If y - x2 - x - 1x2 - x - 1- then x2 - x - 12y1 -

The correct option is A 2(1 x^2) y = x^2 + x + 1x^2 x 1 y' = ( x^2 x 1)( 2x + 1) ( 2x 1)( x^2 + x + 1)( x^2 x 1)^2 y' = ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

If y = x2 +...

Question

If $y = \frac{x^{2} + x + 1}{x^{2} - x - 1}$ , then $(x^{2} - x - 1)^{2} y_{1} =$

2 (1 - x^{2})

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1 - x^{2}

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x^{2} - 1

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2 (x - 1)

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Solution

The correct option is A $2 (1 - x^{2})$
$y = \frac{x^{2} + x + 1}{x^{2} - x - 1}$
$y^{'} = \frac{(x^{2} - x - 1) (2 x + 1) - (2 x - 1) (x^{2} + x + 1)}{{(x^{2} - x - 1)}^{2}}$
$y^{'} = \frac{- 2 x (x + 1)}{{(x^{2} - x - 1)}^{2}}$
$(x^{2} - x - 1) y^{'} = - 2 x (x + 1) = 2 (1 - x^{2})$

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If y=x2+x+1x2−x−1, then (x2−x−1)2y1=

The correct option is A 2(1−x2)y=x2+x+1x2−x−1y′=(x2−x−1)(2x+1)−(2x−1)(x2+x+1)(x2−x−1)2y′=−2x(x+1)(x2−x−1)2(x2−x−1)y′=−2x(x+1)=2(1−x2)

If $y = \frac{x^{2} + x + 1}{x^{2} - x - 1}$ , then $(x^{2} - x - 1)^{2} y_{1} =$

The correct option is A $2 (1 - x^{2})$
$y = \frac{x^{2} + x + 1}{x^{2} - x - 1}$
$y^{'} = \frac{(x^{2} - x - 1) (2 x + 1) - (2 x - 1) (x^{2} + x + 1)}{{(x^{2} - x - 1)}^{2}}$
$y^{'} = \frac{- 2 x (x + 1)}{{(x^{2} - x - 1)}^{2}}$
$(x^{2} - x - 1) y^{'} = - 2 x (x + 1) = 2 (1 - x^{2})$