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

Answer

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

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

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we will use the chain rule. The derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \), where \( u = \frac{1 + x^2}{1 - x^2} \).

Thus, we have:
\[ \frac{dy}{dx} = \cos\left(\frac{1 + x^2}{1 - x^2}\right) \cdot \frac{d}{dx}\left(\frac{1 + x^2}{1 - x^2}\right) \]

### Step 3: Differentiate the inner function using the quotient rule
Now we need to differentiate \( u = \frac{1 + x^2}{1 - x^2} \) using the quotient rule. The quotient rule states that if \( u = \frac{p}{q} \), then:
\[ \frac{du}{dx} = \frac{p' \cdot q - p \cdot q'}{q^2} \]
where \( p = 1 + x^2 \) and \( q = 1 - x^2 \).

Calculating \( p' \) and \( q' \):
- \( p' = 2x \)
- \( q' = -2x \)

Now applying the quotient rule:
\[ \frac{du}{dx} = \frac{(2x)(1 - x^2) - (1 + x^2)(-2x)}{(1 - x^2)^2} \]

### Step 4: Simplify the expression
Now simplify 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: Combine the results
Now substitute \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \cos\left(\frac{1 + x^2}{1 - x^2}\right) \cdot \frac{4x}{(1 - x^2)^2} \]

### Final Result
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{4x \cos\left(\frac{1 + x^2}{1 - x^2}\right)}{(1 - x^2)^2} \]

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