Math, asked by mademonikalyani, 9 months ago


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

Answers

Answered by rishu6845
4

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}  } }

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