 + x^2)^2} \]

### Step 5: Substitute back into the chain rule
Now substituting back into the chain rule:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{-4x}{(1 + x^2)^2} \]

### Step 6: Calculate \( 1 - u^2 \)
Next, we need to compute \( 1 - u^2 \):
\[ u^2 = \left(\frac{1 - x^2}{1 + x^2}\right)^2 = \frac{(1 - x^2)^2}{(1 + x^2)^2} \]
Thus,
\[ 1 - u^2 = 1 - \frac{(1 - x^2)^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - (1 - x^2)^2}{(1 + x^2)^2} \]
Calculating the numerator:
\[ (1 + x^2)^2 - (1 - x^2)^2 = (1 + 2x^2 + x^4) - (1 - 2x^2 + x^4) = 4x^2 \]
So,
\[ 1 - u^2 = \frac{4x^2}{(1 + x^2)^2} \]
Thus,
\[ \sqrt{1 - u^2} = \frac{2x}{1 + x^2} \]

### Step 7: Substitute \( \sqrt{1 - u^2} \) into \( \frac{dy}{dx} \)
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{\frac{2x}{1 + x^2}} \cdot \frac{-4x}{(1 + x^2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{(1 + x^2)}{2x} \cdot \frac{-4x}{(1 + x^2)^2} = \frac{-4}{2(1 + x^2)} = \frac{-2}{1 + x^2} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{-2}{1 + x^2} \]