Show that the derivative of the function \( y=4 \) arctan \( \sqrt{1+4 x} \) is \( \frac{4}{(1+2 x) \sqrt{1+4 x}} \)

To find the derivative of the function \( y = 4 \arctan(\sqrt{1 + 4x}) \), we can apply the chain rule and the derivative formula for the arctan function. First, we recognize that the derivative of \( \arctan(u) \) is \( \frac{1}{1 + u^2} \), where \( u = \sqrt{1 + 4x} \). We also need the derivative of \( u \) with respect to \( x \): 1. Differentiate \( u = \sqrt{1 + 4x} \): \[ \frac{du}{dx} = \frac{1}{2\sqrt{1 + 4x}} \cdot (4) = \frac{2}{\sqrt{1 + 4x}} \] 2. Differentiate \( y \): \[ \frac{dy}{dx} = 4 \cdot \frac{1}{1 + (\sqrt{1 + 4x})^2} \cdot \frac{du}{dx} \] Now, \( (\sqrt{1 + 4x})^2 = 1 + 4x \), so: \[ 1 + (\sqrt{1 + 4x})^2 = 1 + (1 + 4x) = 2 + 4x \] 3. Substitute into the derivative: \[ \frac{dy}{dx} = 4 \cdot \frac{1}{2 + 4x} \cdot \frac{2}{\sqrt{1 + 4x}} = \frac{8}{(2 + 4x) \sqrt{1 + 4x}} \] 4. Simplify the denominator: \[ 2 + 4x = 2(1 + 2x) \] Thus, we have: \[ \frac{dy}{dx} = \frac{8}{2(1 + 2x)\sqrt{1 + 4x}} = \frac{4}{(1 + 2x)\sqrt{1 + 4x}} \] This shows that the derivative of \( y = 4 \arctan(\sqrt{1 + 4x}) \) is indeed \( \frac{4}{(1 + 2x) \sqrt{1 + 4x}} \).