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Notely Maths 12
Class 12 Maths
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2.3 Q-13

Question Statement

Differentiate the following function with respect to xx:
y=x2+1x2−1y = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 - 1}}

Background and Explanation

This question involves differentiating a fraction of two square root functions. To solve it, we use:

  1. Quotient Rule: For derivatives of fractions.
  2. Chain Rule: To handle the square root functions.

Key Rules:

  1. Chain Rule:
    If f(x)=g(x)f(x) = \sqrt{g(x)}, then:
    ddxg(x)=12g(x)⋅dg(x)dx\frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}
  2. Quotient Rule:
    If f(x)=uvf(x) = \frac{u}{v}, then:
    ddx(uv)=v⋅dudx−u⋅dvdxv2\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

Solution

Step 1: Rewrite the function

Let: y=x2+1x2−1y = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 - 1}}

Here:

  • u=x2+1u = \sqrt{x^2 + 1}
  • v=x2−1v = \sqrt{x^2 - 1}

Step 2: Differentiate using the Quotient Rule

Using the quotient rule: dydx=v⋅dudx−u⋅dvdxv2\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

Step 2.1: Differentiate u=x2+1u = \sqrt{x^2 + 1}

Using the chain rule: dudx=12x2+1⋅ddx(x2+1)=12x2+1⋅2x=xx2+1\frac{du}{dx} = \frac{1}{2\sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x^2 + 1) = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}

Step 2.2: Differentiate v=x2−1v = \sqrt{x^2 - 1}

Using the chain rule: dvdx=12x2−1⋅ddx(x2−1)=12x2−1⋅2x=xx2−1\frac{dv}{dx} = \frac{1}{2\sqrt{x^2 - 1}} \cdot \frac{d}{dx}(x^2 - 1) = \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}}

Step 3: Substitute into the Quotient Rule

Now substitute back: dydx=x2−1⋅xx2+1−x2+1⋅xx2−1(x2−1)2\frac{dy}{dx} = \frac{\sqrt{x^2 - 1} \cdot \frac{x}{\sqrt{x^2 + 1}} - \sqrt{x^2 + 1} \cdot \frac{x}{\sqrt{x^2 - 1}}}{(\sqrt{x^2 - 1})^2}

Step 4: Simplify the numerator

Simplify the terms in the numerator:

  1. Combine the fractions: xx2−1x2+1−xx2+1x2−1\frac{x \sqrt{x^2 - 1}}{\sqrt{x^2 + 1}} - \frac{x \sqrt{x^2 + 1}}{\sqrt{x^2 - 1}}

  2. Write as a single fraction: =x(x2−1)2−x(x2+1)2x2+1⋅x2−1= \frac{x (\sqrt{x^2 - 1})^2 - x (\sqrt{x^2 + 1})^2}{\sqrt{x^2 + 1} \cdot \sqrt{x^2 - 1}}

  3. Expand the squares: =x(x2−1)−x(x2+1)x2+1⋅x2−1= \frac{x (x^2 - 1) - x (x^2 + 1)}{\sqrt{x^2 + 1} \cdot \sqrt{x^2 - 1}}

  4. Simplify: =x3−x−x3−xx2+1⋅x2−1= \frac{x^3 - x - x^3 - x}{\sqrt{x^2 + 1} \cdot \sqrt{x^2 - 1}} =−2xx2+1⋅x2−1= \frac{-2x}{\sqrt{x^2 + 1} \cdot \sqrt{x^2 - 1}}

Step 5: Simplify the denominator

The denominator of the quotient rule simplifies to: (x2−1)2=x2−1(\sqrt{x^2 - 1})^2 = x^2 - 1

Thus, the derivative becomes: dydx=−2x(x2−1)3/2\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^{3/2}}

Key Formulas or Methods Used

  1. Quotient Rule:
    ddx(uv)=v⋅dudx−u⋅dvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
  2. Chain Rule for Square Roots:
    ddxg(x)=12g(x)⋅dg(x)dx\frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}

Summary of Steps

  1. Let y=x2+1x2−1y = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 - 1}} and identify uu and vv.
  2. Apply the quotient rule to differentiate yy.
  3. Use the chain rule to find dudx\frac{du}{dx} and dvdx\frac{dv}{dx}.
  4. Simplify the numerator by combining and reducing terms.
  5. Simplify the denominator to express the final derivative: dydx=−2x(x2−1)3/2\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^{3/2}}