Maths › Differentiation › Implicit and parametric differentiation

Implicit and parametric differentiation

Plenty of curves have no equation of the form y = f(x), circles first among them, and calculus still applies. Differentiate an equation exactly as it stands, remembering that y depends on x, or divide two parametric rates, and the gradient follows without any rearranging.

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 given as x in terms of y.
  • Find gradients, tangents and normals on parametric curves by dividing rates.

COMMON MISCONCEPTION

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

Differentiating in place

The chain rule quietly powers a much bigger idea. Since y depends on x, any term in y differentiates through y, so the derivative of y2 with respect to x is 2y × dy/dx, the outer square and then the inner dependence. Do that across a whole equation and you are performing implicit differentiation. The equation never gets rearranged, and never needs to be.

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 that same tangent from the perpendicular radius. Two methods agreeing on −3/4 is no accident, and a question will accept either.

The circle x squared plus y squared equals 25 with its tangent at 3 comma 4: implicit differentiation gives gradient minus x over y, minus three quarters there(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.

GUIDED PRACTICE

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, and 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 you divide. That factorised bracket is where the method marks sit.

When x is given in terms of y

One special case earns its own heading. A relation like x = sin y gives the derivative the wrong way round, so differentiate with respect to y and then flip it. From dx/dy = cos y,

dydx=1dx/dy=1cosy\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{\cos y}

and the flip itself has to appear as a line of working. Write cos y as √(1 − x2) on the principal range, where cos y is positive, and you have just derived the gradient of arcsin. Edexcel does not require that last step at A level, so treat it as a free bonus and revise nothing extra. The x = sin y and x = 3 tan 2y style is what the papers actually set.

Dividing parametric rates

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

The parametric parabola x equals t squared, y equals 2t: dividing the parametric rates gives gradient 1 over t, so the tangent at 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.

INDEPENDENT PRACTICE

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 final line, and that discipline is what these questions reward. Converting to y2 = 4x first is legal, slower and riskier.

ASSESSMENT FOCUS

  • Differentiate implicit equations term by term. Every y-term picks up a dy/dx through the chain rule.
  • Product-rule any xy term, 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, showing the flip as its own line.
  • Parametric gradients divide dy/dt by dx/dt. Find dy/dx in general first; substituting the value of t too early throws away the general expression a later part will want.

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. Check the point satisfies the original equation, 4 + 4 = 8, before trusting any calculus done on it.

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.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

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Answer book

PDF · 38 KB · fully worked · navy
8 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the implicit and parametric differentiation questions page.

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 given as x in terms of y.
  • Find gradients, tangents and normals on parametric curves by dividing rates.

Open the full revision checklist to see every objective in the course in one place.