Question

$d \div dx{ > \tan( { - 1}^)x \div \sqrt{a {}^{2} - x ^{2} }^{2} }$
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Answer 1

Answer:

$\bold{ \dfrac{1}{ \sqrt{ {a}^{2} - {x}^{2} } }}$

Step-by-step explanation:

$\bold{Correct \: question}\longrightarrow \\ derivative \: of \\ {tan}^{ - 1} ( \dfrac{x}{ \sqrt{ {a}^{2} - {x}^{2} } } )$

$\bold{Concept \: used}\longrightarrow \\ \dfrac{d}{dx} ( {sin}^{ - 1} x) = \dfrac{1}{ \sqrt{1 - {x}^{2} } } \\ \dfrac{d}{dx} (x) = 1$

$\bold{Solution}\longrightarrow \\ let \\ y = {tan}^{ - 1} (\dfrac{x}{ \sqrt{ {a}^{2} - {x}^{2} } } ) \\ let \: \: x = a \: sin \alpha \\ = > sin \alpha = \dfrac{x}{a} \\ = > \alpha = {sin}^{ - 1} ( \dfrac{x}{a} )$

$= > y = {tan}^{ - 1} \dfrac{a \: sin \alpha }{ \sqrt{ {a}^{2} - {(a \: sin \alpha )}^{2} } }$

$= > y = {tan}^{ - 1} \dfrac{a \: sin \alpha }{ \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} \alpha } }$

$= > y = {tan}^{ - 1} \dfrac{a \: sin \alpha }{ \sqrt{ {a}^{2} (1 - {sin}^{2} \alpha }) }$

$= > y = {tan}^{ - 1} ( \dfrac{a \: sin \alpha }{ \sqrt{ {a}^{2} \: {cos}^{2} \alpha } } )$

$= > y = {tan}^{ - 1} ( \dfrac{a \: sin \alpha }{a \: cos \alpha } )$

$= > y = {tan}^{ - 1} ( \dfrac{sin \alpha }{cos \alpha } )$

$= > y = {tan}^{ - 1} (tan \alpha )$

$= > y = \alpha$

$= > y = {sin}^{ - 1} ( \dfrac{x}{a} )$

$differentiating \: with \: respect \: to \: x$

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} ( {sin}^{ - 1} \dfrac{x}{a} )$

$= \dfrac{1}{ \sqrt{ 1 - {( \dfrac{x}{a} )}^{2} } } \dfrac{d}{dx} ( \dfrac{x}{a} )$

$= \dfrac{1}{ \sqrt{1 - \dfrac{ {x}^{2} }{ {a}^{2} } } } ( \dfrac{1}{a})$

$= \dfrac{1}{ \sqrt{ \dfrac{ {a}^{2} - {x}^{2} }{ {a}^{2} } } } ( \dfrac{1}{a} )$

$= \dfrac{a}{ \sqrt{ {a}^{2} - {x}^{2} } } ( \dfrac{1}{a} )$

$= \dfrac{1}{ \sqrt{ {a}^{2} - {x}^{2} } }$