Differentiate implicit curves using product and chain rules
Treat y as changing with x, collect every dy/dx term and use the supplied curve point to choose a local gradient.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Use product and chain rules with a changing inner variable.
- Collect a common derivative factor and build tangent and normal lines.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is d(y²)/dx when y depends on x?
What you will learn
- Find a first derivative from an implicit relation, including powers, products and natural-function terms in y.
- Verify the supplied point and find tangent and normal lines on its local branch.
Every occurrence of y carries its rate of change
An implicit equation relates x and y without necessarily solving for one global function y = f(x). On a differentiable local branch, treat y as a function of x. Thus d(y²)/dx = 2y(dy/dx) and d(eʸ)/dx = eʸ(dy/dx) by the chain rule. A product xy contributes y + x(dy/dx) because both its factors vary with x. Treating y as a constant would delete its changing contribution.
For x² + y² = xy + 7, differentiation gives 2x + 2y y′ = y + x y′. Collect to (2y − x)y′ = y − 2x, hence y′ = (y − 2x)/(2y − x) where 2y − x ≠ 0. Verify that the chosen point belongs to the original curve before substituting it. At (1, 3), 1 + 9 = 3 + 7 and the gradient is (3 − 2)/(6 − 1) = 1/5.
Choose the local branch using both coordinates
The tangent at (1, 3) is y − 3 = (x − 1)/5, or x − 5y + 14 = 0. The normal gradient is −5, giving 5x + y = 8. At the same x = 1, the original curve also includes (1, −2), which has a different gradient 4/5. A derivative expression involving both coordinates needs the supplied point, not x alone, to choose the local branch. A zero denominator requires a separate tangent check rather than an automatic finite gradient.
For eʸ + x² = e + 1, differentiation gives eʸ y′ + 2x = 0, so y′ = −2x/eʸ. At (1, 1) the gradient is −2/e. Eliminating y gives y = ln(e + 1 − x²) where |x| < √(e + 1), and its chain derivative agrees. This check preserves the original positive exponential quantity; inputs where e + 1 − x² is non-positive cannot lie on that real curve.
Pause and explain
On x² + y² = xy + 7, what must be used to evaluate a local gradient at x = 1?
Worked example
For x² + y² = xy + 7, find dy/dx and tangent and normal equations at (1, 3).
Show the worked solution
- Differentiate to 2x + 2y y′ = y + x y′, using both chain and product rules.
- Collect derivative terms to obtain y′ = (y − 2x)/(2y − x); the point satisfies the original curve and has gradient 1/5.
- Use the point-gradient lines: tangent x − 5y + 14 = 0 and normal 5x + y = 8.
Answer y′ = (y − 2x)/(2y − x), where 2y − x ≠ 0; at (1, 3), tangent x − 5y + 14 = 0 and normal 5x + y = 8.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For x² + y² = 25, find the gradient, tangent and normal at (3, 4).
Give me a hint
Differentiate implicitly and verify the point before evaluating its local gradient.
Compare my reasoning
- The point is on the circle because 9 + 16 = 25; differentiation gives 2x + 2y y′ = 0 and y′ = −x/y for y ≠ 0.
- At (3, 4), the gradient is −3/4, so the tangent is y − 4 = −3(x − 3)/4, or 3x + 4y = 25.
- The normal gradient is 4/3, giving 4x − 3y = 0 through the same point.
Gradient −3/4; tangent 3x + 4y = 25; normal 4x − 3y = 0.
Look for these in your work
- Use the chain factor when differentiating y².
- Verify the point and both perpendicular line equations.
For eʸ + x² = e + 1, find tangent and normal equations at (1, 1), then check the derivative using an explicit logarithmic form.
Give me a hint
Use the exponential chain derivative eʸ y′ and retain the real curve's positive log argument after eliminating y.
Compare my reasoning
- The point satisfies the original equation; differentiation gives eʸ y′ + 2x = 0, so the gradient there is −2/e.
- The tangent is y − 1 = −(2/e)(x − 1); the normal is y − 1 = (e/2)(x − 1).
- The explicit form is y = ln(e + 1 − x²) for |x| < √(e + 1); differentiating it gives −2x/(e + 1 − x²), agreeing with −2x/eʸ on the curve.
Tangent y − 1 = −(2/e)(x − 1); normal y − 1 = (e/2)(x − 1); explicit check y = ln(e + 1 − x²), |x| < √(e + 1).
Look for these in your work
- Differentiate the changing y inside the exponential.
- Check the derivative and real domain against the equivalent explicit function.
For x² + xy + y² = 7, find tangent and normal equations at (1, 2). Explain how the product term changes the derivative.
Give me a hint
The xy derivative contributes y + x y′; collect all derivative terms after differentiating the full relation.
Compare my reasoning
- The point satisfies 1 + 2 + 4 = 7; differentiation gives 2x + y + x y′ + 2y y′ = 0.
- Hence y′ = −(2x + y)/(x + 2y), giving −4/5 at (1, 2), where its denominator is 5.
- The tangent is 4x + 5y = 14 and the normal is 5x − 4y = −3; both pass through the original point.
Gradient −4/5; tangent 4x + 5y = 14; normal 5x − 4y = −3.
Look for these in your work
- Include both contributions from the changing xy product.
- Collect the derivative coefficient and verify both lines at a valid point.
Common mistakes
- A y power or composite needs its inner derivative dy/dx.
- The derivative of xy has both y and x(dy/dx) terms.
- Verify the original point and use both coordinates for the local branch; a derivative denominator must be checked.
What can you explain now?
What is d(xy)/dx when y depends on x?
Compare with the explanation
y + x(dy/dx).
The product rule differentiates x in one contribution and the changing y in the other.
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Make it stick
Differentiate an implicit relation with a y square, an xy product and an exponential in y. Explain why the original point and a non-zero derivative coefficient are needed before finding its local lines.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.