Question

If \( y = \sqrt{\tan\sqrt{x}} \), prove that: \[ \sqrt{x} \frac{dy}{dx} = \frac{1 + y^4}{4y}. \]

Hint:

To check differentiability at a point, examine both left-hand and right-hand derivatives. A function is not differentiable where there is a sharp corner, cusp, or discontinuity.

Answer 1

Differentiating \( y = \sqrt{\tan\sqrt{x}} \) with respect to \( x \) yields: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan\sqrt{x}}} \cdot \sec^2\sqrt{x} \cdot \frac{1}{2\sqrt{x}}. \] Simplifying this expression gives: \[ \frac{dy}{dx} = \frac{\sec^2\sqrt{x}}{4\sqrt{x}\sqrt{\tan\sqrt{x}}}. \] Since \( \sec^2\sqrt{x} = 1 + \tan^2\sqrt{x} \) and \( y = \sqrt{\tan\sqrt{x}} \), we have \( \tan\sqrt{x} = y^2 \), so \( \sec^2\sqrt{x} = 1 + (y^2)^2 = 1 + y^4 \). Substituting this into the derivative equation: \[ \frac{dy}{dx} = \frac{1 + y^4}{4\sqrt{x}y}. \] Multiplying both sides by \( \sqrt{x} \): \[ \sqrt{x} \frac{dy}{dx} = \frac{1 + y^4}{4y}. \]