Implicit Differentiation
Learn to differentiate equations where y is not explicitly written as a function of x, a key technique for curves and relations.
What is Implicit Differentiation?
In many functions, y is written explicitly in terms of x, for example y = x2 + 3x. But sometimes x and y are mixed together in an equation like x2 + y2 = 25 (a circle). When we cannot easily isolate y, we use implicit differentiation.
Key Idea: Differentiate both sides with respect to x. Whenever you differentiate a term containing y, multiply by dy/dx (chain rule).
This works because y is implicitly a function of x. Every time we differentiate y with respect to x, the chain rule tells us to include the factor dy/dx.
Example: x2 + y2 = 25
Step 1: Differentiate each term: 2x + 2y · dy/dx = 0
Step 2: Solve for dy/dx: 2y · dy/dx = −2x
Result: dy/dx = −x/y
The Process Step by Step
Differentiate every term on both sides of the equation with respect to x.
Apply the chain rule to any term containing y — attach dy/dx as a factor.
Collect all dy/dx terms on one side of the equation.
Factor out dy/dx and solve for it.
Remember: For product terms like xy, you need to use the product rule: d/dx(xy) = y + x · dy/dx.
Finding Tangent Lines to Implicit Curves
Once you find dy/dx, you can evaluate it at a specific point to find the gradient of the tangent line. This is commonly required in HSC problems involving circles, ellipses, and other curves.
Example: Find the tangent to x2 + y2 = 25 at (3, 4)
Step 1: We found dy/dx = −x/y.
Step 2: At (3, 4): dy/dx = −3/4.
Step 3: Using point-gradient form: y − 4 = −3/4(x − 3).
Result: y = −3x/4 + 25/4, or equivalently 3x + 4y = 25.
Key Vocabulary
Implicit Relation
An equation where y is not isolated on one side, e.g. x2 + y2 = 1.
Explicit Function
A function where y is written directly in terms of x, e.g. y = 3x + 2.
dy/dx
The derivative of y with respect to x, representing the rate of change or gradient.
Tangent Line
A straight line that touches a curve at exactly one point and has the same gradient as the curve at that point.
Worked Examples
Find dy/dx given x3 + y3 = 6xy.
Step 1: Differentiate both sides: 3x2 + 3y2 · dy/dx = 6y + 6x · dy/dx
Step 2: Collect dy/dx terms: 3y2 · dy/dx − 6x · dy/dx = 6y − 3x2
Step 3: Factor: dy/dx(3y2 − 6x) = 6y − 3x2
Answer: dy/dx = (6y − 3x2) / (3y2 − 6x) = (2y − x2) / (y2 − 2x)
Find dy/dx given x2y + xy2 = 6.
Step 1: Use product rule on each term:
d/dx(x2y) = 2xy + x2 · dy/dx
d/dx(xy2) = y2 + 2xy · dy/dx
Step 2: 2xy + x2 · dy/dx + y2 + 2xy · dy/dx = 0
Step 3: dy/dx(x2 + 2xy) = −2xy − y2
Answer: dy/dx = −(2xy + y2) / (x2 + 2xy) = −y(2x + y) / x(x + 2y)
Find the gradient of the tangent to x2 + xy + y2 = 7 at the point (1, 2).
Step 1: Differentiate: 2x + y + x · dy/dx + 2y · dy/dx = 0
Step 2: dy/dx(x + 2y) = −2x − y, so dy/dx = −(2x + y)/(x + 2y)
Step 3: At (1, 2): dy/dx = −(2 + 2)/(1 + 4) = −4/5
Answer: The gradient of the tangent at (1, 2) is −4/5.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
Find dy/dx if x2 + y2 = 16.
Question 2
When differentiating the term 3y4 implicitly with respect to x, the result is:
Question 3
Find dy/dx if xy = 10.
Question 4
For the circle x2 + y2 = 25, what is the gradient at the point (3, −4)?
Question 5
Find dy/dx if sin(y) = x.
Key Concepts Summary
- ● Implicit differentiation is used when y cannot easily be isolated as a function of x.
- ● Differentiate every term with respect to x. Attach dy/dx whenever you differentiate a y-term.
- ● Use the product rule for mixed terms like xy or x2y.
- ● Collect dy/dx terms, factor, and solve to find dy/dx.
- ● Substitute a specific point to find the gradient of the tangent at that point.