Implicit differentiation

Implicit differentiation differentiates an identity

F (x, y (x)) = 0

along a differentiable branch and uses the chain rule to solve for its derivative:

F_{x} + F_{y} y^{'} = 0, y^{'} = - \frac{F _{x}}{F _{y}} (F_{y} \neq = 0) .

(1)

The partial derivatives

F_{x}, F_{y}

treat

x, y

as independent coordinates before evaluating on the branch. The implicit function theorem justifies the branch's local existence and regularity under its hypotheses; differentiating alone does not establish either. For the unit circle at

(3/5, 4/5)

, this calculation yields the tangent line

y - 4/5 = - (3/4) (x - 3/5)

Table of contents
- Reciprocity rule for constrained partial derivatives Implicit differentiation
- Cyclic partial derivative identity Implicit differentiation

Reciprocity rule for constrained partial derivatives ( $(\partial y / \partial x)_{z} (\partial x / \partial y)_{z} = 1$ )

 0  0

On a regular level set

F (x, y, z) = 0

, holding

z

fixed gives

F_{x} d x + F_{y} d y = 0

. If

F

is continuously differentiable and both

F_{x}

and

F_{y}

are nonzero, the implicit function theorem gives

(\partial y / \partial x)_{z} = - F_{x} / F_{y}

and

(\partial x / \partial y)_{z} = - F_{y} / F_{x}

. Multiplication proves the identity. The same held-fixed variable is essential, and a zero denominator invalidates the corresponding quotient rather than creating an infinite factor that may be cancelled formally.

Cyclic partial derivative identity ( $(\partial x / \partial y)_{z} (\partial y / \partial z)_{x} (\partial z / \partial x)_{y} = - 1$ )

 0  0

On a level set

f (x, y, z) = c

, implicit differentiation gives the three ratios

- f_{y} / f_{x}

- f_{z} / f_{y}

and

- f_{x} / f_{z}

. Their product is

- 1

wherever all the required coordinate representations are differentiable with nonzero denominators. At coordinate singularities the quotient product need not be defined.