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

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