Implicit and Parametric Differentiation: Methods, Examples
When y Is Not Alone
Most differentiation starts with y written explicitly as a function of x, such as y = x³ − 4x. But many curves are not given that way. The circle x² + y² = 25 mixes x and y together, and a projectile's path or a planet's orbit is most naturally described with both x and y depending on a third variable, time. To find gradients, tangents and turning points on such curves, you need implicit and parametric differentiation. This tutorial covers both methods for further mathematics and first-year university calculus (MTH 102 and similar): differentiating implicitly with the chain rule, tangents and normals to implicit curves, logarithmic differentiation, derivatives of inverse trigonometric functions, second derivatives, parametric equations, the gradient dy/dx = (dy/dt)/(dx/dt), and second derivatives of parametric curves, with worked examples. It assumes you know the rules from the tutorial on differentiation.
Part 1: Implicit Differentiation
An equation such as x² + y² = 25 defines y implicitly as a function of x. You could solve for y = ±√(25 − x²), but that is messy, and for many curves it is impossible. Instead, differentiate both sides with respect to x, treating y as an unknown function of x.
The key rule
By the chain rule, any expression in y differentiates to its derivative with respect to y, multiplied by dy/dx:
- d/dx (y²) = 2y (dy/dx)
- d/dx (y³) = 3y² (dy/dx)
- d/dx (sin y) = cos y (dy/dx)
- d/dx (e^y) = e^y (dy/dx)
Terms that mix x and y, such as xy, need the product rule: d/dx (xy) = y + x (dy/dx).
The method
- Differentiate every term on both sides with respect to x.
- Collect all the terms containing dy/dx on one side.
- Factorise out dy/dx and divide.
Implicit Differentiation: Worked Examples
Worked example 1: a circle
Find dy/dx for x² + y² = 25, and the equation of the tangent at (3, 4).
Differentiating: 2x + 2y (dy/dx) = 0, so dy/dx = −x/y.
At (3, 4): dy/dx = −3/4.
Tangent: y − 4 = −¾(x − 3). Multiplying by 4: 4y − 16 = −3x + 9, so the tangent is 3x + 4y = 25.
The result dy/dx = −x/y shows the tangent is perpendicular to the radius, whose gradient is y/x, which is a familiar circle theorem.
Worked example 2: a product term
Find the gradient of the curve x²y + xy² = 6 at the point (1, 2).
Use the product rule on both terms:
(2xy + x² dy/dx) + (y² + 2xy dy/dx) = 0
Collect terms: dy/dx (x² + 2xy) = −(2xy + y²), so dy/dx = −(2xy + y²) ÷ (x² + 2xy).
At (1, 2): dy/dx = −(4 + 4) ÷ (1 + 4) = −8/5.
Worked example 3: the folium of Descartes
Find dy/dx for x³ + y³ = 6xy, and the gradient at (3, 3).
3x² + 3y² (dy/dx) = 6y + 6x (dy/dx).
So dy/dx (3y² − 6x) = 6y − 3x², and dividing by 3: dy/dx = (2y − x²) ÷ (y² − 2x).
At (3, 3), which lies on the curve because 27 + 27 = 54 = 6 × 9: dy/dx = (6 − 9) ÷ (9 − 6) = −1.
Logarithmic Differentiation
For functions with a variable in the exponent, such as x^x, or long products and quotients, take natural logarithms of both sides first and then differentiate implicitly.
Worked example 4
Differentiate y = x^x.
Take logs: ln y = x ln x.
Differentiate implicitly: (1/y)(dy/dx) = ln x + x × (1/x) = ln x + 1.
So dy/dx = x^x (ln x + 1).
The same method shows that d/dx (aˣ) = aˣ ln a for any positive constant a. For example, d/dx (2ˣ) = 2ˣ ln 2.
Derivatives of Inverse Trigonometric Functions
Implicit differentiation gives the derivatives of inverse functions neatly.
Worked example 5
Show that if y = sin⁻¹x, then dy/dx = 1/√(1 − x²).
Rewrite as sin y = x. Differentiate: cos y (dy/dx) = 1, so dy/dx = 1/cos y.
Since cos²y = 1 − sin²y = 1 − x², and cos y is positive for −π/2 < y < π/2, we get dy/dx = 1/√(1 − x²).
In the same way, y = tan⁻¹x gives sec²y (dy/dx) = 1, and since sec²y = 1 + tan²y = 1 + x², dy/dx = 1/(1 + x²).
Second Derivatives by Implicit Differentiation
Differentiate the expression for dy/dx again, using the quotient rule, then substitute for dy/dx.
Worked example 6
For x² + y² = 25, find d²y/dx².
From dy/dx = −x/y, the quotient rule gives d²y/dx² = −[y − x(dy/dx)] ÷ y².
Substitute dy/dx = −x/y: d²y/dx² = −[y + x²/y] ÷ y² = −(x² + y²) ÷ y³.
Since x² + y² = 25, d²y/dx² = −25/y³. On the upper half of the circle (y > 0) this is negative, so the curve is concave down, as expected.
Part 2: Parametric Differentiation
In parametric equations, both x and y are given as functions of a third variable, called the parameter, usually t or θ. For example, x = 3 cos θ, y = 3 sin θ describes a circle of radius 3, and x = t², y = 2t describes a parabola.
To find the gradient, use the chain rule:
dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0
- The tangent is horizontal (a stationary point) where dy/dt = 0.
- The tangent is vertical where dx/dt = 0.
Parametric Differentiation: Worked Examples
Worked example 7: a parabola
A curve has parametric equations x = t², y = 2t. Find dy/dx, the gradient at t = 2, and the Cartesian equation.
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 1/2.
Eliminate t: t = y/2, so x = y²/4, giving the Cartesian equation y² = 4x.
Worked example 8: a circle
For x = 3 cos θ, y = 3 sin θ, find dy/dx and the gradient at θ = π/4.
dx/dθ = −3 sin θ and dy/dθ = 3 cos θ, so dy/dx = 3 cos θ ÷ (−3 sin θ) = −cot θ.
At θ = π/4: dy/dx = −cot(π/4) = −1.
Worked example 9: stationary points and vertical tangents
For x = t³ − 3t, y = t², find dy/dx and the points where the tangent is horizontal or vertical.
dx/dt = 3t² − 3 and dy/dt = 2t, so dy/dx = 2t ÷ (3t² − 3).
Horizontal tangent: dy/dt = 0 when t = 0, giving the point (0, 0).
Vertical tangents: dx/dt = 0 when t = ±1. At t = 1 the point is (−2, 1), and at t = −1 it is (2, 1).
Second Derivatives of Parametric Curves
dy/dx is a function of t, so to differentiate it again with respect to x, use the chain rule once more:
d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt)
A common mistake is to differentiate dy/dx with respect to t and stop there. You must divide by dx/dt.
Worked example 10
For x = t², y = 2t, find d²y/dx².
dy/dx = 1/t, so d/dt (dy/dx) = −1/t². Dividing by dx/dt = 2t gives d²y/dx² = −1/(2t³).
Implicit and Parametric Differentiation: Practice Questions
- Find dy/dx for x² + 3xy + y² = 5, and its value at (1, 1).
- Differentiate y = 2ˣ using logarithmic differentiation.
- Show that if y = tan⁻¹x, then dy/dx = 1/(1 + x²).
- For x = 2t + 1, y = t² − 3, find dy/dx and its value at t = 3.
- For the ellipse x = 2 cos t, y = 5 sin t, find dy/dx in terms of t.
Answers: 1. 2x + 3y + 3x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + 3y)/(3x + 2y). At (1, 1): −1. 2. ln y = x ln 2, so dy/dx = 2ˣ ln 2. 3. tan y = x gives sec²y (dy/dx) = 1, and sec²y = 1 + x². 4. dy/dx = 2t/2 = t, which is 3 at t = 3. 5. dy/dx = 5 cos t ÷ (−2 sin t) = −(5/2) cot t.
Conclusion
Implicit differentiation handles equations where y is mixed with x: differentiate every term with respect to x, remembering that each y term picks up a factor of dy/dx, then solve for dy/dx. Logarithmic differentiation extends the idea to functions like x^x, and it also gives the derivatives of inverse trigonometric functions. For parametric curves, divide dy/dt by dx/dt to get the gradient, and divide d/dt (dy/dx) by dx/dt again for the second derivative. Together, these methods let you find tangents, normals and turning points on almost any curve.