Implicit Differentiation Formula
Implicit differentiation explained with step-by-step examples.
Find dy/dx when y cannot be isolated, using the chain rule on both sides.
The Method
Apply chain rule to any term containing y: d/dx[f(y)] = f'(y) · dy/dx
Solve for dy/dx.
Implicit differentiation is used when a curve is defined by an equation relating x and y together — and it is difficult or impossible to isolate y explicitly. Instead of rewriting y = f(x), you differentiate the entire equation term by term with respect to x, remembering that y is a function of x and applying the chain rule accordingly.
Key Rule
| Expression | Derivative with respect to x | Reason |
|---|---|---|
| x² | 2x | Normal power rule |
| y² | 2y · dy/dx | Chain rule (y depends on x) |
| xy | y + x · dy/dx | Product rule |
| sin(y) | cos(y) · dy/dx | Chain rule |
Example 1 — Circle
x² + y² = 25 (circle of radius 5)
Step 1: Differentiate both sides → 2x + 2y · dy/dx = 0
Step 2: Solve for dy/dx → 2y · dy/dx = −2x
dy/dx = −x/y
Example 2 — Cubic Curve
x³ + y³ = 6xy (folium of Descartes)
Step 1: Differentiate → 3x² + 3y² · dy/dx = 6y + 6x · dy/dx
Step 2: Group dy/dx terms → 3y² · dy/dx − 6x · dy/dx = 6y − 3x²
Step 3: Factor → dy/dx(3y² − 6x) = 6y − 3x²
dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x)
Example 3 — Finding the Tangent Line
Find the tangent slope to x² + xy + y² = 7 at point (1, 2)
Differentiate: 2x + y + x · dy/dx + 2y · dy/dx = 0
Plug in x=1, y=2: 2 + 2 + 1·dy/dx + 4·dy/dx = 0 → 4 + 5·dy/dx = 0
dy/dx = −4/5 (slope of tangent at that point)
When to Use Implicit Differentiation
- Finding slopes on circles, ellipses, hyperbolas, and other conic sections
- Differentiating inverse functions (e.g. deriving d/dx[arcsin(x)])
- Related rates problems where both variables change with time
- Any equation where solving for y explicitly would be very messy
- Curves defined parametrically or in polar form