Question

If $\sf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2})$ , then show that $\sf \dfrac{dy}{dx} = \dfrac{cos^{-1}x}{(1-x^2)^{\frac{3}{2}}}$

Continuity and Differentiability, Class 12 ​

Answer 1

REFER TO THE ATTACHMENT!

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Answer 2

GIVEN :–

$\\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2})$

TO PROVE :–

$\\ \implies\bf \dfrac{dy}{dx} = \dfrac{cos^{-1}x}{(1-x^2)^{\frac{3}{2}}}$

SOLUTION :–

$\\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2})$

• We should write this as –

$\\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log( \{1 - x^2 \}^{ \frac{1}{2} } )$

$\\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - \dfrac{1}{2} log(1 - x^2) \: \: \: \: \: \: \: \: \: [ \: \because \: \: log( {a}^{b} ) = b log(a) ]$

$\\ \implies \bf y =cos^{-1}x. \left( \dfrac{x}{\sqrt{1 - x^2}} \right)- \dfrac{1}{2} log(1 - x^2)$

• Now Differentiate with respect to 'x' –

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

• Using identity –

$\\ \implies \bf \dfrac{d(u.v)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

• And –

$\\ \implies \bf \dfrac{d \left( \dfrac{u}{v} \right)}{dx} = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx} }{ {v}^{2} }$

• So that –

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

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

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

$\\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \dfrac{ {( \sqrt{1 - {x}^{2} } )}^{2} + {x}^{2} }{ \sqrt{1 - {x}^{2} } } }{1 - x^2}\right) + \left( \dfrac{0}{1 - x^2}\right)$

$\\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \dfrac{ {1 - {x}^{2} }+ {x}^{2}}{ \sqrt{1 - {x}^{2} } } }{1 - x^2}\right) + 0$

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

$\\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{1}{(1 - x^2 )^{ \frac{3}{2} } }\right)$

$\\ \implies \bf \dfrac{dy}{dx} =\dfrac{cos^{-1}x}{(1 - x^2 )^{ \frac{3}{2} } }$

$\\ \bf \:\:\: {\underline {\underline { \bf Hence \:\: Proved }}$