MathsCBSE Class 12 · Continuity and Differentiability

Implicit Differentiation: Finding dy/dx Without Solving for y (Class 12)

Q: What is an implicit function?

A function y of x defined by an equation that connects x and y without giving y alone on one side, such as x² + xy + y² = 7. Implicit differentiation finds dy/dx directly from such an equation.

Q: Why do y-terms get a dy/dx when differentiated?

Because y depends on x. By the chain rule, the derivative of f(y) with respect to x is f′(y) times dy/dx.

Q: Can the answer contain both x and y?

Yes. That is normal for implicit differentiation. If a point on the curve is given, substitute both coordinates to get a number.

Q: Which NCERT section covers implicit differentiation?

Section 5.3.2 (Derivatives of Implicit Functions) of Chapter 5, Continuity and Differentiability, in NCERT Mathematics Part I for Class 12.

Q: Are tangents and normals needed with this topic?

No. Tangents and normals were dropped from the 2026-27 syllabus. Questions ask for dy/dx, its value at a point, or a proof involving it.

Updated 15 October 2026 · Fact-checked

In implicit differentiation you differentiate both sides of an equation in x and y with respect to x, treating y as a function of x. Every y-term picks up a factor dy/dx by the chain rule, for example d/dx(y²) = 2y dy/dx and d/dx(xy) = x dy/dx + y. Then collect the dy/dx terms and solve.

Notice that every y-term gets a y′ from the chain rule, then you collect the y′ terms, factor and divide to get the slope.

Illustration drawn for QuestionBuddy.

Understand Implicit Differentiation

Many curves are given by an equation that mixes $x$ and $y$ , such as $x^{2} + x y + y^{2} = 7$ or $x sin y + y cos x = π$ . Solving for $y$ is awkward or impossible, but we can still find the slope $\frac{d y}{d x}$ . The idea: near a point, the equation still defines $y$ as a function of $x$ , so differentiate the whole equation with respect to $x$ and let the chain rule handle $y$ .

The one rule to remember. If $f$ is any function, then $\frac{d}{d x} f (y) = f^{'} (y) \frac{d y}{d x}$ . So

$\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ , $\frac{d}{d x} (sin y) = cos y \frac{d y}{d x}$ , $\frac{d}{d x} (e^{y}) = e^{y} \frac{d y}{d x}$ ;
products need the product rule: $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$ , $\frac{d}{d x} (x^{2} y^{3}) = 2 x y^{3} + 3 x^{2} y^{2} \frac{d y}{d x}$ ;
composite expressions need the chain rule: $\frac{d}{d x} sin (x y) = cos (x y) (x \frac{d y}{d x} + y)$ .

The method. Differentiate each term on both sides; write $y^{'}$ for $\frac{d y}{d x}$ to save space. Move all $y^{'}$ terms to one side and everything else to the other, factor out $y^{'}$ , and divide. For $x^{2} + x y + y^{2} = 7$ : $2 x + (x y^{'} + y) + 2 y y^{'} = 0$ , so $y^{'} (x + 2 y) = - (2 x + y)$ and $y^{'} = - \frac{2 x + y}{x + 2 y}$ . The answer usually contains both $x$ and $y$ ; to get a number, substitute a point that lies on the curve.

Smart moves.

*Simplify first.* If the equation can be rearranged (for example, an infinite nested root $y = x + x + \dots$ becomes $y^{2} = x + y$ ), do it before differentiating.
*Substitute to remove roots.* Expressions like $1 - x^{2}$ and $1 - y^{2}$ suggest $x = sin A$ , $y = sin B$ ; a trigonometric identity can then turn the equation into something like $A + B = \frac{π}{2}$ , which differentiates easily using the derivative of $sin^{- 1}$ .
*Eliminate a parameter.* If two equations share a letter $t$ , combine them to remove $t$ , then differentiate implicitly.

Implicit differentiation also prepares you for logarithmic differentiation and second derivatives of implicit curves. It is NCERT Section 5.3.2 (Chapter 5) and has been the most frequently asked part of this chapter in recent board papers, as 2- and 3-mark questions and inside 5-mark ones.

Implicit Differentiation: formulas to remember

Chain rule for y-terms: $\frac{d}{d x} f (y) = f^{'} (y) \frac{d y}{d x}$; Every function of y picks up a factor $y^{'}$ .
Powers of y: $\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$; Do not forget the $\frac{d y}{d x}$ .
Product of x and y: $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$; Product rule; the $y$ term has no $y^{'}$ .
Trig and exponential of y: $\frac{d}{d x} sin y = cos y y^{'}, \frac{d}{d x} e^{y} = e^{y} y^{'}$; Same pattern for every function of y.
Derivatives of inverse trigonometric functions: $\frac{d}{d x} sin^{- 1} x = \frac{1}{1 - x ^{2}}, \frac{d}{d x} cos^{- 1} x = - \frac{1}{1 - x ^{2}}, \frac{d}{d x} tan^{- 1} x = \frac{1}{1 + x ^{2}}$; Often needed after a trigonometric substitution.
Implicit slope from F(x, y) = c: $\frac{d y}{d x} = - \frac{(terms from differentiating in x )}{(coefficient of y ^{'} )}$; Collect, factor and divide.

How to solve Implicit Differentiation questions

A five-step routine that works for every implicit-differentiation question.

1Simplify the equation if possible (clear fractions, square both sides of a nested root, or substitute $x = sin A$ for square roots like $1 - x^{2}$ ).
2Differentiate every term on both sides with respect to $x$ , using the product and chain rules, and attach $\frac{d y}{d x}$ to every derivative of a $y$ -expression.
3Collect all terms containing $\frac{d y}{d x}$ on one side and the rest on the other.
4Factor out $\frac{d y}{d x}$ and divide to get the formula.
5If a value is asked, check that the point lies on the curve, then substitute; if a result is to be proved, simplify using the original equation.

Implicit Differentiation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Implicit slope as a trade-off rate

Suppose a small factory's output is $Q = x y$ , where $x$ is labour hours and $y$ is machine hours. The manager wants to keep output fixed at $Q = 20$ while changing the mix.

Find $\frac{d y}{d x}$ along the curve $x y = 20$ , and its value when $x = 16$ . Interpret the sign and size.

Try it first, then show the solutionSolution

$x y = 20 \Rightarrow x y = 400$ .
Differentiate: $x \frac{d y}{d x} + y = 0$ , so $\frac{d y}{d x} = - \frac{y}{x}$ .
At $x = 16$ : $y = \frac{400}{16} = 25$ , so $\frac{d y}{d x} = - \frac{25}{16} \approx - 1.56$ .
Negative: more labour means fewer machine hours are needed. Size: one extra labour hour saves about 1.56 machine hours at this mix.

Answer: $\frac{d y}{d x} = - \frac{y}{x}$ ; at $(16, 25)$ it is $- \frac{25}{16}$ , so each extra labour hour replaces about 1.56 machine hours.

Memory hook: Keep output fixed, trade labour for machines: the rate is the implicit slope.

Slope of a curve given implicitly

Imagine a circular running track modelled by $x^{2} + y^{2} = 400$ (metres), with the centre at the origin. A runner is at the point $(12, 16)$ .

Find the slope of the track at the runner's position without solving for $y$ , and say in which direction $y$ changes as $x$ increases there.

Try it first, then show the solutionSolution

Differentiate: $2 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{x}{y}$ .
Check the point: $144 + 256 = 400$ .
At $(12, 16)$ : $\frac{d y}{d x} = - \frac{12}{16} = - \frac{3}{4}$ .
Negative slope: as $x$ increases, $y$ decreases along this part of the track.

Answer: Slope $= - \frac{3}{4}$ ; moving in the positive x-direction, the runner's y-coordinate falls by 0.75 m per metre.

Memory hook: On a circle, the slope is minus x over y.

Tricks to remember Implicit Differentiation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Every y wears a y′ badge
Whenever you differentiate something containing $y$ , pin a $y^{'}$ on it. No badge, no marks.
Use it when: Avoiding the most common slip.
Product rule for mixed terms
$x y$ is a product: 'first times derivative of second plus second times derivative of first' gives $x y^{'} + y$ .
Use it when: Terms like $x y$ , $x^{2} y$ , $x sin y$ .
Collect, factor, divide
After differentiating: all $y^{'}$ terms to the left, factor out $y^{'}$ , divide by the bracket.
Use it when: The final algebra in every question.
Square roots of 1 − x² call for sine
See $1 - x^{2}$ and $1 - y^{2}$ together? Put $x = sin A$ , $y = sin B$ .
Use it when: Proof-type implicit questions.

Implicit Differentiation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A quadratic curve · 2 marks · Very short answerBoard level

If $x^{2} - x y + 2 y^{2} = 8$ , find $\frac{d y}{d x}$ and its value at $(2, 2)$ .

Try it first, then show the solutionSolution

Differentiate: $2 x - (x \frac{d y}{d x} + y) + 4 y \frac{d y}{d x} = 0$ . (1 mark)
$(4 y - x) \frac{d y}{d x} = y - 2 x$ , so $\frac{d y}{d x} = \frac{y - 2 x}{4 y - x}$ . (½ mark)
$(2, 2)$ lies on the curve ( $4 - 4 + 8 = 8$ ); $\frac{d y}{d x} = \frac{2 - 4}{8 - 2} = - \frac{1}{3}$ . (½ mark)

Answer: $\frac{d y}{d x} = \frac{y - 2 x}{4 y - x}$ ; at $(2, 2)$ it is $- \frac{1}{3}$ .

Example 2: Trigonometric terms in x and y · 3 marks · Short answerBoard level

Find $\frac{d y}{d x}$ if $x sin y + y cos x = π$ .

Try it first, then show the solutionSolution

$\frac{d}{d x} (x sin y) = sin y + x cos y \frac{d y}{d x}$ . (1 mark)
$\frac{d}{d x} (y cos x) = cos x \frac{d y}{d x} - y sin x$ . (1 mark)
So $sin y + x cos y y^{'} + cos x y^{'} - y sin x = 0$ , giving $y^{'} (x cos y + cos x) = y sin x - sin y$ . (½ mark)
$\frac{d y}{d x} = \frac{y s i n x - s i n y}{x c o s y + c o s x}$ . (½ mark)

Answer: $\frac{d y}{d x} = \frac{y s i n x - s i n y}{x c o s y + c o s x}$

Example 3: An infinite nested root · 3 marks · Short answerHarder board level

If $y = x + x + x + \dots to \infty$ , prove that $\frac{d y}{d x} = \frac{1}{2 y - 1}$ .

Try it first, then show the solutionSolution

The expression under the first root is $x + y$ (the same infinite expression), so $y = x + y$ . (1 mark)
Square: $y^{2} = x + y$ . (½ mark)
Differentiate: $2 y \frac{d y}{d x} = 1 + \frac{d y}{d x}$ . (1 mark)
$(2 y - 1) \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} = \frac{1}{2 y - 1}$ . (½ mark)

Answer: $\frac{d y}{d x} = \frac{1}{2 y - 1}$

Example 4: A substitution that removes the roots · 3 marks · Short answerHarder board level

If $y 1 - x^{2} + x 1 - y^{2} = 1$ , where $0 < x, y < 1$ , prove that $\frac{d y}{d x} = - \frac{1 - y ^{2}}{1 - x ^{2}}$ .

Try it first, then show the solutionSolution

Put $x = sin A$ , $y = sin B$ with $A, B \in (0, \frac{π}{2})$ , so $1 - x^{2} = cos A$ and $1 - y^{2} = cos B$ . (1 mark)
The equation becomes $sin B cos A + sin A cos B = 1$ , i.e. $sin (A + B) = 1$ , so $A + B = \frac{π}{2}$ . (1 mark)
Hence $sin^{- 1} x + sin^{- 1} y = \frac{π}{2}$ . Differentiating: $\frac{1}{1 - x ^{2}} + \frac{1}{1 - y ^{2}} \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{1 - y ^{2}}{1 - x ^{2}}$ . (1 mark)

Answer: $\frac{d y}{d x} = - \frac{1 - y ^{2}}{1 - x ^{2}}$

Example 5: Eliminating a parameter first · 3 marks · Short answerHOTS / competency

If $x^{2} + y^{2} = t + \frac{1}{t}$ and $x^{4} + y^{4} = t^{2} + \frac{1}{t ^{2}}$ , prove that $x^{3} y \frac{d y}{d x} = - 1$ .

Try it first, then show the solutionSolution

Square the first equation: $x^{4} + y^{4} + 2 x^{2} y^{2} = t^{2} + 2 + \frac{1}{t ^{2}}$ . (1 mark)
Subtract the second: $2 x^{2} y^{2} = 2$ , so $x^{2} y^{2} = 1$ . (1 mark)
Differentiate: $2 x y^{2} + 2 x^{2} y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{y}{x}$ . Then $x^{3} y \frac{d y}{d x} = - x^{2} y^{2} = - 1$ . (1 mark)

Answer: $x^{3} y \frac{d y}{d x} = - x^{2} y^{2} = - 1$

Example 6: Case study: an elliptical cycling track · 4 marks · Case-basedHOTS / competency

Suppose a cycling track in a park is shaped like the ellipse $x^{2} + 4 y^{2} = 100$ (in metres), with the centre of the park at the origin.
(i) Find $\frac{d y}{d x}$ for the track. (1 mark)
(ii) Find the slope at the point $(6, 4)$ . (1 mark)
(iii) At which points of the track is $\frac{d y}{d x} = 0$ ? (1 mark)
(iv) At which points is $\frac{d y}{d x}$ undefined, and why? (1 mark)

Try it first, then show the solutionSolution

(i) $2 x + 8 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{x}{4 y}$ . (1 mark)
(ii) $(6, 4)$ : $36 + 64 = 100$ , on the track; slope $= - \frac{6}{16} = - \frac{3}{8}$ . (1 mark)
(iii) $\frac{d y}{d x} = 0$ when $x = 0$ : then $4 y^{2} = 100$ , so $(0, 5)$ and $(0, - 5)$ . (1 mark)
(iv) Undefined when $y = 0$ : then $x^{2} = 100$ , so $(10, 0)$ and $(- 10, 0)$ ; there the track runs vertically. (1 mark)

Answer: (i) $- \frac{x}{4 y}$ (ii) $- \frac{3}{8}$ (iii) $(0, \pm 5)$ (iv) $(\pm 10, 0)$

Where students lose marks in Implicit Differentiation

The slips examiners see most often, and what to write instead.

Mistake: Forgetting dy/dx on a y-term
y depends on x, so its derivative needs the chain rule.
Wrong working
$\frac{d}{d x} (y^{3}) = 3 y^{2}$ .
Right working
$\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}$ .
Do this: After differentiating, scan each y-term for its $y^{'}$ .
Usually costs: Usually the whole answer.
Mistake: Not using the product rule on xy
$x y$ is a product of two functions of $x$ .
Wrong working
$\frac{d}{d x} (x y) = y^{'}$ or $\frac{d}{d x} (x y) = 1 \cdot y^{'}$ .
Right working
$\frac{d}{d x} (x y) = x y^{'} + y$ .
Do this: Underline mixed terms before differentiating and treat them as products.
Usually costs: 1 to 2 marks.
Mistake: Substituting a point that is not on the curve
The formula for $y^{'}$ is valid only at points of the curve.
Wrong working
Curve $x^{2} + x y + y^{2} = 7$ ; 'slope at $(1, 1)$ ' $= - 1$ .
Right working
$(1, 1)$ gives $1 + 1 + 1 = 3 \neq = 7$ , so it is not on the curve; use a point like $(1, 2)$ .
Do this: Check the point in the original equation first.
Usually costs: The value mark.
Mistake: Sign errors when moving terms
Terms change sign when moved across the equals sign.
Wrong working
$sin y + x cos y y^{'} + cos x y^{'} - y sin x = 0 \Rightarrow y^{'} (x cos y + cos x) = sin y - y sin x$ .
Right working
$y^{'} (x cos y + cos x) = y sin x - sin y$ .
Do this: Move one term at a time and rewrite the full line.
Usually costs: The final answer mark.

Exam tips

Implicit differentiation was asked in the 2024, 2025 and 2026 board papers, often inside a 5-mark question; show the differentiated equation on a separate line before rearranging.
Leave the answer in terms of both $x$ and $y$ unless a point is given or the question asks you to simplify.
For 'prove that' questions, use the original equation to replace awkward expressions at the end.
When a substitution like $x = sin A$ is used, state the interval for $A$ so that $1 - x^{2} = cos A$ is valid.
Write $\frac{d y}{d x}$ in full in the final answer; $y^{'}$ is fine in working.

Practice questions on Implicit Differentiation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $x^{2} + y^{2} = 25$ , then $\frac{d y}{d x}$ at $(3, 4)$ is:
(a) $\frac{3}{4}$
(b) $- \frac{3}{4}$
(c) $\frac{4}{3}$
(d) $- \frac{4}{3}$
Hint
$\frac{d y}{d x} = - \frac{x}{y}$ .
Answer
(b) $- \frac{3}{4}$ .
Q2 · 2 marks · Very short answerBoard level
If $x^{2} y + x y^{2} = 6$ , find $\frac{d y}{d x}$ at $(1, 2)$ .
Hint
Use the product rule on both terms.
Answer
$2 x y + x^{2} y^{'} + y^{2} + 2 x y y^{'} = 0 \Rightarrow y^{'} = - \frac{2 x y + y ^{2}}{x ^{2} + 2 x y}$ . At $(1, 2)$ : $y^{'} = - \frac{4 + 4}{1 + 4} = - \frac{8}{5}$ .
Q3 · 2 marks · Very short answerBoard level
Find $\frac{d y}{d x}$ if $x y + y^{2} = tan x + y$ .
Hint
Differentiate both sides; $y$ on the right also gives $y^{'}$ .
Answer
$y + x y^{'} + 2 y y^{'} = sec^{2} x + y^{'}$ , so $y^{'} (x + 2 y - 1) = sec^{2} x - y$ and $\frac{d y}{d x} = \frac{s e c ^{2} x - y}{x + 2 y - 1}$ .
Q4 · 3 marks · Short answerHarder board level
Find $\frac{d y}{d x}$ if $tan (x + y) + tan (x - y) = 1$ .
Hint
$\frac{d}{d x} tan (x - y) = sec^{2} (x - y) (1 - y^{'})$ .
Answer
$sec^{2} (x + y) (1 + y^{'}) + sec^{2} (x - y) (1 - y^{'}) = 0$ , so $y^{'} [sec^{2} (x + y) - sec^{2} (x - y)] = - [sec^{2} (x + y) + sec^{2} (x - y)]$ and $\frac{d y}{d x} = \frac{s e c ^{2} ( x + y ) + s e c ^{2} ( x - y )}{s e c ^{2} ( x - y ) - s e c ^{2} ( x + y )}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $y^{2} = 4 x$ , then $\frac{d y}{d x} = \frac{2}{y}$ .
Reason (R): $\frac{d}{d x} (y^{2}) = 2 y$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Where is the $\frac{d y}{d x}$ in R?
Answer
(c). $2 y \frac{d y}{d x} = 4$ gives $\frac{d y}{d x} = \frac{2}{y}$ , so A is true. R is false: $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ , not $2 y$ .
Q6 · 3 marks · Short answerHarder board level
If $y = sin x + sin x + sin x + \dots to \infty$ , prove that $\frac{d y}{d x} = \frac{c o s x}{2 y - 1}$ .
Hint
Write $y^{2} = sin x + y$ .
Answer
$y^{2} = sin x + y \Rightarrow 2 y y^{'} = cos x + y^{'} \Rightarrow y^{'} (2 y - 1) = cos x \Rightarrow \frac{d y}{d x} = \frac{c o s x}{2 y - 1}$ .
Q7 · 4 marks · Short answerHOTS / competency
For the curve $x^{3} + y^{3} = 3 x y$ : (i) find $\frac{d y}{d x}$ ; (ii) find its value at $(\frac{3}{2}, \frac{3}{2})$ ; (iii) find the point in the first quadrant, other than the origin, where $\frac{d y}{d x} = 0$ .
Hint
$\frac{d y}{d x} = 0$ means the numerator $y - x^{2} = 0$ ; substitute $y = x^{2}$ into the curve.
Answer
(i) $3 x^{2} + 3 y^{2} y^{'} = 3 y + 3 x y^{'}$ , so $\frac{d y}{d x} = \frac{y - x ^{2}}{y ^{2} - x}$ . (ii) $\frac{1.5 - 2.25}{2.25 - 1.5} = - 1$ . (iii) $y = x^{2}$ in the curve: $x^{3} + x^{6} = 3 x^{3}$ , so $x^{3} = 2$ , giving $(2^{1/3}, 2^{2/3})$ ; the denominator there is $2^{4/3} - 2^{1/3} \neq = 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is $\frac{d}{d x} (y^{2})$ ?Show answer
$2 y \frac{d y}{d x}$ .
2.What is $\frac{d}{d x} (x y)$ ?Show answer
$x \frac{d y}{d x} + y$ .
3.Find $\frac{d y}{d x}$ for $x^{2} + y^{2} = r^{2}$ .Show answer
$- \frac{x}{y}$ .
4.What is $\frac{d}{d x} (cos y)$ ?Show answer
$- sin y \frac{d y}{d x}$ .
5.If $y = x + y$ , what is $\frac{d y}{d x}$ ?Show answer
$\frac{1}{2 y - 1}$ .
6.What is $\frac{d}{d x} (sin^{- 1} x)$ ?Show answer
$\frac{1}{1 - x ^{2}}$ .

Implicit Differentiation: frequently asked questions

What is an implicit function?

A function y of x defined by an equation that connects x and y without giving y alone on one side, such as x² + xy + y² = 7. Implicit differentiation finds dy/dx directly from such an equation.

Why do y-terms get a dy/dx when differentiated?

Because y depends on x. By the chain rule, the derivative of f(y) with respect to x is f′(y) times dy/dx.

Can the answer contain both x and y?

Yes. That is normal for implicit differentiation. If a point on the curve is given, substitute both coordinates to get a number.

Which NCERT section covers implicit differentiation?

Section 5.3.2 (Derivatives of Implicit Functions) of Chapter 5, Continuity and Differentiability, in NCERT Mathematics Part I for Class 12.

Are tangents and normals needed with this topic?

No. Tangents and normals were dropped from the 2026-27 syllabus. Questions ask for dy/dx, its value at a point, or a proof involving it.