Exams Joint Entrance Examination Mathematics Differentiation Question 15 Question 15 of 67 Medium MCQ
y = log e ( 1 − x 2 1 + x 2 ) , − 1 < x < 1 x = 1 2 225 ( y ′ − y ′ ′ ) y={\log }_{e}(\frac{1-{x}^{2}}{1+{x}^{2}}),-1<x<1x=\frac{1}{2}225({y}^{'}-{y}^{''}) y = log e ( 1 + x 2 1 − x 2 ) , − 1 < x < 1 x = 2 1 225 ( y ′ − y ′′ )
Solutions · 1 Post your solution Official solution
y = log e ( 1 − x 2 1 + x 2 ) d y d x = y ′ = − 4 x 1 − x 4 \begin{gathered} y={\log }_{e}(\frac{1-{x}^{2}}{1+{x}^{2}}) \\ \frac{dy}{dx}={y}^{'}=\frac{-4x}{1-{x}^{4}} \end{gathered} y = log e ( 1 + x 2 1 − x 2 ) d x d y = y ′ = 1 − x 4 − 4 x
Again,
d 2 y d x 2 = y ′ ′ = − 4 ( 1 + 3 x 4 ) ( 1 − x 4 ) 2 \frac{{d}^{2}y}{d{x}^{2}}={y}^{''}=\frac{-4(1+3{x}^{4})}{{(1-{x}^{4})}^{2}} d x 2 d 2 y = y ′′ = ( 1 − x 4 ) 2 − 4 ( 1 + 3 x 4 )
Again
y ′ − y ′ ′ = − 4 x 1 − x 4 + 4 ( 1 + 3 x 4 ) ( 1 − x 4 ) 2 {y}^{'}-{y}^{''}=\frac{-4x}{1-{x}^{4}}+\frac{4(1+3{x}^{4})}{{(1-{x}^{4})}^{2}} y ′ − y ′′ = 1 − x 4 − 4 x + ( 1 − x 4 ) 2 4 ( 1 + 3 x 4 )
at x = 1 2 x=\frac{1}{2} x = 2 1 ,
y ′ − y ′ ′ = 736 225 {y}^{'}-{y}^{''}=\frac{736}{225} y ′ − y ′′ = 225 736
Thus 225 ( y ′ − y ′ ′ ) = 225 × 736 225 = 736 225({y}^{'}-{y}^{''})=225\times \frac{736}{225}=736 225 ( y ′ − y ′′ ) = 225 × 225 736 = 736 Answer the question to unlock