use the first principle y=ln (√(cos x))


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Question Number 43854 by peter frank last updated on 16/Sep/18

$u s e t h e f i r s t p r i n c i p l e y = \ln \sqrt{\cos x}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18
	$y+△y=ln(√(cos(x+△x))) △y=(1/2){ln(cos(x+△x)}−(1/2){lncosx} △y=(1/2)ln{((cos(x+△x))/(cosx))} (dy/dx)=lim_(△x→0) ((△y)/(△x)) =(1/2)lim_(△x→0) (1/2)×((ln{1+((cos(x+△x))/(cosx))−1})/({((cos(x+△x))/(cosx))−1}))×((cos(x+△x)−cosx)/(cosx×△x)) let t=((cos(x+△x))/(cosx))−1 when △x→0 t→0 so (1/2)lim_(t→0) ×((ln(1+t))/t)×lim_(△x→0) ((2sin(x+((△x)/2))sin((−((△x)/2)))/(cosx×(((−△x)/2))×((−2)/))) (1/2)×1×((2sinx×1)/(−2cosx))=(1/2)×−tanx recheck ((d{(1/2)lncosx})/dx)=(1/2)×−tanx$
	$y + △ y = l n \sqrt{c o s (x + △ x)}$ $△ y = \frac{1}{2} {l n (c o s (x + △ x)} - \frac{1}{2} {l n c o s x}$ $△ y = \frac{1}{2} l n {\frac{c o s (x + △ x)}{c o s x}}$ $\frac{d y}{d x} = \lim_{△ x \to 0} \frac{△ y}{△ x}$ $= \frac{1}{2} \lim_{△ x \to 0} \frac{1}{2} \times \frac{l n {1 + \frac{c o s (x + △ x)}{c o s x} - 1}}{{\frac{c o s (x + △ x)}{c o s x} - 1}} \times \frac{c o s (x + △ x) - c o s x}{c o s x \times △ x}$ $l e t t = \frac{c o s (x + △ x)}{c o s x} - 1$ $w h e n △ x \to 0 t \to 0$ $s o$ $\frac{1}{2} \lim_{t \to 0} \times \frac{l n (1 + t)}{t} \times \lim_{△ x \to 0} \frac{2 s i n (x + \frac{△ x}{2}) s i n ((- \frac{△ x}{2})}{c o s x \times (\frac{- △ x}{2}) \times \frac{- 2}{}}$ $\frac{1}{2} \times 1 \times \frac{2 s i n x \times 1}{- 2 c o s x} = \frac{1}{2} \times - t a n x$ $r e c h e c k$ $\frac{d {\frac{1}{2} l n c o s x}}{d x} = \frac{1}{2} \times - t a n x$
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