differenciate square root of tan x with respect to x from 1st principal method - Maths - Statistics

Question

differenciate square root of tan x with respect to x from 1st
principal method

Gursheen kaur. · Accepted Answer

The given function is $f (x) = \sqrt{\tan x}$

$f (x) = \sqrt{\tan x} ∴ f (x + h) = \sqrt{\tan (x + h)} \frac{d}{d x} [f (x)] = f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan (x + h)} + \sqrt{\tan x}}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} = \lim_{h \to 0} \frac{(\sqrt{\tan (x + h)})^{2} - (\sqrt{\tan x})^{2}}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h} \times \lim_{h \to 0} \frac{1}{(\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \frac{1}{(\sqrt{\tan x} + \sqrt{\tan x})} \lim_{h \to 0} \frac{1}{h} (\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}) = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{1}{h} [\frac{\sin (x + h) \cos x - \cos (x + h) \sin x}{\cos x \cos (x + h)}] = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{\sin (x + h - x)}{h \cos x \cos (x + h)} [∵ \sin (A - B) = \sin A \cos B - \cos A \sin B] = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{\sin h}{h} \times \lim_{h \to 0} \frac{1}{\cos x \cos (x + h)} = \frac{1}{2 \sqrt{\tan x}} \times 1 \times \frac{1}{\cos^{2} x} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{\sec^{2} x}{2 \sqrt{\tan x}}$

differenciate square root of tan x with respect to x from 1st principal method.