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Derivative First Principles Simplified Revision Notes

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Derivative First Principles

Introduction

  • Limit: A limit in calculus signifies the value a function approaches as the input draws near a specific point. Limits are vital for deriving derivatives, which measure instantaneous rates of change. For example, imagine a car slowing gradually to a stop at traffic lights—its speed reducing to zero akin to a function approaching a limit.

  • Derivative: The gradient of the tangent line to the curve of a function at a specific point, serving as a bridge between algebra and geometry.

infoNote

Real-world Applications:

  • Derivatives are used to determine rates of change and to optimise systems in fields such as engineering and economics.

The Derivative Quotient

  • Formal Definition: limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

  • Significance: Used to predict trends in graphs, model biological processes, and measure athletic acceleration.

  • Misconceptions about Limits:

    • Limits are dynamic entities, not static.
    • Example: Evaluate limx2x24x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}, even though it is undefined at x=2x=2.
chatImportant

Key Connection: A solid understanding of limits is essential for comprehending broader calculus topics such as continuity and integrals.

Finding Derivatives from First Principles

  • Utilise the limit definition of a derivative: limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} to find the derivative of function: f(x)=x2f(x) = x^2.

Step-by-Step Derivation Example

  • Initial Setup:

    • Expression: f(x+h)=(x+h)2f(x+h) = (x+h)^2.
  • Expand & Simplify:

    • Expand: (x+h)2=x2+2xh+h2(x+h)^2 = x^2 + 2xh + h^2.
    • Simplify: (x2+2xh+h2)x2h=2xh+h2h=2x+h\frac{(x^2 + 2xh + h^2) - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h.
  • Apply the Limit:

    • Result: f(x)=limh0(2x+h)=2xf'(x) = \lim_{h \to 0} (2x + h) = 2x.
chatImportant

Pitfalls to Avoid:

  • Missteps in expansion or limit application.
  • Verify each step of your calculation thoroughly.

Numerical and Graphical Illustrations

Numerical

  • Approach: Numerical calculations of f(x+h)f(x)h\frac{f(x+h)-f(x)}{h} become more accurate as hh decreases, providing improved approximations.

    Table demonstrating improvements in numerical derivative approximations with smaller h.

Graphical

  • Visualise: Secant lines gradually converge to tangent lines as h0h \to 0.

    Graphical illustration of secant lines becoming tangent by reducing h.

Numerical Estimation

  • Numerical Methods: Provide approximations for derivatives when analytical solutions are not feasible.
  • Finite Difference Methods:
    • Forward: f(x)f(x+h)f(x)hf'(x) \approx \frac{f(x+h) - f(x)}{h}.
    • Backward: f(x)f(x)f(xh)hf'(x) \approx \frac{f(x) - f(x-h)}{h}.
    • Central: f(x)f(x+h)f(xh)2hf'(x) \approx \frac{f(x+h) - f(x-h)}{2h}.
    Interactive graph showing comparison of finite difference methods with pros and cons.

Error Analysis

  • Types:

    • Truncation errors arise from approximations.
    • Round-off errors result from the finite precision of computing.
    • Mitigation: Careful selection of interval sizes and strategic error reduction methods.

    Flowchart illustrating types of numerical errors such as truncation and round-off.

Practice Problems with Solutions

Problem 1: Derive f(x)=x3f(x) = x^3 from first principles.
Solution:

  1. Begin with the definition: f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
  2. Substitute f(x)=x3f(x) = x^3: f(x)=limh0(x+h)3x3hf'(x) = \lim_{h \to 0} \frac{(x+h)^3-x^3}{h}
  3. Expand (x+h)3=x3+3x2h+3xh2+h3(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
  4. Simplify: f(x)=limh0x3+3x2h+3xh2+h3x3h=limh03x2h+3xh2+h3h=limh0(3x2+3xh+h2)f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2)
  5. Apply the limit: f(x)=3x2f'(x) = 3x^2

Problem 2: Use the limit definition to derive f(x)=1xf(x) = \frac{1}{x}. Solution:

  1. Begin with the definition: f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
  2. Substitute f(x)=1xf(x) = \frac{1}{x}: f(x)=limh01x+h1xhf'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}
  3. Find a common denominator: f(x)=limh0x(x+h)x(x+h)h=limh0hx(x+h)hf'(x) = \lim_{h \to 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}
  4. Simplify: f(x)=limh01x(x+h)=1x2f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x^2}
  5. Therefore, f(x)=1x2f'(x) = -\frac{1}{x^2}

Animation showing secant lines becoming tangent lines.

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