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 Fy0 ,

dydx=FxFy.

Conditions / Assumptions

Worked Example

Circle

From x2+y2=100 ,

2x+2yy=0y=xy(y0).

Exponential relation (correct isolation)

From exy+x2y2=5 ,

exy(y+xy)+2x2yy=0.

Collect y terms:

exyy+2x+y(xexy2y)=0,

so

y=exyy2xxexy2y,

provided the denominator is nonzero. Equivalently, y=Fx/Fy with F=exy+x2y2 .

Implicit partial derivatives

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

zx=(x2+2yz)z2+2xy,zy=(y2+2xz)z2+2xy,

when z2+2xy0 .

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 ↩︎