Find `dy/dx` if :
`y=(3-v)/(2+v),v=(4x)/(1-x^(2))`

Answer

To find \(\frac{dy}{dx}\) given \(y = \frac{3 - v}{2 + v}\) and \(v = \frac{4x}{1 - x^2}\), we can use the chain rule. Here’s a step-by-step solution:

### Step 1: Differentiate \(y\) with respect to \(v\)

Given:
\[ y = \frac{3 - v}{2 + v} \]

Using the quotient rule:
\[ \frac{dy}{dv} = \frac{(2 + v)(-1) - (3 - v)(1)}{(2 + v)^2} \]

Simplifying the numerator:
\[ = \frac{-(2 + v) - (3 - v)}{(2 + v)^2} = \frac{-2 - v - 3 + v}{(2 + v)^2} = \frac{-5}{(2 + v)^2} \]

### Step 2: Differentiate \(v\) with respect to \(x\)

Given:
\[ v = \frac{4x}{1 - x^2} \]

Using the quotient rule again:
\[ \frac{dv}{dx} = \frac{(1 - x^2)(4) - (4x)(-2x)}{(1 - x^2)^2} \]

Simplifying the numerator:
\[ = \frac{4(1 - x^2) + 8x^2}{(1 - x^2)^2} = \frac{4 + 4x^2}{(1 - x^2)^2} = \frac{4(1 + x^2)}{(1 - x^2)^2} \]

### Step 3: Apply the chain rule

Now, using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \left(\frac{-5}{(2 + v)^2}\right) \cdot \left(\frac{4(1 + x^2)}{(1 - x^2)^2}\right) \]

### Step 4: Substitute \(v\) back into the equation

Now, substitute \(v = \frac{4x}{1 - x^2}\):
\[ \frac{dy}{dx} = \frac{-5 \cdot 4(1 + x^2)}{(2 + \frac{4x}{1 - x^2})^2 \cdot (1 - x^2)^2} \]

### Step 5: Simplify the expression

Now, simplify \(2 + \frac{4x}{1 - x^2}\):
\[ = \frac{2(1 - x^2) + 4x}{1 - x^2} = \frac{2 - 2x^2 + 4x}{1 - x^2} = \frac{2 + 4x - 2x^2}{1 - x^2} \]

Thus,
\[ \frac{dy}{dx} = \frac{-20(1 + x^2)}{\left(\frac{2 + 4x - 2x^2}{1 - x^2}\right)^2 \cdot (1 - x^2)^2} \]

### Final Result

This simplifies to:
\[ \frac{dy}{dx} = \frac{-20(1 + x^2)(1 - x^2)^2}{(2 + 4x - 2x^2)^2} \]