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NCERT Exemplar · Q70

Q.If y=1+1x21−1x2y = \dfrac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}, then dydx\dfrac{dy}{dx} is
(A) −4x(x2−1)2\dfrac{-4x}{(x^2 - 1)^2}
(B) −4xx2−1\dfrac{-4x}{x^2 - 1}
(C) 1−x24x\dfrac{1 - x^2}{4x}
(D) 4xx2−1\dfrac{4x}{x^2 - 1}

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Simplify the compound fraction first, then differentiate using the quotient rule; the derivative is dydx=−4x(x2−1)2\dfrac{dy}{dx} = \dfrac{-4x}{(x^2 - 1)^2}.

The expression looks intimidating with fractions nested inside fractions, but the smart move is to simplify before differentiating. Once we have a clean quotient, the quotient rule will give us the derivative efficiently.

The quotient rule states that for y=uvy = \dfrac{u}{v}, we have:

dydx=v⋅dudx−u⋅dvdxv2\dfrac{dy}{dx} = \dfrac{v \cdot \dfrac{du}{dx} - u \cdot \dfrac{dv}{dx}}{v^2}

The key insight: the numerator of the derivative involves the bottom times the derivative of the top minus the top times the derivative of the bottom.

Simplifying the expression

  1. Rewrite with a common denominator inside each fraction.

    The numerator is 1+1x2=x2+1x21 + \dfrac{1}{x^2} = \dfrac{x^2 + 1}{x^2}.

    The denominator is 1−1x2=x2−1x21 - \dfrac{1}{x^2} = \dfrac{x^2 - 1}{x^2}.

  2. Divide the two fractions.

    When dividing fractions, we multiply by the reciprocal:

y=x2+1x2x2−1x2=x2+1x2⋅x2x2−1=x2+1x2−1y = \dfrac{\frac{x^2 + 1}{x^2}}{\frac{x^2 - 1}{x^2}} = \dfrac{x^2 + 1}{x^2} \cdot \dfrac{x^2}{x^2 - 1} = \dfrac{x^2 + 1}{x^2 - 1}

Now we have a much cleaner form: y=x2+1x2−1y = \dfrac{x^2 + 1}{x^2 - 1}.

Applying the quotient rule

  1. Identify uu and vv.

    Let u=x2+1u = x^2 + 1 and v=x2−1v = x^2 - 1.

  2. Find the derivatives.

    dudx=2x\dfrac{du}{dx} = 2x

    dvdx=2x\dfrac{dv}{dx} = 2x …

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