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

To find the derivative of \( f(x) = \sqrt{\tan(x)} \) from first principles, we will use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] First, we need to compute \( f(x+h) \): \[ f(x+h) = \sqrt{\tan(x+h)} \] Now, we can substitute this into the derivative formula: \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\tan(x+h)} - \sqrt{\tan(x)}}{h} \] To simplify this expression, we can multiply the numerator and the denominator by the conjugate of the numerator: \[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{\tan(x+h)} - \sqrt{\tan(x)}\right)\left(\sqrt{\tan(x+h)} + \sqrt{\tan(x)}\right)}{h\left(\sqrt{\tan(x+h)} + \sqrt{\tan(x)}\right)} \] This simplifies the numerator: \[ \sqrt{\tan(x+h)} - \sqrt{\tan(x)} = \frac{\tan(x+h) - \tan(x)}{\sqrt{\tan(x+h)} + \sqrt{\tan(x)}} \] Thus, we have: \[ f'(x) = \lim_{h \to 0} \frac{\tan(x+h) - \tan(x)}{h\left(\sqrt{\tan(x+h)} + \sqrt{\tan(x)}\right)} \] Using the derivative of \( \tan(x) \), which is \( \sec^2(x) \), we can express the limit: \[ \lim_{h \to 0} \frac{\tan(x+h) - \tan(x)}{h} = \sec^2(x) \] Now substituting this back into our expression for \( f'(x) \): \[ f'(x) = \frac{\sec^2(x)}{\sqrt{\tan(x+h)} + \sqrt{\tan(x)}} \] As \( h \to 0 \), \( \tan(x+h) \to \tan(x) \), so \( \sqrt{\tan(x+h)} \to \sqrt{\tan(x)} \). Therefore, we can simplify the denominator: \[ \sqrt{\tan(x+h)} + \sqrt{\tan(x)} \to 2\sqrt{\tan(x)} \] Thus, we have: \[ f'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}} \] Finally, we can express \( \sec^2(x) \) in terms of \( \tan(x) \): \[ \sec^2(x) = 1 + \tan^2(x) \] So the derivative of \( f(x) = \sqrt{\tan(x)} \) is: \[ f'(x) = \frac{1 + \tan^2(x)}{2\sqrt{\tan(x)}} \] In conclusion, the derivative of \( \sqrt{\tan(x)} \) is: \[ \boxed{\frac{1 + \tan^2(x)}{2\sqrt{\tan(x)}}} \]