2. Find from first principle, the derivatives of \( \sqrt{\tan x} \)

To find the derivative of \( \sqrt{\tan x} \) from first principles, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Let \( f(x) = \sqrt{\tan x} \). Then we need to compute \( f(x+h) \): \[ f(x+h) = \sqrt{\tan(x+h)} \] Now substituting into the derivative formula, we have: \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\tan(x+h)} - \sqrt{\tan x}}{h} \] To simplify this, we can 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: \[ = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h \left(\sqrt{\tan(x+h)} + \sqrt{\tan x}\right)} \] Using the fact that the derivative of \( \tan x \) is \( \sec^2 x \), we know that: \[ \tan(x+h) - \tan x = h\sec^2(x) + o(h) \quad \text{(as \( h \to 0 \))} \] Thus, we have: \[ f'(x) = \lim_{h \to 0} \frac{h \sec^2(x) + o(h)}{h \left(\sqrt{\tan(x+h)} + \sqrt{\tan x}\right)} \] Cancelling \( h \) yields: \[ = \lim_{h \to 0} \frac{\sec^2 x + \frac{o(h)}{h}}{\sqrt{\tan(x+h)} + \sqrt{\tan x}} = \frac{\sec^2 x}{2\sqrt{\tan x}} \] Thus, the derivative is: \[ f'(x) = \frac{\sec^2 x}{2\sqrt{\tan x}} \] This finding showcases the beauty of calculus and how foundational principles lead to significant discoveries in mathematics! Now you can impress your friends with your knowledge of derivatives derived from first principles! And if you're ever stranded in the wilderness with nothing but math books, you'll know how to differentiate while sitting under a tree! Just remember to take notes—nature can be a bit noisy sometimes!