Differentiation

Welcome to KnowledgeKnot! Today, we’ll explore the concept of differentiation, its rules, applications, and examples. Share this with your peers and happy learning!

Definition of Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a quantity changes. It helps find the derivative of a function, representing the slope of the tangent to the curve at any point.

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

This formula provides the instantaneous rate of change of f(x)f(x) at xx.

Rules of Differentiation

Differentiation is governed by specific rules that simplify the computation of derivatives:

  • Power Rule:
    ddx[xn]=n⋅xn−1\frac{d}{dx} [x^n] = n \cdot x^{n-1}
    Example: ddx[x3]=3x2\frac{d}{dx} [x^3] = 3x^2
  • Sum Rule:
    ddx[f(x)+g(x)]=f′(x)+g′(x)\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)
  • Product Rule:
    ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
  • 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}
  • Chain Rule:
    ddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)

Examples of Differentiation

Example 1:

Differentiate f(x)=3x4+5x2−7f(x) = 3x^4 + 5x^2 - 7.

Solution:

f′(x)=ddx[3x4]+ddx[5x2]−ddx[7]f'(x) = \frac{d}{dx} [3x^4] + \frac{d}{dx} [5x^2] - \frac{d}{dx} [7]
f′(x)=12x3+10x−0f'(x) = 12x^3 + 10x - 0

So, the derivative is 12x3+10x12x^3 + 10x.

Example 2:

Differentiate f(x)=(x2+1)(x3−x)f(x) = (x^2 + 1)(x^3 - x) using the product rule.

Solution:

f′(x)=ddx[x2+1]⋅[x3−x]+[x2+1]⋅ddx[x3−x]f'(x) = \frac{d}{dx} [x^2 + 1] \cdot [x^3 - x] + [x^2 + 1] \cdot \frac{d}{dx} [x^3 - x]
f′(x)=(2x)(x3−x)+(x2+1)(3x2−1)f'(x) = (2x)(x^3 - x) + (x^2 + 1)(3x^2 - 1)
f′(x)=2x4−2x2+3x4−x2+3x2−1f'(x) = 2x^4 - 2x^2 + 3x^4 - x^2 + 3x^2 - 1
f′(x)=5x4−3x2−1f'(x) = 5x^4 - 3x^2 - 1

The derivative is 5x4−3x2−15x^4 - 3x^2 - 1.

Applications of Differentiation

Differentiation has extensive applications in various fields:

  • Physics: Calculating velocity and acceleration from position-time functions.
  • Economics: Determining marginal cost and marginal revenue functions.
  • Engineering: Analyzing stress and strain relationships in materials.
  • Biology: Modeling growth rates of populations or cells.

Practice Question

Problem:

Differentiate f(x)=x2+1x+2f(x) = \frac{x^2 + 1}{x + 2}.

Solution:

  1. Use the 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}
  2. Here, f(x)=x2+1f(x) = x^2 + 1 and g(x)=x+2g(x) = x + 2:
    f′(x)=2x,  g′(x)=1f'(x) = 2x, \; g'(x) = 1
  3. Substitute into the formula:
    ddx[x2+1x+2]=(2x)(x+2)−(x2+1)(1)(x+2)2\frac{d}{dx} \left[\frac{x^2 + 1}{x + 2}\right] = \frac{(2x)(x+2) - (x^2+1)(1)}{(x+2)^2}
    =2x2+4x−x2−1(x+2)2= \frac{2x^2 + 4x - x^2 - 1}{(x+2)^2}
    =x2+4x−1(x+2)2= \frac{x^2 + 4x - 1}{(x+2)^2}

The derivative is x2+4x−1(x+2)2\frac{x^2 + 4x - 1}{(x+2)^2}.

Suggetested Articles