Differentiate-tan-1-1-x-2-1-x-without-using-any-trigonometric-substitution-

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 \dots

t h e n w i t h o u t t r i g o \dots

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 \dots

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

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})}