Implicit differentiation finds dy/dx for curves that cannot easily be written as y = f(x), such as x² + y² = 25. Differentiate every term with respect to x. Any term in y is differentiated as normal and then multiplied by dy/dx, by the chain rule, and a term like xy needs the product rule. Finally, collect the dy/dx terms and rearrange.
Implicit differentiation rules
- d/dx (yⁿ) = nyⁿ⁻¹ dy/dx
- d/dx (sin y) = cos y dy/dx and d/dx (cos y) = −sin y dy/dx
- d/dx (e^y) = e^y dy/dx and d/dx (ln y) = (1/y) dy/dx
- d/dx (xy) = y + x dy/dx
Every term containing y picks up a factor of dy/dx.
What is implicit differentiation?
Some curves are defined by an equation linking x and y in which y is not the subject, and often cannot be made the subject. Implicit differentiation lets you find the gradient anyway, by differentiating both sides of the equation with respect to x term by term.
Differentiating a term in y
A function of y is differentiated with the chain rule: differentiate it as if y were the variable, then multiply by dy/dx. So y² gives 2y dy/dx, sin y gives cos y dy/dx, and e^y gives e^y dy/dx. Terms in x only are differentiated as normal, and constants give 0.
Forgetting the dy/dx factor on a y-term is the most common mistake in the topic.
Terms with both x and y
A term like xy is a product, so it needs the product rule, with the y-part still picking up dy/dx: d/dx (xy) = y + x dy/dx. Similarly, x²y gives 2xy + x² dy/dx. Treat these carefully; they are where most marks are lost.
Rearranging for dy/dx
Once every term is differentiated, dy/dx usually appears in more than one place. Collect all the dy/dx terms on one side and everything else on the other, factorise dy/dx out, and divide. The answer normally contains both x and y.
Gradients and tangents
To find the gradient at a point, substitute both coordinates into the expression for dy/dx, not just the x-value. Then use y − y₁ = m(x − x₁) for a tangent, or the negative reciprocal gradient for a normal. To find where the tangent is horizontal, set the numerator of dy/dx equal to zero and solve it together with the curve's equation.
Worked examples
Example 1. Find dy/dx for x² + y² = 25.
- Differentiate each term: 2x + 2y dy/dx = 0.
- Rearrange: 2y dy/dx = −2x.
Answer: dy/dx = −x/y.
Marks: M1 for 2x + 2y dy/dx = 0, A1 for −x/y.
Example 2. Find dy/dx for x² + xy = 6.
- Product rule on xy: 2x + (y + x dy/dx) = 0.
- So x dy/dx = −(2x + y).
Answer: dy/dx = −(2x + y)/x.
Marks: M1 for the product rule on xy, A1 for the answer.
Example 3. Find dy/dx for x² + y² + xy = 7.
- Differentiate: 2x + 2y dy/dx + y + x dy/dx = 0.
- Collect: (2y + x) dy/dx = −(2x + y).
Answer: dy/dx = −(2x + y)/(x + 2y).
Marks: M1 for differentiating every term, M1 for factorising dy/dx, A1 for the result.
Example 4. Find the equation of the tangent to x² + y² = 25 at (−4, 3).
- dy/dx = −x/y, so at (−4, 3) the gradient is −(−4)/3 = 4/3.
- Tangent: y − 3 = (4/3)(x + 4).
- Multiply by 3: 3y − 9 = 4x + 16.
Answer: 4x − 3y + 25 = 0.
Marks: M1 for the gradient 4/3, M1 for the line, A1 for 4x − 3y + 25 = 0.
Example 5. Find dy/dx for y = x + cos y.
- Differentiate: dy/dx = 1 − sin y dy/dx.
- Collect: dy/dx(1 + sin y) = 1.
Answer: dy/dx = 1/(1 + sin y).
Marks: M1 for −sin y dy/dx, M1 for collecting, A1 for the result.
Common mistakes
- Missing the dy/dx on a y-term. y² differentiates to 2y dy/dx, not 2y.
- Not using the product rule on xy. d/dx (xy) = y + x dy/dx, not just dy/dx or x dy/dx.
- Substituting only x. Implicit gradients contain x and y, so substitute both coordinates.
- Differentiating the constant. A constant on the right-hand side differentiates to 0.
- Not factorising dy/dx. Collect every dy/dx term, then factorise before dividing.
How implicit differentiation is examined
Implicit differentiation appears on WMA14, the Pure 4 paper for Edexcel International A-Level; 9MA0 students meet it in their second year. Questions usually ask you to:
- show that dy/dx equals a given expression
- find the gradient, tangent or normal at a given point
- find the points where the tangent is horizontal or vertical
- handle curves involving trig, exponential or log terms in y
Practise with the Pure 4 differentiation exam questions, which include implicit questions with full mark schemes. The Pure 4 differentiation worksheet and the differentiation revision engine give more practice.
This uses the chain rule and product rule together. Its companion topic is parametric differentiation.
Frequently asked questions
When do you use implicit differentiation?
When a curve's equation mixes x and y and y is not, or cannot easily be made, the subject.
Why does a y-term get multiplied by dy/dx?
Because y is a function of x, so differentiating a function of y with respect to x needs the chain rule: d/dx f(y) = f′(y) × dy/dx.
How do you find where a curve has a horizontal tangent?
Set dy/dx = 0, which usually means setting its numerator to zero, then solve that together with the curve's equation.