Find \( \frac{d y}{d x} \) if \( y=\left(\frac{1+2 x}{1+x}\right)^{2} \)

To find \( \frac{dy}{dx} \) for the function \( y = \left( \frac{1 + 2x}{1 + x} \right)^2 \), we will use the chain rule and the quotient rule. First, let's define the inner function: \[ u = \frac{1 + 2x}{1 + x} \] Then, we can express \( y \) in terms of \( u \): \[ y = u^2 \] Next, we will find \( \frac{dy}{du} \): \[ \frac{dy}{du} = 2u \] Now, we need to find \( \frac{du}{dx} \). We will apply the quotient rule, which states that if \( u = \frac{f(x)}{g(x)} \), then: \[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] In our case, \( f(x) = 1 + 2x \) and \( g(x) = 1 + x \). Calculating the derivatives: \[ f'(x) = 2, \quad g'(x) = 1 \] Now applying the quotient rule: \[ \frac{du}{dx} = \frac{(2)(1 + x) - (1 + 2x)(1)}{(1 + x)^2} \] Simplifying the numerator: \[ = \frac{2 + 2x - 1 - 2x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \] Now we have both \( \frac{dy}{du} \) and \( \frac{du}{dx} \). We can use the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \frac{1}{(1 + x)^2} \] Substituting back for \( u \): \[ \frac{dy}{dx} = 2 \left( \frac{1 + 2x}{1 + x} \right) \cdot \frac{1}{(1 + x)^2} \] Now, simplifying this expression: \[ \frac{dy}{dx} = \frac{2(1 + 2x)}{(1 + x)(1 + x)^2} = \frac{2(1 + 2x)}{(1 + x)^3} \] Thus, the final result is: \[ \frac{dy}{dx} = \frac{2(1 + 2x)}{(1 + x)^3} \]