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Properties Derived from Trigonometric Identities

i tan 2 xii t...

Question

(i) tan² x

(ii) tan (2x + 1)

(iii) tan 2x

(iv) $\sqrt{\tan x}$

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Solution

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan^{2} (x + h) - \tan^{2} x}{h} = \lim_{h \to 0} \frac{[\tan (x + h) + \tan x] [\tan (x + h) - \tan x]}{h} = \lim_{h \to 0} \frac{[\frac{\sin (x + h)}{\cos (x + h)} + \frac{\sin x}{\cos x}] [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}]}{h} = \lim_{h \to 0} \frac{[\sin (x + h) \cos x + \cos (x + h) \sin x] [\sin (x + h) \cos x - \cos (x + h) \sin x]}{h \cos^{2} x \cos^{2} (x + h)} = \lim_{h \to 0} \frac{[\sin (2 x + h)] [\sin h]}{h \cos^{2} x \cos^{2} (x + h)} = \frac{1}{\cos^{2} x} \lim_{h \to 0} \sin (2 x + h) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\cos^{2} (x + h)} = \frac{1}{\cos^{2} x} \sin (2 x) (1) \frac{1}{\cos^{2} x} = \frac{1}{\cos^{2} x} 2 \sin x \cos x \frac{1}{\cos^{2} x} = 2 \times \frac{\sin x}{\cos x} \times \frac{1}{\cos^{2} x} = 2 \tan x \sec^{2} x$

$(ii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan (2 x + 2 h + 1) - \tan (2 x + 1)}{h} = \lim_{h \to 0} \frac{\frac{si n (2 x + 2 h + 1)}{\cos (2 x + 2 h + 1)} - \frac{\sin (2 x + 1)}{\cos (2 x + 1)}}{h} = \lim_{h \to 0} \frac{si n (2 x + 2 h + 1) \cos (2 x + 1) - \cos (2 x + 2 h + 1) \sin (2 x + 1)}{h \cos (2 x + 2 h + 1) \cos (2 x + 1)} = \lim_{h \to 0} \frac{\sin (2 x + 2 h + 1 - 2 x - 1)}{h \cos (2 x + 2 h + 1) \cos (2 x + 1)} = \frac{1}{\cos (2 x + 1)} \lim_{h \to 0} \frac{\sin (2 h)}{2 h} \times 2 \lim_{h \to 0} \frac{1}{\cos (2 x + 2 h + 1)} = \frac{1}{\cos (2 x + 1)} \times 2 \times \frac{1}{\cos (2 x + 1)} = \frac{2}{\cos^{2} (2 x + 1)} = 2 \sec^{2} (2 x + 1)$

$(iii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan (2 x + 2 h) - \tan (2 x)}{h} = \lim_{h \to 0} \frac{\frac{si n (2 x + 2 h)}{\cos (2 x + 2 h)} - \frac{\sin (2 x)}{\cos (2 x)}}{h} = \lim_{h \to 0} \frac{si n (2 x + 2 h) \cos (2 x) - \cos (2 x + 2 h) \sin (2 x)}{h \cos (2 x + 2 h) \cos (2 x)} = \lim_{h \to 0} \frac{\sin (2 x + 2 h - 2 x)}{h \cos (2 x + 2 h) \cos (2 x)} = \frac{1}{\cos 2 x} \lim_{h \to 0} \frac{\sin (2 h)}{2 h} \times 2 \times \lim_{h \to 0} \frac{1}{\cos (2 x + 2 h)} = \frac{1}{\cos 2 x} \times 2 \times \frac{1}{\cos 2 x} = \frac{2}{\cos^{2} (2 x)} = 2 \sec^{2} (2 x)$

$(iv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan (x + h)} + \sqrt{\tan x}}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\sin (x + h) \cos x - \cos (x + h) \sin x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = \lim_{h \to 0} \frac{\sin h}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{(\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = (1) \frac{1}{2 \sqrt{\tan x} \cos^{2} x} = \frac{\sec^{2} x}{2 \sqrt{\tan x}}$

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