Implicit and Parametric Differentiation

A-Level Maths · Pure Mathematics

Implicit and Parametric Differentiation

Not all curves can be written as y = f(x). Implicit differentiation handles equations where y is tangled with x, while parametric differentiation handles curves defined through a third variable (the parameter).

Implicit Differentiation

An implicitly defined curve is one where x and y are mixed together in one equation, such as x² + y² = 25 (a circle). To find dy/dx:

1. Differentiate every term with respect to x

2. When differentiating a term involving y, apply the chain rule: d/dx[f(y)] = f'(y) × dy/dx

3. Collect all dy/dx terms on one side and solve for dy/dx

Worked Example 1: Find dy/dx for x² + y² = 25.

Differentiate each term with respect to x:

2x + 2y·(dy/dx) = 0

2y·(dy/dx) = -2x

dy/dx = -x/y

This gives the gradient at any point (x, y) on the circle. At (3, 4): dy/dx = -3/4.

Worked Example 2: Find dy/dx for x³ + 3xy + y³ = 8.

Differentiate term by term. The term 3xy requires the product rule:

d/dx(3xy) = 3[x·(dy/dx) + y·1] = 3x(dy/dx) + 3y

So the full differentiation gives:

3x² + 3x(dy/dx) + 3y + 3y²(dy/dx) = 0

Collect dy/dx terms: (3x + 3y²)(dy/dx) = -3x² - 3y

dy/dx = -(x² + y) / (x + y²)

Tangents and Normals from Implicit Equations

Once you have dy/dx in terms of x and y, substitute the coordinates of the point to find the gradient. Then use y - y₁ = m(x - x₁) for the tangent, and the negative reciprocal gradient for the normal.

Worked Example: Find the equation of the tangent to x² + xy + y² = 7 at (1, 2).

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

(x + 2y)(dy/dx) = -2x - y

dy/dx = -(2x + y)/(x + 2y)

At (1, 2): dy/dx = -(2 + 2)/(1 + 4) = -4/5

Tangent: y - 2 = -4/5(x - 1), which gives 4x + 5y = 14.

Parametric Equations

A curve can be defined by expressing x and y separately in terms of a parameter (usually t or θ):

x = f(t), y = g(t)

Common examples include:

  • Circle: x = rcosθ, y = rsinθ
  • Parabola: x = at², y = 2at
  • Ellipse: x = acosθ, y = bsinθ

Parametric Differentiation

To find dy/dx for parametric curves, use the chain rule:

dy/dx = (dy/dt) / (dx/dt)

This is valid provided dx/dt ≠ 0.

Worked Example 1: A curve is defined by x = t² + 1, y = t³ - t. Find dy/dx.

dx/dt = 2t and dy/dt = 3t² - 1

dy/dx = (3t² - 1) / (2t)

At t = 2: dy/dx = (12 - 1)/4 = 11/4

Worked Example 2: A curve is defined by x = 3cosθ, y = 3sinθ. Find dy/dx.

dx/dθ = -3sinθ and dy/dθ = 3cosθ

dy/dx = 3cosθ / (-3sinθ) = -cosθ/sinθ = -cotθ

Second Derivatives with Parameters

To find d²y/dx², differentiate dy/dx with respect to t, then divide by dx/dt:

d²y/dx² = (d/dt(dy/dx)) / (dx/dt)

Worked Example: For x = t², y = t³, find d²y/dx².

dy/dx = 3t²/2t = 3t/2

d/dt(dy/dx) = d/dt(3t/2) = 3/2

d²y/dx² = (3/2) / (2t) = 3/(4t)

Converting Between Parametric and Cartesian Forms

To convert parametric equations to Cartesian form, eliminate the parameter:

  • Rearrange one equation for t
  • Substitute into the other
  • For trig parameters, use sin²θ + cos²θ = 1

Example: x = 2t, y = t² becomes t = x/2, so y = (x/2)² = x²/4.

Example: x = 5cosθ, y = 5sinθ. Then x² + y² = 25cos²θ + 25sin²θ = 25.

Exam Tips

  • In implicit differentiation, never forget dy/dx when differentiating y terms — this is the most common error.
  • Products of x and y (like 3xy) always need the product rule.
  • For parametric tangent/normal questions, find the value of the parameter at the given point first, then compute dy/dx.
  • When finding where a parametric curve crosses the x-axis, set y = 0 (not t = 0).
  • Check whether a question asks for the answer in terms of the parameter or in Cartesian form.
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