Implicit Differentiation

Summary

When a relation F(x,y)=C defines y as a function of x only implicitly, differentiate both sides with respect to x and solve for y′ . Use the chain rule on every term involving y .

Prerequisites

Derivatives, Chain Rules

Procedure

  1. Differentiate each term of F(x,y)=C with respect to x , treating y=y(x) .
  2. Collect all terms that contain y′ .
  3. Solve linearly for y′ .

For a level set F(x,y)=C with Fy≠0 ,

dydx=−FxFy.

Conditions / Assumptions

Worked Example

Circle

From x2+y2=100 ,

2x+2yy′=0⟹y′=−xy(y≠0).

Exponential relation (correct isolation)

From exy+x2−y2=5 ,

exy(y+xy′)+2x−2yy′=0.

Collect y′ terms:

exyy+2x+y′(xexy−2y)=0,

so

y′=−exyy−2xxexy−2y,

provided the denominator is nonzero. Equivalently, y′=−Fx/Fy with F=exy+x2−y2 .

Implicit partial derivatives

For x3+y3+z3+6xyz=1 with z=z(x,y) ,

∂z∂x=−(x2+2yz)z2+2xy,∂z∂y=−(y2+2xz)z2+2xy,

when z2+2xy≠0 .

Common Mistakes

Connections

References

Implicit differentiation is developed in OpenStax Calculus Volume 1; the multivariable form uses the implicit function theorem setup in Volume 3.[1]


  1. OpenStax, Calculus Volume 1, Section 3.8; Calculus Volume 3, Section 4.8, https://openstax.org/details/books/calculus-volume-1 ↩︎