Question

➟ Differentiate
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Answer 1

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

Given that

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{ {2}^{x +1} \: . \: {3}^{x} }{1 + {(36)}^{x} } \bigg)$

can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{ {2}^{x} \times 2 \: . \: {3}^{x} }{1 + {(6 \times 6)}^{x} } \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: {a}^{x + y} = {a}^{x} \times {a}^{y} }}$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{ {(2 \times 3)}^{x} \times 2 }{1 + {( {6}^{2} )}^{x} } \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: {(ab)}^{x} = {a}^{x} \times {b}^{x} }}$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{ {(6)}^{x} \times 2 }{1 + {( {6}^{x})}^{2} } \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: {( {a}^{y} )}^{x} = {( {a}^{x} )}^{y}}}$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{ 2 \times {(6)}^{x}}{1 + {( {6}^{x})}^{2} } \bigg)$

We know,

$\green{\boxed{ \bf\: {2tan}^{ - 1}x = {sin}^{ - 1} \: \dfrac{2x}{1 + {x}^{2} } }}$

Using this formula, we get

$\rm :\longmapsto\:y = 2 \: {tan}^{ - 1} ({6}^{x} )$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} \: 2 \: {tan}^{ - 1} ({6}^{x} )$

$\rm :\longmapsto\:\dfrac{dy}{dx}= 2 \: \dfrac{d}{dx} \: {tan}^{ - 1} ({6}^{x} )$

$\: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: \dfrac{d}{dx} \: k \: f(x) \: = \: k \: \dfrac{d}{dx} \: f(x)}}$

$\rm :\longmapsto\:\dfrac{dy}{dx}= 2 \: \times \dfrac{1}{1 + {( {6}^{x}) }^{2} } \: \dfrac{d}{dx} \: ({6}^{x} )$

$\: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: \dfrac{d}{dx} \: {tan}^{ - 1}x \: = \: \dfrac{1}{1 + {x}^{2} }}}$

$\rm :\longmapsto\:\dfrac{dy}{dx}= \dfrac{2}{1 + {( {36}^{x}) }} \: {6}^{x} \: log(6)$

$\: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: \dfrac{d}{dx} \: {a}^{ x} \: = \: {a}^{x} \: log(a) }}$

Hence,

$\red{\bf :\longmapsto\:\dfrac{dy}{dx}= \dfrac{2}{1 + {( {36}^{x}) }} \: {6}^{x} \: log(6)}$

Additional Information :-

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {sin}^{ - 1}x = \dfrac{1}{ \sqrt{1 - {x}^{2} } }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {cos}^{ - 1}x = \dfrac{ - \: 1}{ \sqrt{1 - {x}^{2} } }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {sec}^{ - 1}x = \dfrac{ \: 1}{ x\sqrt{{x}^{2} - 1 } }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1}x = \dfrac{ - \: 1}{ x\sqrt{{x}^{2} - 1 } }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k = 0}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = \: {sec}^{2} x}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = \: { - \: cosec}^{2} x}$