Exams Joint Entrance Examination Mathematics Differentiation Question 6 Question 6 of 67 Medium MCQ
If log e y = 3 sin − 1 x {\log }_{e}y=3{\sin }^{-1}x log e y = 3 sin − 1 x , then ( 1 − x 2 ) y ′ ′ − x y ′ (1-{x}^{2}){y}^{''}-x{y}^{'} ( 1 − x 2 ) y ′′ − x y ′ at x = 1 2 x=\frac{1}{2} x = 2 1 is equal to
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log e y = 3 sin − 1 x y = e 3 sin − 1 x d y d x = e 3 sin − 1 x ⋅ 3 1 − x 2 \begin{gathered} {\log }_{e}y=3{\sin }^{-1}x \\ y={e}^{3{\sin }^{-1}x} \\ \frac{dy}{dx}={e}^{3{\sin }^{-1}x}\cdot \frac{3}{\sqrt{1-{x}^{2}}} \end{gathered} log e y = 3 sin − 1 x y = e 3 s i n − 1 x d x d y = e 3 s i n − 1 x ⋅ 1 − x 2 3
1 − x 2 d y d x = 3 y \sqrt{1-{x}^{2}}\frac{dy}{dx}=3y 1 − x 2 d x d y = 3 y
Again differentiate
1 − x 2 ⋅ y ′ ′ − 2 x 2 1 − x 2 y ′ = 3 y ′ ( 1 − x ) 2 y ′ ′ − x y ′ = 3 y ′ ( 1 − x 2 ) \begin{gathered} \sqrt{1-{x}^{2}}\cdot {y}^{''}-\frac{2x}{2\sqrt{1-{x}^{2}}}{y}^{'}=3{y}^{'} \\ (1-x{)}^{2}{y}^{''}-x{y}^{'}=3{y}^{'}(\sqrt{1-{x}^{2}}) \end{gathered} 1 − x 2 ⋅ y ′′ − 2 1 − x 2 2 x y ′ = 3 y ′ ( 1 − x ) 2 y ′′ − x y ′ = 3 y ′ ( 1 − x 2 )
So value of 3 y ′ ( 1 − x 2 ) 3{y}^{'}(\sqrt{1-{x}^{2}}) 3 y ′ ( 1 − x 2 ) at x = 1 2 x=\frac{1}{2} x = 2 1
3 ⋅ 3 1 − x 2 e sin − 1 x ( 1 − x 2 ) = 9 e 3 π 6 = 9 e π 2 \begin{gathered} 3\cdot \frac{3}{\sqrt{1-{x}^{2}}}{e}^{{\sin }^{-1}x}(\sqrt{1-{x}^{2}}) \\ =9{e}^{3\frac{\pi }{6}}=9{e}^{\frac{\pi }{2}} \end{gathered} 3 ⋅ 1 − x 2 3 e s i n − 1 x ( 1 − x 2 ) = 9 e 3 6 π = 9 e 2 π Answer the question to unlock