← Unit 3: Differentiation — Composite, Implicit & Inverse Functions
Implicit Differentiation
When y is defined implicitly by an equation in x and y (not solved for y), differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to every y-term.
Differentiating both sides
Take d/dx of every term in the equation. Any term with y requires the chain rule: d/dx[y²] = 2y·(dy/dx), since y is implicitly a function of x.
After differentiating, the equation will contain dy/dx (often written y'). Collect every term with dy/dx on one side, factor it out, and solve algebraically.
Worked Example
Find dy/dx if x² + y² = 25.
- Differentiate both sides with respect to x: d/dx[x²] + d/dx[y²] = d/dx[25]
- 2x + 2y·(dy/dx) = 0 (chain rule applied to y²; derivative of the constant 25 is 0)
- Solve for dy/dx: 2y·(dy/dx) = -2x, so dy/dx = -x/y
Answer: dy/dx = -x/y
Key terms (1)
- Implicit differentiation
- Differentiating both sides of an equation relating x and y, applying the chain rule to every y term to solve for dy/dx.
Practice Quiz
Question 1 of 3Score so far: 0/0
Find dy/dx for xy = 6.