Question

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Answer 1

QUESTION :–

• Differentiate the following w.r.t. 'x' –

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$\bf \implies { \sin}^{ - 1} \left( \sqrt{ \dfrac{1 + {x}^{2} }{2}} \right)$

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ANSWER :–

GIVEN :–

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$\bf \implies { \sin}^{ - 1} \left( \sqrt{ \dfrac{1 + {x}^{2} }{2}} \right)$

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TO FIND :–

• Differentiate form = ?

SOLUTION :–

• Let the function –

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$\bf \implies y = { \sin}^{ - 1} \left( \sqrt{ \dfrac{1 + {x}^{2} }{2}} \right)$

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• Now Differentiate with respect to 'x' –

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$\bf \implies \dfrac{dy}{dx} = \dfrac{ d \left \{{ \sin}^{ - 1} \left( \sqrt{ \dfrac{1 + {x}^{2} }{2}} \right) \right \}}{dx}$

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• We know that –

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$\bf \implies \dfrac{ d \left \{{ \sin}^{ - 1}(x) \right \}}{dx} = \dfrac{1}{ \sqrt{1 + {x}^{2} } }$

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$\bf \implies \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 + { \left( \sqrt{\dfrac{1 + {x}^{2} }{2}} \right)}^{2} } } \times \dfrac{1}{2{{\left( \sqrt{\dfrac{1 + {x}^{2} }{2}} \right)}} } \times \dfrac{2x}{2}$

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$\bf \implies \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 + { \left( \sqrt{\dfrac{1 + {x}^{2} }{2}} \right)}^{2} } } \times \dfrac{x}{ \sqrt{2( 1 + {x}^{2} )} }$

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$\bf \implies \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 + \dfrac{1 + {x}^{2} }{2} } } \times \dfrac{x}{ \sqrt{2( 1 + {x}^{2} )} }$

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$\bf \implies \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{ \dfrac{2 + 1 +{x}^{2}}{2} } } \times \dfrac{x}{ \sqrt{2( 1 + {x}^{2} )} }$

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$\bf \implies \dfrac{dy}{dx} = \dfrac{ \sqrt{2} }{ \sqrt{3+{x}^{2}} } \times \dfrac{x}{ \sqrt{2} \sqrt{( 1 + {x}^{2} )} }$

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$\bf \implies \large { \boxed{ \bf\dfrac{dy}{dx} = \dfrac{x}{ \sqrt{( 1 + {x}^{2} )(3 + {x}^{2} )}}}}$