Find `(dy)/(dx)`, when :
`y=sin((1+x^(2))/(1-x^(2)))`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sin\left(\frac{1 + x^2}{1 - x^2}\right)\), we will use the chain rule and the quotient rule for differentiation. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let:
- \(u = \frac{1 + x^2}{1 - x^2}\) (inner function)
- \(y = \sin(u)\) (outer function)

### Step 2: Differentiate the outer function
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
We know that:
\[ \frac{dy}{du} = \cos(u) \]

### Step 3: Differentiate the inner function \(u\)
To differentiate \(u = \frac{1 + x^2}{1 - x^2}\), we will use the quotient rule:
\[ \frac{du}{dx} = \frac{(1 - x^2)(2x) - (1 + x^2)(-2x)}{(1 - x^2)^2} \]

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

### Step 5: Combine the results
Now substituting back into the chain rule:
\[ \frac{dy}{dx} = \cos\left(\frac{1 + x^2}{1 - x^2}\right) \cdot \frac{4x}{(1 - x^2)^2} \]

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

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