Differentiate tan^(−1) ((((√(1+x^2 ))−1)/x)) without using any trigonometric substitution !


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Question Number 38232 by rahul 19 last updated on 23/Jun/18

$Differentiate$ $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$ $w i t h o u t u s i n g a n y t r i g o n o m e t r i c$ $s u b s t i t u t i o n!$
Commented by math khazana by abdo last updated on 23/Jun/18

$w e h a v e f (x) = a r c t a n (u (x)) w i t h$ $u (x) = \frac{\sqrt{1 + x^{2}} - 1}{x} = \frac{1 + x^{2} - 1}{x (\sqrt{1 + x^{2}} + 1)} = \frac{x}{\sqrt{1 + x^{2}} + 1}$ $(x \neq 0) \Rightarrow f^{^{'}} (x) = \frac{u^{^{'}} (x)}{1 + {(u (x))}^{2}} b u t$ $u^{^{'}} (x) = \frac{\sqrt{1 + x^{2}} + 1 - x (\frac{x}{\sqrt{1 + x^{2}}})}{{(\sqrt{1 + x^{2}} + 1)}^{2}}$ $= \frac{1 + x^{2} + \sqrt{1 + x^{2}} - x^{2}}{\sqrt{1 + x^{2}} {(\sqrt{1 + x^{2}} + 1)}^{2}} = \frac{1 + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} {(\sqrt{1 + x^{2}} + 1)}^{2}} \Rightarrow$ $f^{^{'}} (x) = \frac{1 + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} {(\sqrt{1 + x^{2}} + 1)}^{2}} . \frac{1}{1 + {(\frac{x}{\sqrt{1 + x^{2}}} + 1)}^{2}}$ $= \frac{1 + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2} {(\sqrt{1 + x^{2}} + 1)}^{2}}} \frac{1 + x^{2}}{1 + x^{2} + {(x + \sqrt{1 + x^{2}})}^{2}}$ $= \frac{\sqrt{1 + x^{2}} + 1 + x^{2}}{{(\sqrt{1 + x^{2}} + 1)}^{2} (1 + x^{2} + {(x + \sqrt{1 + x^{2}})}^{2}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

	$l e t f i n d t h e a n s w e r u s i n g t r i g o . . .$ $t h e n w i t h o u t t r i g o . . .$ $y = {t a n}^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$ $x = t a n k \frac{d x}{d k} = {s e c}^{2} k = 1 + x^{2}$ $y = {t a n}^{- 1} (\frac{s e c k - 1}{t a n k})$ $y = {t a n}^{-} (\frac{1 - c o s k}{s i n k})$ $y = {t a n}^{- 1} (t a n \frac{k}{2})$ $y = \frac{k}{2} \frac{d y}{d k} = \frac{1}{2}$ $\frac{d y}{d x} = \frac{d y}{d k} \times \frac{d k}{d x} = \frac{1}{2} \times \frac{1}{1 + x^{2}}$
	Answered by MJS last updated on 23/Jun/18

	$\frac{d}{d x} [\arctan x] = \frac{1}{x^{2} + 1}$ $this is a standard derivate . . .$ $we get it this way :$ $\arctan x = f (x)$ $x = \tan (f (x))$ $\frac{d}{d x} [x] = \frac{d}{d x} [\tan (f (x))]$ $1 = \sec^{2} (f (x)) f^{'} (x)$ $1 = \sec^{2} (\arctan x) f^{'} (x)$ $f^{'} (x) = \frac{1}{\sec^{2} (\arctan x)}$ $\sec^{2} α = \frac{1}{\cos^{2} α} = \frac{\sin^{2} α + \cos^{2} α}{\cos^{2} α} = \tan^{2} α + 1$ $f^{'} (x) = \frac{1}{\tan^{2} (\arctan x) + 1} = \frac{1}{x^{2} + 1}$ $\frac{d}{d x} [\arctan \frac{\sqrt{1 + x^{2}} - 1}{x}] =$ $= \frac{\frac{d}{d x} [\frac{\sqrt{1 + x^{2}} - 1}{x}]}{{(\frac{\sqrt{1 + x^{2}} - 1}{x})}^{2} + 1} = \frac{\frac{\sqrt{1 + x^{2}} - 1}{x^{2} \sqrt{1 + x^{2}}}}{2 \frac{1 + x^{2} - \sqrt{1 + x^{2}}}{x^{2}}} =$ $= \frac{1}{2 (1 + x^{2})}$
	Answered by ajfour last updated on 23/Jun/18

	$l e t x = \tan θ$ $y = \tan^{- 1} (\frac{1 - \cos θ}{\sin θ})$ $= \tan^{- 1} (\tan \frac{θ}{2}) = \frac{θ}{2}$ $y = \frac{1}{2} \tan^{- 1} x$ $\frac{d y}{d x} = \frac{1}{2 (1 + x^{2})} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18
	$tany=(((√(1+x^2 )) −1)/x) xtany+1=(√(1+x^2 )) x^2 tan^2 y+2xtany+1=1+x^2 x^2 tan^2 y+2xtany−x^2 =0 x^2 (tan^2 y−1)+2xtany=0 x{x(tan^2 y−1)+2tany}=0 x(tan^2 y−1)+2tany=0 x=((2tany)/(1−tan^2 y))=((2sinycosy)/(cos^2 y−sin^2 y))=tan2y (dx/dy)=2sec^2 2y^ =2(1+x^2 ) (dy/dx)=(1/(2(1+x^2 )))$
	$t a n y = \frac{\sqrt{1 + x^{2}} - 1}{x}$ $x t a n y + 1 = \sqrt{1 + x^{2}}$ $x^{2} {t a n}^{2} y + 2 x t a n y + 1 = 1 + x^{2}$ $x^{2} {t a n}^{2} y + 2 x t a n y - x^{2} = 0$ $x^{2} ({t a n}^{2} y - 1) + 2 x t a n y = 0$ $x {x ({t a n}^{2} y - 1) + 2 t a n y} = 0$ $x ({t a n}^{2} y - 1) + 2 t a n y = 0$ $x = \frac{2 t a n y}{1 - {t a n}^{2} y} = \frac{2 s i n y c o s y}{{c o s}^{2} y - {s i n}^{2} y} = t a n 2 y$ $\frac{d x}{d y} = 2 {s e c}^{2} 2 \overset{}{y} = 2 (1 + x^{2})$ $\frac{d y}{d x} = \frac{1}{2 (1 + x^{2})}$
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