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

Answer

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

### Step 1: Differentiate the outer function
We start by differentiating the outer function, which is \(\cos(u)\), where \(u = \frac{1 - x^2}{1 + x^2}\). The derivative of \(\cos(u)\) is \(-\sin(u)\).

\[ \frac{dy}{dx} = -\sin\left(\frac{1 - x^2}{1 + x^2}\right) \cdot \frac{du}{dx} \]

### Step 2: Differentiate the inner function
Next, we need to find \(\frac{du}{dx}\) where \(u = \frac{1 - x^2}{1 + x^2}\). We will use the quotient rule for differentiation, which states that if \(u = \frac{f(x)}{g(x)}\), then:

\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]

Here, \(f(x) = 1 - x^2\) and \(g(x) = 1 + x^2\).

### Step 3: Calculate \(f'(x)\) and \(g'(x)\)
Now we calculate the derivatives:

- \(f'(x) = -2x\)
- \(g'(x) = 2x\)

### Step 4: Apply the quotient rule
Now we can apply the quotient rule:

\[ \frac{du}{dx} = \frac{(-2x)(1 + x^2) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

### Step 5: Simplify the expression
Let's 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 6: Substitute back into the derivative
Now substituting \(\frac{du}{dx}\) back into the expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\sin\left(\frac{1 - x^2}{1 + x^2}\right) \cdot \frac{-4x}{(1 + x^2)^2} \]
\[ = \frac{4x \sin\left(\frac{1 - x^2}{1 + x^2}\right)}{(1 + x^2)^2} \]

### Final Result
Thus, the final result for \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{4x \sin\left(\frac{1 - x^2}{1 + x^2}\right)}{(1 + x^2)^2} \]

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