Question

Solve the given problem in attachment .

Answer 1

$\huge{\underline{\underline{ \bf{⎆Question:-}}}}$

$\underline{ \boxed{ \sf{ ➥ To \: \: differentiate \: \: ➭ \: {cos}^{ - 1 } ( \frac{2x}{1 + {x}^{2} } )}}}$

$\huge{\underline{\underline{ \bf{☂Solution:-}}}}$

$\large\underline { {{\bold{\bold{Formula \: used-}}}}}$

$\small{\underbrace{{\color{purple}{ ⇝sin2θ \: = \: \frac{2tanθ }{1 \: + \: {tan}^{2} θ } }}}}$

$\small{\underbrace{{\color{purple}{ ⇝ \frac{d( {tan}^{ - 1}) }{dx} \: = \frac{1}{1 - {x}^{2} } }}}}$

$\green{ \small \underline{ \mathbb{\underline{ \:➱ Explanation : }}}}$

≛ Let :

$\longmapsto \: y \: = {cos}^{ - 1} ( \frac{1}{1 + {x}^{2} } )$

≛ Put x = tanθ

$\large\longmapsto \: y \: = \: \frac{2tanθ}{1 + {tan}^{2}θ }$

$\large\longmapsto \: y \: = \: {cos}^{ - 1} (sin2θ)$

$\longmapsto \: y \: = \: {cos}^{ - 1} (cos ( \frac{\pi}{2} - 2 θ ))$

$\longmapsto \: y \: = \frac{\pi}{2} - 2 θ$

$∵ \: x \: = \: tanθ \\ ∴θ = {tan}^{ - 1}$

$☆Put \: \: , \: \: θ = {tan}^{ - 1}$

$⇝y = \large\frac{d (\frac{\pi}{2} - 2 {tan}^{ - 1}x)}{dx}$

$⇝y = \frac{\pi}{2} - 2 {tan}^{ - 1}$

$⇝ \large\frac{dy}{dx} = - 2 \times ( \frac{1}{1 + {x}^{2} } )$

Hence differentiation is :

$\large{\underbrace{\overbrace{\color{red}{ ⛧ - 2 \times ( \frac{1}{1 + {x}^{2} } ) }}}}$