Question

Question
$\rm \to y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x } ,\: Find\: \bigg(\dfrac{dy}{dx}\bigg)\:\: at\:\:x =-1$

NOTE
ANSWER is 2/π

Answer 1

$\large{\bold{Gɪᴠᴇɴ :}}$

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$\rm \:\:●\:y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x }$

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$\large{\bold{Tᴏ \:Fɪɴᴅ :}}$

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$\rm \:\:●\:Find\: \bigg(\dfrac{dy}{dx}\bigg)\:\: at\:\:x =-1$

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$\large{\bold{Sᴏʟᴜᴛɪᴏɴ :}}$

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➪ $\rm y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x }$

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We know that,

$\sf{tan^{-1}x+cot^{-1}x = \dfrac{\pi}{2} \:\:∀x}$

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➪ $\rm y = \dfrac{tan^{-1}x-cot^{-1}x }{\dfrac{\pi}{2}}$

➪ $\rm y = \dfrac{2}{\pi} (tan^{-1}x-cot^{-1}x)$

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By appling derivative,

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➪ $\rm \dfrac{dy}{dx} = \dfrac{2}{\pi} .\dfrac{d}{dx} (tan^{-1}x-cot^{-1}x)$

➪ $\rm \dfrac{dy}{dx} = \dfrac{2}{\pi}\left[ \dfrac{d}{dx} tan^{-1}x-\dfrac{d}{dx} cot^{-1}x\right]$

➪ $\rm \dfrac{dy}{dx} = \dfrac{2}{\pi}\left[ \dfrac{1}{1+x^2} -\left(-\dfrac{1}{1+x^2}\right)\right]$

➪ $\rm \dfrac{dy}{dx} = \dfrac{2}{\pi}. \dfrac{2}{1+x^2}$

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Given, x = -1

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➪ $\rm \dfrac{dy}{dx} = \dfrac{4}{\pi(1+(-1)^2)}$

➪ $\rm \dfrac{dy}{dx} = \dfrac{\cancel{4}}{\cancel{2}\pi } =$$\rm{\red{ \dfrac{2}{\pi}}}$

Answer 2

Given Question :-

$\rm \to y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x } ,\: Find\: \bigg(\dfrac{dy}{dx}\bigg)\:\: at\:\:x =-1$

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{ y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x } } \\ &\sf{and \: x \: = \: - \: 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{\dfrac{dy}{dx} \: at \: x \: = \: - \: 1}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$(1). \: \boxed{ \bf \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2}}$

$(2). \: \boxed{ \bf \: \dfrac{d}{dx} {tan}^{ - 1}x = \dfrac{1}{1 + {x}^{2} } }$

$\large\underline{\bold{Solution - }}$

Given that

$\rm :\longmapsto\: y = \dfrac{tan^{-1}x-cot^{-1}x }{tan^{-1}x+cot^{-1}x }$

$\rm :\longmapsto\: y = \dfrac{tan^{-1}x- \bigg(\dfrac{\pi}{2} - tan^{-1}x \bigg)}{\dfrac{\pi}{2} }$

$\rm :\longmapsto\:y = \dfrac{2}{\pi} \bigg(2 {tan}^{ - 1}x - \dfrac{\pi}{2} \bigg)$

$\rm :\longmapsto\:y \: = \: \dfrac{4}{\pi} {tan}^{ - 1} x - 1$

• Now, Differentiate w. r. t. x, we get

$\rm :\implies\:\dfrac{dy}{dx} = \dfrac{4}{\pi} \dfrac{d}{dx} {tan}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{4}{\pi} \times \dfrac{1}{1 + {x}^{2} }$

$\rm :\implies\:\dfrac{dy}{dx} = \dfrac{4}{\pi \: (1 + {x}^{2}) }$

• Now, put x = - 1, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{4}{\pi \: (1 + 1)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{4}{2\pi \: }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ \: 2 \: }{ \: \pi \:}$

Additional Information :-

$(1). \: \boxed{ \bf \: \dfrac{d}{dx} {tan}^{ - 1}x = \dfrac{1}{1 + {x}^{2} } }$

$(2). \: \boxed{ \bf \: \dfrac{d}{dx} {cot}^{ - 1}x = - \: \dfrac{1}{1 + {x}^{2} } }$

$(3). \: \boxed{ \bf \: \dfrac{d}{dx} {sin}^{ - 1}x = \dfrac{1}{ \sqrt{1 - {x}^{2} } } }$

$(4). \: \boxed{ \bf \: \dfrac{d}{dx} {cos}^{ - 1}x = \dfrac{ - \: 1}{ \sqrt{1 - {x}^{2} } } }$

$(5). \: \boxed{ \bf \: \dfrac{d}{dx} {sec}^{ - 1}x = \dfrac{1}{ x\sqrt{{x}^{2} - 1 } } }$

$(6). \: \boxed{ \bf \: \dfrac{d}{dx} {cosec}^{ - 1}x = \dfrac{ - \: 1}{ x\sqrt{{x}^{2} - 1 } } }$