Question

Solve the following

$y = {tan}^{ - 1} \frac{5x}{1 - {6x}^{2} } \: find \frac{dy}{dx}$
​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {tan}^{ - 1}\bigg[\dfrac{5x}{1 - {6x}^{2} } \bigg]$

can be rewritten as

$\rm :\longmapsto\:y = {tan}^{ - 1}\bigg[\dfrac{3x + 2x}{1 - (3x)(2x) } \bigg]$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {tan}^{ - 1}\bigg[\dfrac{x + y}{1 - xy} \bigg] = {tan}^{ - 1}x + {tan}^{ - 1}y \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:y = {tan}^{ - 1}3x + {tan}^{ - 1}2x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}[ {tan}^{ - 1}3x + {tan}^{ - 1}2x]$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}{tan}^{ - 1}x = \frac{1}{1 + {x}^{2} } \: }}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {(3x)}^{2} }\dfrac{d}{dx}(3x) + \dfrac{1}{1 + {(2x)}^{2} }\dfrac{d}{dx}(2x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{3}{1 + {9x}^{2} } + \dfrac{2}{1 + {4x}^{2} }$

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LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

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

Given function is

$</p><p>\rm :\longmapsto\:y = {tan}^{ - 1}\bigg[\dfrac{5x}{1 - {6x}^{2} } \bigg] \\ \\ can \: \: \: be \: \: \: rewritten \: \: \: as \\ \\ \rm :\longmapsto\:y = {tan}^{ - 1}\bigg[\dfrac{3x + 2x}{1 - (3x)(2x) } \bigg]$

We know,

$\begin{gathered}\purple{\rm :\longmapsto\:\boxed{\tt{ {tan}^{ - 1}\bigg[\dfrac{x + y}{1 - xy} \bigg] = {tan}^{ - 1}x + {tan}^{ - 1}y \: }}} \\ \end{gathered}$

So\:\:\:, using\:\:\: this \:\:\:identity,\:\:\: we\:\:\: get

$\rm :\longmapsto\:y = {tan}^{ - 1}3x + {tan}^{ - 1}2x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}[ {tan}^{ - 1}3x + {tan}^{ - 1}2x] \\ \\ We \: \: \: know, \\ \\ \begin{gathered}\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}{tan}^{ - 1}x = \frac{1}{1 + {x}^{2} } \: }}} \\ \end{gathered} \\ \\ So, using this result, we get \\ \\ \rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {(3x)}^{2} }\dfrac{d}{dx}(3x) + \dfrac{1}{1 + {(2x)}^{2} }\dfrac{d}{dx}(2x) \\ \\ \rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{3}{1 + {9x}^{2} } + \dfrac{2}{1 + {4x}^{2} }</p><p> </p><p>$

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