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Exploring Functions in Advanced Mathematics

Implicit Functions

Definition

Definition of Implicit Functions

A functional relationship determined by the equation F(x,y)=0F(x, y) = 0 is called an implicit function.

Characteristics

Implicit functions have the following characteristics:

  • Non-explicit: Cannot directly solve for an explicit expression of y in terms of x
  • Equation-determined: The functional relationship needs to be determined through the equation
  • Multi-valuedness: Multiple function values may correspond to the same independent variable value
  • Locality: Implicit functions are usually valid only in certain regions

Common Examples

Circle Equation

x2+y2=1x^2 + y^2 = 1 determines an implicit function, which can be solved as:

  • y=1−x2y = \sqrt{1 - x^2} (upper semicircle)
  • y=−1−x2y = -\sqrt{1 - x^2} (lower semicircle)

Ellipse Equation

x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 determines an implicit function

Other Examples

  • x3+y3=3xyx^3 + y^3 = 3xy (Folium of Descartes)
  • ex+ey=1e^x + e^y = 1
  • sin⁡(x+y)=x+y\sin(x + y) = x + y

Implicit Function Differentiation Methods

Basic Steps

  1. Differentiate both sides of the equation with respect to x
  2. Use the chain rule to handle the derivative of y
  3. Solve for dydx\frac{dy}{dx}

Specific Method

For the equation F(x,y)=0F(x, y) = 0:

  1. Differentiate both sides with respect to x: ∂F∂x+∂F∂y⋅dydx=0\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0

  2. Solve for dydx\frac{dy}{dx}: dydx=−∂F∂x∂F∂y\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}

Notes

  • Denominator not zero: ∂F∂y≠0\frac{\partial F}{\partial y} \neq 0
  • Locality: The result is valid only near certain points
  • Multi-valuedness: May need to consider multiple branches

Examples of Implicit Function Differentiation

Example 1: Derivative of Circle

For x2+y2=1x^2 + y^2 = 1:

  1. Differentiate both sides: 2x+2y⋅dydx=02x + 2y \cdot \frac{dy}{dx} = 0
  2. Solve for the derivative: dydx=−xy\frac{dy}{dx} = -\frac{x}{y}

Example 2: Derivative of Ellipse

For x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1:

  1. Differentiate both sides: 2xa2+2yb2⋅dydx=0\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0
  2. Solve for the derivative: dydx=−b2xa2y\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}

Exercises

Exercise 1

Find the derivative of the implicit function x2+y2=4x^2 + y^2 = 4 at the point (1,3)(1, \sqrt{3}).

implicit functiondifferentiation

Approach: Differentiate both sides of the equation with respect to x, then solve for dydx\frac{dy}{dx}.

Steps:

  1. Differentiate both sides of x2+y2=4x^2 + y^2 = 4 with respect to x: 2x+2y⋅dydx=02x + 2y \cdot \frac{dy}{dx} = 0
  2. Solve for dydx\frac{dy}{dx}: 2y⋅dydx=−2x2y \cdot \frac{dy}{dx} = -2x dydx=−xy\frac{dy}{dx} = -\frac{x}{y}
  3. Substitute the point (1,3)(1, \sqrt{3}): dydx=−13=−33\frac{dy}{dx} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}

Answer:

The derivative at point (1,3)(1, \sqrt{3}) is −33-\frac{\sqrt{3}}{3}.

Exercise 2

Find the derivative of the implicit function x3+y3=3xyx^3 + y^3 = 3xy.

implicit functiondifferentiation

Approach: Differentiate both sides of the equation with respect to x, then solve for dydx\frac{dy}{dx}.

Steps:

  1. Differentiate both sides of x3+y3=3xyx^3 + y^3 = 3xy with respect to x: 3x2+3y2⋅dydx=3y+3x⋅dydx3x^2 + 3y^2 \cdot \frac{dy}{dx} = 3y + 3x \cdot \frac{dy}{dx}
  2. Rearrange the equation: 3y2⋅dydx−3x⋅dydx=3y−3x23y^2 \cdot \frac{dy}{dx} - 3x \cdot \frac{dy}{dx} = 3y - 3x^2 (3y2−3x)⋅dydx=3y−3x2(3y^2 - 3x) \cdot \frac{dy}{dx} = 3y - 3x^2
  3. Solve for dydx\frac{dy}{dx}: dydx=3y−3x23y2−3x=y−x2y2−x\frac{dy}{dx} = \frac{3y - 3x^2}{3y^2 - 3x} = \frac{y - x^2}{y^2 - x}

Answer:

dydx=y−x2y2−x\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}.

Exercise 3

Find the derivative of the implicit function ex+ey=1e^x + e^y = 1.

implicit functiondifferentiation

Approach: Differentiate both sides of the equation with respect to x, then solve for dydx\frac{dy}{dx}.

Steps:

  1. Differentiate both sides of ex+ey=1e^x + e^y = 1 with respect to x: ex+ey⋅dydx=0e^x + e^y \cdot \frac{dy}{dx} = 0
  2. Solve for dydx\frac{dy}{dx}: ey⋅dydx=−exe^y \cdot \frac{dy}{dx} = -e^x dydx=−exey=−ex−y\frac{dy}{dx} = -\frac{e^x}{e^y} = -e^{x-y}

Answer:

dydx=−ex−y\frac{dy}{dx} = -e^{x-y}.

Summary

Symbols Appearing in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
F(x,y)F(x, y)Mathematical symbolF of x and yFunction of two variables
F(x,y)=0F(x, y) = 0Mathematical symbolF of x and y equals zeroImplicit function equation
dydx\frac{dy}{dx}Mathematical symboldy by dxDerivative of y with respect to x
∂F∂x\frac{\partial F}{\partial x}Mathematical symbolpartial F by partial xPartial derivative of FF with respect to xx
∂F∂y\frac{\partial F}{\partial y}Mathematical symbolpartial F by partial yPartial derivative of FF with respect to yy
exe^xMathematical symbole to the power xExponential function with base ee
eye^yMathematical symbole to the power yExponential function with base ee
sin⁡(x+y)\sin(x + y)Mathematical symbolsine of x plus ySine function
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Parametric Functions

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