Pure Mathematics · 3rd secondary · First Term

Calculus - Differentiation and its applications — Implicit and parametric differentiation

About this lesson

This unit covers implicit differentiation and parametric differentiation in calculus. It begins by contrasting explicit functions (y = f(x)) with implicit relations where x and y are mixed in an equation, such as xy + y - 4 = 0 or x² + y² = 9. The text explains that some implicit relations can be rewritten as explicit functions, while others, like a circle, define multiple functions or are difficult to solve for y. The method of implicit differentiation is introduced: differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule, then solve for dy/dx. Examples include polynomial and trigonometric implicit equations. The unit also covers parametric differentiation, where x and y are each expressed as functions of a parameter t (or θ). The derivative dy/dx is found by dividing dy/dt by dx/dt. Worked examples show how to compute dy/dx at specific parameter values, and how to find the derivative of one function with respect to another using parametric differentiation. The unit includes practice exercises and problems involving slopes of tangents and horizontal/vertical tangents.

Main topics in this lesson

  • Implicit differentiation
  • Parametric differentiation
  • Derivatives of implicit functions
  • Derivatives of parametric curves
  • Finding slopes of tangents
  • Derivative of one function with respect to another

Key terms and vocabulary

  • implicit differentiation
  • explicit function
  • implicit function
  • implicit relation
  • parametric differentiation
  • parameter
  • chain rule
  • horizontal tangent
  • vertical tangent

Questions and answers on Calculus - Differentiation and its applications — Implicit and parametric differentiation - practice and revision

Multiple choice questions on this lesson in Pure Mathematics for 3rd secondary, First Term, with practice exercises for revision and exam preparation.

1

إذا كانت x = θ - sin θ و y = 1 - cos θ، فإن dy/dx تُساوي:

  • 1 sin θ/(1 - cos θ)
  • 2 (1 - cos θ)/sin θ
  • 3 cos θ/(1 - sin θ)
  • 4 (1 - sin θ)/cos θ
2

إذا كانت العلاقة الضمنية x² + y² + 2x = 0، فإن dy/dx تُساوي:

  • 1 -(x+1)/y
  • 2 (x+1)/y
  • 3 -x/(y+1)
  • 4 x/(y+1)
3

إذا كانت x = t² + 1 و y = 2t، فإن قيمة dy/dx عند t = 1 تُساوي:

  • 1 1
  • 2 2
  • 3 1/2
  • 4 4
4

إذا كانت العلاقة الضمنية xy = 8، فإن dy/dx تُساوي:

  • 1 -y/x
  • 2 y/x
  • 3 -x/y
  • 4 x/y
5

إذا كانت x = 3cos θ و y = 3sin θ، فإن dy/dx تُساوي:

  • 1 -cot θ
  • 2 cot θ
  • 3 -tan θ
  • 4 tan θ
6

إذا كانت العلاقة الضمنية x² - y² = 16، فإن dy/dx تُساوي:

  • 1 x/y
  • 2 -x/y
  • 3 y/x
  • 4 -y/x

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