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Implicit Differentiation

Implicit differentiation is used when x and y are related by an equation and y is not first written explicitly as a function of x.

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Student-friendly explanation

Differentiate both sides with respect to x, treating y as a function of x. Every derivative of a y-term must include dy/dx through the chain rule. After differentiating, collect all dy/dx terms and solve for dy/dx.

How to write this in exams

  1. 1

    Start with the exact idea

    Implicit differentiation is used when x and y are related by an equation and y is not first written explicitly as a function of x.

  2. 2

    Then show how to use it

    Differentiate every term with respect to x. Use product rule where x and y are multiplied. Use chain rule for powers or functions of y. Bring all dy/dx terms to one side. Factor dy/dx and divide by its coefficient.

  3. 3

    Add one concrete example

    For x^2 + y^2 = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

  4. 4

    Avoid this incomplete answer

    Forgetting product rule in d/dx(xy), and writing it as x dy/dx only instead of x dy/dx + y.

Definition

Implicit differentiation is used when x and y are related by an equation and y is not first written explicitly as a function of x.

Example

For x^2 + y^2 = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

Rule to remember

Key condition: y is treated as a differentiable function of x. Product rule may be needed for terms like xy. Chain rule gives d/dx[y^n] = ny^(n-1) dy/dx.

Memory hook

A y differentiated with respect to x carries dy/dx.

Examples and method

Worked example

Find dy/dx for x^2y + y^3 = 10. Differentiate with respect to x: d/dx(x^2y) + d/dx(y^3) = 0. Using product rule, x^2(dy/dx) + 2xy + 3y^2(dy/dx) = 0. Collect terms: (x^2 + 3y^2)dy/dx = -2xy. Therefore dy/dx = -2xy/(x^2 + 3y^2).

Method to apply

Differentiate every term with respect to x. Use product rule where x and y are multiplied. Use chain rule for powers or functions of y. Bring all dy/dx terms to one side. Factor dy/dx and divide by its coefficient.

Diagram support

A curve may be represented by the implicit equation, but the exam calculation does not require drawing it unless specifically asked.

How CBSE asks it

Questions give equations such as x^2 + y^2 = a^2, xy = c, or expressions with powers and products, then ask for dy/dx.

Avoid common mistakes

Common confusion

Students differentiate y^2 as 2y instead of 2y dy/dx, which ignores that y depends on x.

Common wrong answer

Forgetting product rule in d/dx(xy), and writing it as x dy/dx only instead of x dy/dx + y.

Exam tip

Whenever differentiating a y-term with respect to x, attach dy/dx unless the term is a constant.

Quick check

Find dy/dx if xy + y^2 = 3.

Differentiate: x(dy/dx) + y + 2y(dy/dx) = 0. Hence dy/dx = -y/(x + 2y).

Answer writing and exam use

1-mark answer

Implicit differentiation is used when x and y are related by an equation and y is not first written explicitly as a function of x.

2-mark answer

Implicit differentiation is used when x and y are related by an equation and y is not first written explicitly as a function of x. Key condition: y is treated as a differentiable function of x. Product rule may be needed for terms like xy. Chain rule gives d/dx[y^n] = ny^(n-1) dy/dx. For x^2 + y^2 = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

3-mark answer

Differentiate both sides with respect to x, treating y as a function of x. Every derivative of a y-term must include dy/dx through the chain rule. After differentiating, collect all dy/dx terms and solve for dy/dx. Key condition: y is treated as a differentiable function of x. Product rule may be needed for terms like xy. Chain rule gives d/dx[y^n] = ny^(n-1) dy/dx. Find dy/dx for x^2y + y^3 = 10. Differentiate with respect to x: d/dx(x^2y) + d/dx(y^3) = 0. Using product rule, x^2(dy/dx) + 2xy + 3y^2(dy/dx) = 0. Collect terms: (x^2 + 3y^2)dy/dx = -2xy. Therefore dy/dx = -2xy/(x^2 + 3y^2). Questions give equations such as x^2 + y^2 = a^2, xy = c, or expressions with powers and products, then ask for dy/dx. Forgetting product rule in d/dx(xy), and writing it as x dy/dx only instead of x dy/dx + y.
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