Implicit Differentiation Class 12 Mathematics | Continuity and Differentiability

Understand it

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.

Example

For x² + y² = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

Formula and rule

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[yⁿ] = nyⁿ⁻¹ dy/dx.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Find dy/dx for x²y + y³ = 10.

1Differentiate with respect to x: d/dx(x²y) + d/dx(y³) = 0.
2Using product rule, x²(dy/dx) + 2xy + 3y²(dy/dx) = 0.
3Collect terms: (x² + 3y²)dy/dx = -2xy.
4Therefore dy/dx = -2xy/(x² + 3y²).

Method you can reuse

1Differentiate every term with respect to x.
2Use product rule where x and y are multiplied.
3Use chain rule for powers or functions of y.
4Bring all dy/dx terms to one side.
5Factor dy/dx and divide by its coefficient.

Visual tip

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

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Find dy/dx if xy + y² = 3.

Think first, then show answer

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

Common wrong answer

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

Don't mix this up

Students differentiate y² as 2y instead of 2y dy/dx, which ignores that y depends on x.

Exam tip

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

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Questions give equations such as x² + y² = a², xy = c, or expressions with powers and products, then ask for dy/dx.

1
Enough for 1 mark
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
Add for 2 marks
Example: For x² + y² = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.
3
Add for 3 marks
Differentiate both sides with respect to x, treating y as a function of x.

Use the concept graph around this idea instead of revising it in isolation.

Continuity at a Point

A function f is continuous at x = a if f(a) is defined, lim(x → a) f(x) exists, and lim(x → a) f(x) = f(a).

Differentiability and Continuity

A function f is differentiable at x = a if its left-hand derivative and right-hand derivative at a both exist and are equal.

Chain Rule for Composite Functions

The chain rule gives the derivative of a composite function f(g(x)) by differentiating the outer function at g(x) and multiplying by the derivative of g(x).

Derivatives of Inverse Trigonometric, Exponential, and Logarithmic Functions

This concept covers standard derivatives of inverse trigonometric functions, exponential functions, and logarithmic functions, along with their domain conditions.

Previous: Chain Rule for Composite Functions Next: Derivatives of Inverse Trigonometric, Exponential, and Logarithmic Functions Back to full mind map

Implicit Differentiation

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Check yourself

Write it in the exam

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Implicit Differentiation

Understand it

See it worked out

Check yourself

Write it in the exam

Related concepts

Practice this concept with focused MCQs

Help improve this page