Maths › Differentiation › Implicit and parametric differentiation

Implicit and parametric differentiation

Plenty of curves refuse to be written y = f(x), circles first among them, and calculus does not care. Differentiate an equation exactly as it stands, remembering that y depends on x, or divide two parametric rates, and gradients arrive without any rearranging at all.

Year 12-13EDEXCEL 9MA0 7.5

Builds on The product, quotient and chain rules and Parametric equations.

IN THIS TOPIC

  • Differentiate implicit relations term by term, with the chain rule supplying dy/dx.
  • Use dy/dx = 1/(dx/dy) for relations like x = sin y.
  • Find gradients and tangents on parametric curves by dividing rates.

WHAT YOU PROBABLY THINK

You cannot differentiate a curve that will not give you y = f(x).

Differentiating in place

The chain rule quietly powers a bigger idea. Since y depends on x, any term in y differentiates through y: the derivative of y2 with respect to x is 2y × dy/dx, the outer square then the inner dependence. Applied across a whole equation, this is implicit differentiation, and it retires the opening lie: the equation never needs rearranging.

WORKED EXAMPLE

The circle, differentiated as it stands

Find the gradient of x2 + y2 = 25 at the point (3, 4).

Differentiate every term with respect to x: 2x + 2y dy/dx = 0.

Solve: dy/dx = −x/y, and at (3, 4) the gradient is −3/4.

The circles lesson found this same tangent from the perpendicular radius; the two methods agreeing on −3/4 is no accident, and questions accept either.

The circle x squared plus y squared equals 25 with its tangent at the point 3 comma 4: implicit differentiation gives gradient minus x over y, which is minus three quarters there, and the tangent sits perpendicular to the radius(3, 4)slope −¾x² + y² = 25dy/dx = −x/y, no rearranging required
FIG. 1The circle differentiated without rearranging: dy/dx = −x/y gives slope −¾ at (3, 4), perpendicular to the radius as geometry always promised.

YOUR TURN

A mixed term joins in

Find the gradient of the curve x2 + 3xy + y2 = 11 at the point (1, 2), before opening the working.

Show the working

The 3xy term needs the product rule: its derivative is 3y + 3x dy/dx.

Altogether: 2x + 3y + (3x + 2y) dy/dx = 0, so dy/dx = −(2x + 3y)/(3x + 2y).

At (1, 2): −(2 + 6)/(3 + 4) = −8/7.

Collect the dy/dx terms on one side before dividing; the factorised bracket is where the method marks sit.

One special case earns its own line. When a relation gives x in terms of y, like x = sin y, differentiate with respect to y and flip: dx/dy = cos y, so

dydx = 1dx/dy = 1cos y

and writing cos y as √(1 − x2) turns it into the gradient of arcsin, the inverse-function derivative arriving free of charge.

Dividing parametric rates

On a parametric curve both coordinates move with t, and the gradient is one rate over the other: dy/dx = (dy/dt) ÷ (dx/dt). No elimination, no Cartesian conversion, just two derivatives and a division.

The parametric parabola x equals t squared, y equals 2t with its tangent at t equal to 2: dividing the two parametric rates gives gradient 1 over t, so the tangent at the point 4 comma 4 has slope one halft = 2: slope ½x = t², y = 2tdy/dx = (dy/dt) ÷ (dx/dt) = 2/(2t) = 1/t
FIG. 2The traced parabola x = t², y = 2t with its tangent at t = 2: rates 2 and 2t divide to 1/t, so the slope at (4, 4) is ½.

TRY IT UNSEEN

A parametric tangent, start to finish

The curve C has parametric equations x = t2, y = 2t. Find the equation of the tangent to C at the point where t = 2.

Show the working

Rates: dx/dt = 2t and dy/dt = 2, so dy/dx = 2/(2t) = 1/t.

At t = 2 the point is (4, 4) and the gradient is ½.

Tangent: y − 4 = ½(x − 4), that is y = x/2 + 2.

Everything stayed in t until the last line, which is the discipline these questions reward; converting to y2 = 4x first is legal but slower and riskier.

THE EXAM BIT

  • Differentiate implicit equations term by term; every y-term picks up a dy/dx through the chain rule.
  • Product-rule any xy terms, then collect dy/dx on one side and factorise before dividing.
  • For x given in terms of y, find dx/dy and take the reciprocal; state the flip as a line of working.
  • Parametric gradients divide dy/dt by dx/dt; evaluate at the stated t before building tangent or normal.
  • Substituting the point too early wastes the general gradient; find dy/dx in full first, then substitute.

CHECK YOURSELF

The curve C is given by y2 + 2xy = 8. Find dy/dx in terms of x and y, and the gradient of C at (1, 2).

Show a hint

Both terms need the chain rule; one also needs the product rule.

Show the answer

Differentiating: 2y dy/dx + 2y + 2x dy/dx = 0, so (2y + 2x) dy/dx = −2y.

dy/dx = −y/(x + y).

At (1, 2): −2/3, and the point checks in the original equation, 4 + 4 = 8, before any calculus is trusted.

Differentiate equations as they stand; y-terms carry dy/dx through the chain rule.

Parametric gradients divide the two rates; x-in-terms-of-y flips its derivative.

CHECK YOUR PROGRESS

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  • Differentiate implicit relations term by term, with the chain rule supplying dy/dx.
  • Use dy/dx = 1/(dx/dy) for relations like x = sin y.
  • Find gradients and tangents on parametric curves by dividing rates.

No animated video for this topic yet; these notes stand alone.