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    Math Deficiency – II
    MD-002
    Progress0 / 32 topics
    Topics
    1. Complex Numbers2. Arithmetic with Complex Numbers (Add, subtract, multiply and divide complex numbers)3. Trigonometric Polar Form of Complex Numbers4. De Moivre's Theorem and nth Roots5. Recursion6. Sequences and Series7. Sigma Notation8. Arithmetic Series9. Geometric Series (Sum infinite and finite geometric series and categorize geometric series)10. Counting with Permutations and Combinations11. Basic Probability12. Binomial Theorem13. Limit: Notation, Graphs to Find Limits, Tables to Find Limits14. Substitution to Find Limits, Rationalization to Find Limits15. One Sided Limits and Continuity16. Rate of Change: Instantaneous Rate of Change17. Tangent Lines and Rates of Change18. Derivatives: The Derivative Function19. Introduction to Techniques of Differentiation20. The Product and Quotient Rules21. Derivatives of Trigonometric Functions22. The Chain Rule23. Derivatives of Logarithmic Functions24. Derivatives of Exponential and Inverse Trigonometric Functions25. Increase, Decrease, and Concavity26. Relative Extrema, Absolute Maxima and Minima27. Integrals: An Overview of the Area Problem28. Area Under a Curve29. The Indefinite Integral30. Integration by Substitution31. The Definition of Area as a Limit; Sigma Notation32. The Definite Integral
    MD-002›The Product and Quotient Rules
    Math Deficiency – IITopic 20 of 32

    The Product and Quotient Rules

    13 minread
    2,196words
    Intermediatelevel

    The Product and Quotient Rules

    The Product Rule and the Quotient Rule are two important differentiation techniques used when dealing with the derivative of products and quotients of functions, respectively. These rules simplify the process of differentiating complex expressions that involve multiplication or division of functions.


    1. The Product Rule

    The Product Rule is used when differentiating the product of two functions. If you have a function f(x)f(x)f(x) that is the product of two functions, say g(x)g(x)g(x) and h(x)h(x)h(x), the derivative of f(x)=g(x)⋅h(x)f(x) = g(x) \cdot h(x)f(x)=g(x)⋅h(x) is given by:

    f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)

    Explanation:

    • The derivative of the product of two functions is the derivative of the first function, g′(x)g'(x)g′(x), multiplied by the second function, h(x)h(x)h(x), plus the first function, g(x)g(x)g(x), multiplied by the derivative of the second function, h′(x)h'(x)h′(x).
    • This rule ensures that you account for both the rate of change of the first function and the second function.

    Example:

    Let’s say we have the function f(x)=(x2+1)(x3−2)f(x) = (x^2 + 1)(x^3 - 2)f(x)=(x2+1)(x3−2), and we want to differentiate it.

    1. Identify the functions:

      • g(x)=x2+1g(x) = x^2 + 1g(x)=x2+1
      • h(x)=x3−2h(x) = x^3 - 2h(x)=x3−2
    2. Find the derivatives:

      • g′(x)=2xg'(x) = 2xg′(x)=2x
      • h′(x)=3x2h'(x) = 3x^2h′(x)=3x2
    3. Apply the Product Rule:

      f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x) f′(x)=(2x)(x3−2)+(x2+1)(3x2)f'(x) = (2x)(x^3 - 2) + (x^2 + 1)(3x^2)f′(x)=(2x)(x3−2)+(x2+1)(3x2)
    4. Simplify:

      f′(x)=2x(x3−2)+3x2(x2+1)f'(x) = 2x(x^3 - 2) + 3x^2(x^2 + 1)f′(x)=2x(x3−2)+3x2(x2+1) f′(x)=2x4−4x+3x4+3x2f'(x) = 2x^4 - 4x + 3x^4 + 3x^2f′(x)=2x4−4x+3x4+3x2 f′(x)=5x4+3x2−4xf'(x) = 5x^4 + 3x^2 - 4xf′(x)=5x4+3x2−4x

    So, the derivative of f(x)=(x2+1)(x3−2)f(x) = (x^2 + 1)(x^3 - 2)f(x)=(x2+1)(x3−2) is:

    f′(x)=5x4+3x2−4xf'(x) = 5x^4 + 3x^2 - 4xf′(x)=5x4+3x2−4x

    2. The Quotient Rule

    The Quotient Rule is used when differentiating the quotient (division) of two functions. If you have a function f(x)f(x)f(x) that is the quotient of two functions, say g(x)g(x)g(x) and h(x)h(x)h(x), the derivative of f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}f(x)=h(x)g(x)​ is given by:

    f′(x)=h(x)⋅g′(x)−g(x)⋅h′(x)[h(x)]2f'(x) = \frac{h(x) \cdot g'(x) - g(x) \cdot h'(x)}{[h(x)]^2}f′(x)=[h(x)]2h(x)⋅g′(x)−g(x)⋅h′(x)​

    Explanation:

    • The derivative of the quotient is found by taking the derivative of the numerator g(x)g(x)g(x), multiplying it by the denominator h(x)h(x)h(x), minus the numerator g(x)g(x)g(x) multiplied by the derivative of the denominator h′(x)h'(x)h′(x), all divided by the square of the denominator, [h(x)]2[h(x)]^2[h(x)]2.
    • The key idea here is that you have to account for the way both the numerator and denominator change as xxx changes.

    Example:

    Let’s differentiate f(x)=x2+1x3−2xf(x) = \frac{x^2 + 1}{x^3 - 2x}f(x)=x3−2xx2+1​.

    1. Identify the functions:

      • g(x)=x2+1g(x) = x^2 + 1g(x)=x2+1
      • h(x)=x3−2xh(x) = x^3 - 2xh(x)=x3−2x
    2. Find the derivatives:

      • g′(x)=2xg'(x) = 2xg′(x)=2x
      • h′(x)=3x2−2h'(x) = 3x^2 - 2h′(x)=3x2−2
    3. Apply the Quotient Rule:

      f′(x)=h(x)⋅g′(x)−g(x)⋅h′(x)[h(x)]2f'(x) = \frac{h(x) \cdot g'(x) - g(x) \cdot h'(x)}{[h(x)]^2}f′(x)=[h(x)]2h(x)⋅g′(x)−g(x)⋅h′(x)​ f′(x)=(x3−2x)(2x)−(x2+1)(3x2−2)(x3−2x)2f'(x) = \frac{(x^3 - 2x)(2x) - (x^2 + 1)(3x^2 - 2)}{(x^3 - 2x)^2}f′(x)=(x3−2x)2(x3−2x)(2x)−(x2+1)(3x2−2)​
    4. Simplify: First, expand the terms in the numerator:

      f′(x)=2x(x3−2x)−(x2+1)(3x2−2)(x3−2x)2f'(x) = \frac{2x(x^3 - 2x) - (x^2 + 1)(3x^2 - 2)}{(x^3 - 2x)^2}f′(x)=(x3−2x)22x(x3−2x)−(x2+1)(3x2−2)​ f′(x)=2x4−4x2−(3x4−2x2+3x2−2)(x3−2x)2f'(x) = \frac{2x^4 - 4x^2 - (3x^4 - 2x^2 + 3x^2 - 2)}{(x^3 - 2x)^2}f′(x)=(x3−2x)22x4−4x2−(3x4−2x2+3x2−2)​

      Simplifying further:

      f′(x)=2x4−4x2−3x4+5x2+2(x3−2x)2f'(x) = \frac{2x^4 - 4x^2 - 3x^4 + 5x^2 + 2}{(x^3 - 2x)^2}f′(x)=(x3−2x)22x4−4x2−3x4+5x2+2​ f′(x)=−x4+x2+2(x3−2x)2f'(x) = \frac{-x^4 + x^2 + 2}{(x^3 - 2x)^2}f′(x)=(x3−2x)2−x4+x2+2​

    So, the derivative of f(x)=x2+1x3−2xf(x) = \frac{x^2 + 1}{x^3 - 2x}f(x)=x3−2xx2+1​ is:

    f′(x)=−x4+x2+2(x3−2x)2f'(x) = \frac{-x^4 + x^2 + 2}{(x^3 - 2x)^2}f′(x)=(x3−2x)2−x4+x2+2​

    3. Key Points to Remember

    • Product Rule: Used for the derivative of a product of two functions.

      f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)
    • Quotient Rule: Used for the derivative of a quotient of two functions.

      f′(x)=h(x)⋅g′(x)−g(x)⋅h′(x)[h(x)]2f'(x) = \frac{h(x) \cdot g'(x) - g(x) \cdot h'(x)}{[h(x)]^2}f′(x)=[h(x)]2h(x)⋅g′(x)−g(x)⋅h′(x)​
    • Signs in the Quotient Rule: The difference in the numerator of the quotient rule means you subtract the product of the numerator function and the derivative of the denominator from the product of the denominator function and the derivative of the numerator.

    • Order of Application: Both rules involve applying the product or quotient of two functions, so the order of differentiation matters in both cases. Ensure you correctly identify which functions are the numerator and which are the denominator (in the quotient rule).


    4. Summary

    • The Product Rule helps differentiate products of two functions by adding the derivative of the first function times the second function and the first function times the derivative of the second function.
    • The Quotient Rule helps differentiate a quotient of two functions by applying the formula involving the derivatives of both the numerator and the denominator, and dividing by the square of the denominator.

    Both of these rules are critical when dealing with functions that involve multiplication or division, and they allow you to find the derivatives of more complex expressions with ease.

    Previous topic 19
    Introduction to Techniques of Differentiation
    Next topic 21
    Derivatives of Trigonometric Functions

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