Skip to content
Notely Maths 12
Class 12 Maths
🚨 This site is a work in progress. Exciting updates are coming soon!

2.4 Q-5

Question Statement

Differentiate the given expressions with respect to the specified variables.

Background and Explanation

To solve these problems, we use differentiation techniques such as the chain rule and the quotient rule. The chain rule helps us find the derivative of composite functions, while the quotient rule is applied when dealing with ratios of functions.

Solution

Part i: Differentiate x2−1x2x^2 - \frac{1}{x^2} w.r.t. x4x^4

  1. Assign Variables:

    Let:

y=x2−1x2(1)andu=x4(2). y = x^2 - \frac{1}{x^2} \quad \text{(1)} \quad \text{and} \quad u = x^4 \quad \text{(2)}.
  1. Find dydx\frac{dy}{dx}:
dydx=ddx(x2−1x2)=2x+2x3. \frac{dy}{dx} = \frac{d}{dx}\left(x^2 - \frac{1}{x^2}\right) = 2x + \frac{2}{x^3}.
  1. Find dudx\frac{du}{dx}:
dudx=ddx(x4)=4x3. \frac{du}{dx} = \frac{d}{dx}(x^4) = 4x^3.
  1. Apply the Chain Rule:
dydu=dydx⋅1dudx. \frac{dy}{du} = \frac{dy}{dx} \cdot \frac{1}{\frac{du}{dx}}.

Substitute:

dydu=(2x+2x3)⋅14x3. \frac{dy}{du} = \left(2x + \frac{2}{x^3}\right) \cdot \frac{1}{4x^3}.
  1. Simplify:
dydu=x4+12x4. \frac{dy}{du} = \frac{x^4 + 1}{2x^4}.

Answer: dydu=x4+12x4\frac{dy}{du} = \frac{x^4 + 1}{2x^4}.

Part ii: Differentiate (1+x2)n(1 + x^2)^n w.r.t. x2x^2

  1. Assign Variables:

    Let:

y=(1+x2)nandu=x2. y = (1 + x^2)^n \quad \text{and} \quad u = x^2.
  1. Find dydx\frac{dy}{dx}: Using the chain rule:
dydx=n(1+x2)n−1⋅ddx(1+x2)=n(1+x2)n−1⋅2x. \frac{dy}{dx} = n(1 + x^2)^{n-1} \cdot \frac{d}{dx}(1 + x^2) = n(1 + x^2)^{n-1} \cdot 2x.

Simplify:

dydx=2nx(1+x2)n−1. \frac{dy}{dx} = 2nx(1 + x^2)^{n-1}.
  1. Find dydu\frac{dy}{du}: Using dydu=dydx⋅1dudx\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{1}{\frac{du}{dx}}, and noting dudx=2x\frac{du}{dx} = 2x:
dydu=2nx(1+x2)n−12x. \frac{dy}{du} = \frac{2nx(1 + x^2)^{n-1}}{2x}.
  1. Simplify:
dydu=n(1+x2)n−1. \frac{dy}{du} = n(1 + x^2)^{n-1}.

Answer: dydu=n(1+x2)n−1\frac{dy}{du} = n(1 + x^2)^{n-1}.

Part iii: Differentiate x2+1x2−1\frac{x^2 + 1}{x^2 - 1} w.r.t. x−1x+1\frac{x-1}{x+1}

  1. Assign Variables:

    Let:

y=x2+1x2−1andu=x−1x+1. y = \frac{x^2 + 1}{x^2 - 1} \quad \text{and} \quad u = \frac{x-1}{x+1}.
  1. Find dydx\frac{dy}{dx}: Using the quotient rule:
dydx=(2x)(x2−1)−(2x)(x2+1)(x2−1)2. \frac{dy}{dx} = \frac{(2x)(x^2 - 1) - (2x)(x^2 + 1)}{(x^2 - 1)^2}.

Simplify:

dydx=−4x(x2−1)2. \frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2}.
  1. Find dudx\frac{du}{dx}: Using the quotient rule:
dudx=(1)(x+1)−(1)(x−1)(x+1)2. \frac{du}{dx} = \frac{(1)(x+1) - (1)(x-1)}{(x+1)^2}.

Simplify:

dudx=2(x+1)2. \frac{du}{dx} = \frac{2}{(x+1)^2}.
  1. Apply the Chain Rule:
dydu=dydx⋅dxdu. \frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du}.

Substitute:

dydu=−4x(x2−1)2⋅(x+1)22. \frac{dy}{du} = \frac{-4x}{(x^2 - 1)^2} \cdot \frac{(x+1)^2}{2}.
  1. Simplify:
dydu=−2x(x+1)2(x2−1)2. \frac{dy}{du} = \frac{-2x(x+1)^2}{(x^2 - 1)^2}.

Answer: dydu=−2x(x+1)2(x2−1)2\frac{dy}{du} = \frac{-2x(x+1)^2}{(x^2 - 1)^2}.

Key Formulas or Methods Used

  1. Quotient Rule:
ddx(f(x)g(x))=f′(x)g(x)−f(x)g′(x)[g(x)]2. \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.
  1. Chain Rule:
dydu=dydx⋅dxdu. \frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du}.

Summary of Steps

  1. Assign variables to simplify the functions.
  2. Differentiate both numerator and denominator separately.
  3. Apply the chain rule or quotient rule as needed.
  4. Simplify the expressions for clarity.