Maths › Differentiation › Implicit and parametric differentiation
Implicit and parametric differentiation
Many curves, circles among them, have no equation of the form y = f(x), but their gradients are still found by calculus. Implicit differentiation treats y as a function of x and differentiates the equation as it stands, while parametric differentiation divides one rate by another.
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.
- Find tangents and normals on implicit curves from dy/dx at the point.
COMMON MISCONCEPTION
You cannot differentiate a curve that will not give you y = f(x).
Differentiate the equation as it stands. Every y-term picks up a factor dy/dx from the chain rule, so x2 + y2 = 25 gives 2x + 2y dy/dx = 0 and dy/dx = −x/y for y ≠ 0, which is −3/4 at (3, 4). A parametric curve needs no rearranging either: divide dy/dt by dx/dt wherever dx/dt ≠ 0.
Differentiating in place
Where y depends on x, the chain rule differentiates any term in y with respect to x: the derivative of y2 is 2y × dy/dx, the outer square differentiated, then multiplied by dy/dx for the dependence of y on x. Applying that to every term of an equation is implicit differentiation, and it does not require the equation to be rearranged into the form y = f(x).
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 for y ≠ 0, and at (3, 4) the gradient is −3/4.
Circle geometry gives the same tangent, since the tangent is perpendicular to the radius. Either method is a valid answer.
The gradient at a point, together with the point itself, gives the equations of two lines. On the circle at (3, 4) the gradient is −3/4, so the tangent there is y − 4 = −(3/4)(x − 3), that is 3x + 4y = 25. The normal is the perpendicular to the curve through the same point, so its gradient is the negative reciprocal 4/3 and its equation is y − 4 = (4/3)(x − 3), that is 4x − 3y = 0; on a circle the normal passes through the centre, which is a check on the working.
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 and factorise before dividing, and note that the division requires 3x + 2y ≠ 0.
When x is given in terms of y
A relation such as x = sin y gives dx/dy directly. Differentiate with respect to y and take the reciprocal, which is valid where dx/dy ≠ 0. From dx/dy = cos y,
Show the reciprocal step as a line of working. Writing cos y as √(1 − x2) on −π/2 < y < π/2, where cos y is positive, gives the derivative of arcsin x. That step is not required by Edexcel 9MA0; relations of the form x = sin y and x = 3 tan 2y are.
Dividing parametric rates
On a parametric curve both coordinates depend on t, and the chain rule gives dy/dx = (dy/dt) ÷ (dx/dt), valid at points where dx/dt ≠ 0. Where dx/dt = 0 and dy/dt ≠ 0 the tangent is vertical. No elimination or conversion to Cartesian form is required: two derivatives and a division.
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.
The working stays in t until the final line. Converting to y2 = 4x first is valid but longer.
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, where dx/dt ≠ 0. Find dy/dx in terms of t first, then substitute, since a later part may need the general expression.
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 division requires x + y ≠ 0. Check that the point satisfies the original equation, 4 + 4 = 8, before using it.
Differentiate equations as they stand; every y-term carries a factor dy/dx from the chain rule.
Parametric gradients divide dy/dt by dx/dt where dx/dt ≠ 0; for x given in terms of y, dy/dx is the reciprocal of dx/dy where dx/dy ≠ 0.
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.
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.
- Find tangents and normals on implicit curves from dy/dx at the point.
Open the full revision checklist to see every objective in the course in one place.