Find from first principle, the derivatives of \( \sqrt{\text { tan }} \)

To find the derivative of \( f(x) = \sqrt{\tan(x)} \) from first principles, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] Substituting our function into this formula gives: \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\tan(x + h)} - \sqrt{\tan(x)}}{h} \] To simplify, we can use the conjugate to eliminate the square roots. Multiply the numerator and denominator by the conjugate: \[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{\tan(x + h)} - \sqrt{\tan(x)}\right)\left(\sqrt{\tan(x + h)} + \sqrt{\tan(x)}\right)}{h \left(\sqrt{\tan(x + h)} + \sqrt{\tan(x)}\right)} \] This simplifies to: \[ f'(x) = \lim_{h \to 0} \frac{\tan(x + h) - \tan(x)}{h \left(\sqrt{\tan(x + h)} + \sqrt{\tan(x)}\right)} \] The limit of the numerator as \( h \to 0 \) is \( \tan'(x) = \sec^2(x) \). Thus we have: \[ f'(x) = \frac{\sec^2(x)}{\sqrt{\tan(x + h)} + \sqrt{\tan(x)}} \] Taking the limit as \( h \to 0 \) gives us: \[ f'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}} \] So, the derivative of \( \sqrt{\tan(x)} \) is: \[ f'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}} \]