Given \( y = \sqrt{\tan\sqrt{x}} \),
differentiate both sides w.r.t \( x \): \[ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan\sqrt{x}}} \cdot \sec^2\sqrt{x} \cdot \frac{1}{2\sqrt{x}}. \]
Simplify: \[ \frac{dy}{dx} = \frac{\sec^2\sqrt{x}}{4\sqrt{x}\sqrt{\tan\sqrt{x}}}. \]
Substitute \( y = \sqrt{\tan\sqrt{x}} \): \[ \sec^2\sqrt{x} = 1 + \tan^2\sqrt{x} = 1 + y^4. \]
Thus: \[ \frac{dy}{dx} = \frac{1 + y^4}{4\sqrt{x}y}. \]
Multiply both sides by \( \sqrt{x} \): \[ \sqrt{x} \frac{dy}{dx} = \frac{1 + y^4}{4y}. \]