MATH10250MATH10430Chap2DiffImplicit

Introduction to Implicit Differentiation

• Implicit differentiation is used when differentiating equations where the variables are intertwined (not explicitly defined).

• For a function y = y(x):

  • The derivative of y² is given by the chain rule:

    • ( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} )

  • The derivative of ln(y) is:

    • ( \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} )

Key Examples of Implicit Differentiation

Example 1: Circle Equation

• Consider the circle equation: ( x^2 + y^2 = 1 )

• Differentiate both sides:

  • ( \frac{d}{dx}(x^2 + y^2) = 0 )

  • Leading to:
    - ( 2x + 2y \frac{dy}{dx} = 0 )

    • This simplifies to ( \frac{dy}{dx} = -\frac{x}{y} )

Example 2: Folium of Descartes

• Given the curve: ( x^3 + y^3 - 9xy = 0 )

• Differentiate implicitly:

  • ( 3x^2 + 3y^2 \frac{dy}{dx} - 9y - 9x \frac{dy}{dx} = 0 )

  • Rearranging gives:
    - ( (3y^2 - 9x) \frac{dy}{dx} = 9y - 3x^2 )

  • Solving for ( \frac{dy}{dx} ):

    • ( \frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} )

• At point (2, 4), substituting gives:
  - ( \frac{dy}{dx} = \frac{4}{5} )

• The tangent line's equation is derived as:

  • ( y - 4 = \frac{4}{5}(x - 2) )

  • Resulting in: ( 4x - 5y + 12 = 0 )

Example 3: Differentiation with Multiple Variables

• For an equation: ( 2u^2 + uv^2 - 3v^3 = 0 )

• Differentiate implicitly w.r.t. u:

  • ( 2 \frac{d}{du}(u^2) + \frac{d}{du}(uv^2) - 3 \frac{d}{du}(v^3) = 0 )

• Resolving derivatives provides:

  • ( 4u + 2uv \frac{dv}{du} + v^2 - 9v^2 \frac{dv}{du} = 0 )

  • Which can be simplified to find ( \frac{dv}{du} ):

  • ( \frac{dv}{du} = \frac{-4u + v^2}{2uv - 9v^2} )

Example 4: Speed and Acceleration

• Relation: ( e^s = s^2 - 3t^2 )

  • Differentiate w.r.t. time (t):

    • ( e^s \frac{ds}{dt} = 2s \frac{ds}{dt} - 6t )

• Rearranging gives:

  • ( (e^s - 2s) \frac{ds}{dt} = -6t )

    • Resulting in speed: ( \frac{ds}{dt} = \frac{-6t}{e^s - 2s} )

• For acceleration, differentiate ( \frac{ds}{dt} ) again:

  • Use product and quotient rules to derive: ( \frac{d^2s}{dt^2} ).

Exercises

• Exercise 1: Analyze the relation ( s^2 = ts + t^3 ) for speed and acceleration using implicit differentiation.

Notes:

• Implicit differentiation is a critical tool for dealing with complex relationships in calculus, particularly useful for engineers in scenarios where variables are interconnected in a relational format without explicit function representation.