The Central Idea
An explicit relationship solves one variable directly:
\[ y=f(x). \]An implicit relationship instead gives an equation involving the variables:
\[ \boxed{ F(x,y)=0 }. \]The equation may still determine \(y\) locally as a function of \(x\), even when no convenient explicit formula has been written down.
Explicit versus Implicit Form
The equation
\[ y=x^2+1 \]is explicit.
The same relationship can be written implicitly as
\[ y-x^2-1=0. \]Thus,
\[ \boxed{ F(x,y)=y-x^2-1 }. \]Why Implicit Form Matters
Many important relationships are naturally implicit.
Examples include:
- circles and ellipses;
- spheres and other level surfaces;
- constraint equations;
- equilibrium conditions;
- conservation laws;
- systems that are difficult or impossible to solve explicitly in elementary form.
Example: A Circle
The unit circle is
\[ x^2+y^2=1. \]In implicit form,
\[ \boxed{ F(x,y)=x^2+y^2-1=0 }. \]Solving explicitly for \(y\) produces two branches:
\[ y=\pm\sqrt{1-x^2}. \]The implicit equation describes the full circle at once.
Implicit Differentiation
Suppose
\[ F(x,y)=0 \]and \(y\) depends on \(x\).
Differentiate both sides with respect to \(x\), remembering that \(y\) is itself a function of \(x\).
Chain Rule for an Implicit Relationship
Differentiating
\[ F(x,y(x))=0 \]gives
\[ F_x+F_y\frac{dy}{dx}=0. \]Therefore, whenever \(F_y\neq0\),
\[ \boxed{ \frac{dy}{dx} = -\frac{F_x}{F_y} }. \]Example: Slope on a Circle
Let
\[ F(x,y)=x^2+y^2-1. \]Then
\[ F_x=2x, \qquad F_y=2y. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = -\frac{x}{y} } \]wherever \(y\neq0\).
Interpret the Circle Formula
At the point
\[ \left(\frac35,\frac45\right), \]the slope is
\[ \frac{dy}{dx} = -\frac{3/5}{4/5} = -\frac34. \]Therefore,
\[ \boxed{ m_{\text{tangent}}=-\frac34 }. \]Why the Denominator Matters
The formula
\[ \frac{dy}{dx} = -\frac{F_x}{F_y} \]requires
\[ F_y\neq0. \]If \(F_y=0\), the relationship may fail to define \(y\) as a differentiable function of \(x\) near that point.
Vertical Tangents
On the unit circle,
\[ F_y=2y. \]At
\[ (1,0) \]and
\[ (-1,0), \]we have
\[ F_y=0. \]These are precisely the points where the circle has vertical tangent lines and cannot locally be written as a single differentiable function \(y=f(x)\).
Solving Instead for \(x\)
If
\[ F_x\neq0, \]the relationship may instead define \(x\) locally as a function of \(y\).
Differentiating with respect to \(y\) gives
\[ F_x\frac{dx}{dy}+F_y=0. \]Therefore,
\[ \boxed{ \frac{dx}{dy} = -\frac{F_y}{F_x} }. \]Two Local Viewpoints
At many regular points, an implicit curve can be viewed locally in one of two ways:
\[ y=f(x) \]when \(F_y\neq0\), or
\[ x=g(y) \]when \(F_x\neq0\).
The appropriate viewpoint depends on which partial derivative is nonzero.
Gradient Geometry
For the level curve
\[ F(x,y)=0, \]the gradient
\[ \boxed{ \nabla F = \langle F_x,F_y\rangle } \]is normal to the curve wherever
\[ \nabla F\neq\mathbf{0}. \]Tangent Vectors
If
\[ \mathbf{t}=\langle a,b\rangle \]is tangent to the curve, then
\[ \boxed{ \nabla F\cdot\mathbf{t}=0 }. \]Thus the gradient gives the local normal direction while tangent vectors lie in the perpendicular direction.
Recovering the Slope from the Gradient
A tangent direction to the graph \(y=y(x)\) is
\[ \mathbf{t} = \left\langle 1,\frac{dy}{dx} \right\rangle. \]Orthogonality gives
\[ F_x + F_y\frac{dy}{dx} = 0. \]Therefore,
\[ \frac{dy}{dx} = -\frac{F_x}{F_y}. \]The slope formula is therefore another expression of gradient orthogonality.
Example: Ellipse
Consider
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1. \]Define
\[ F(x,y) = \frac{x^2}{9} + \frac{y^2}{4} - 1. \]Then
\[ F_x=\frac{2x}{9} \]and
\[ F_y=\frac{y}{2}. \]Ellipse Slope
Therefore,
\[ \frac{dy}{dx} = - \frac{2x/9}{y/2}. \]Simplifying,
\[ \boxed{ \frac{dy}{dx} = -\frac{4x}{9y} }. \]Implicit Relationships in Three Variables
A relation
\[ \boxed{ F(x,y,z)=0 } \]usually describes a surface in three-dimensional space.
Under suitable conditions, one variable may be solved locally as a function of the other two.
Locally Solving for \(z\)
Suppose
\[ F(x,y,z)=0 \]and locally
\[ z=z(x,y). \]Differentiate with respect to \(x\):
\[ F_x+F_z z_x=0. \]Therefore, if \(F_z\neq0\),
\[ \boxed{ z_x = -\frac{F_x}{F_z} }. \]Partial with Respect to \(y\)
Differentiating the same relationship with respect to \(y\) gives
\[ F_y+F_z z_y=0. \]Therefore,
\[ \boxed{ z_y = -\frac{F_y}{F_z} } \]whenever \(F_z\neq0\).
Example: Sphere as an Implicit Surface
Consider
\[ x^2+y^2+z^2=9. \]Define
\[ F(x,y,z)=x^2+y^2+z^2-9. \]Then
\[ F_x=2x, \qquad F_y=2y, \qquad F_z=2z. \]Implicit Partial Derivatives on the Sphere
Wherever \(z\neq0\),
\[ \boxed{ z_x=-\frac{x}{z} } \]and
\[ \boxed{ z_y=-\frac{y}{z} }. \]Tangent Plane from Implicit Derivatives
If the surface is locally written as
\[ z=z(x,y), \]then the tangent plane at
\[ (a,b,c) \]is
\[ z-c = z_x(a,b)(x-a) + z_y(a,b)(y-b). \]Tangent Plane from the Gradient
The same tangent plane can be obtained more directly from
\[ \nabla F(a,b,c). \]Since the gradient is normal to the level surface,
\[ \boxed{ F_x(a,b,c)(x-a) + F_y(a,b,c)(y-b) + F_z(a,b,c)(z-c) = 0 }. \]Example: Sphere Tangent Plane
On
\[ x^2+y^2+z^2=9, \]consider
\[ (1,2,2). \]The gradient is
\[ \nabla F(1,2,2) = \langle2,4,4\rangle. \]Therefore,
\[ 2(x-1)+4(y-2)+4(z-2)=0. \]Equivalently,
\[ \boxed{ x+2y+2z=9 }. \]Implicit Function Theorem: The Core Idea
The implicit function theorem gives conditions under which an implicit equation can actually be solved locally for one variable as a differentiable function of the others.
In two variables, if
\[ F(a,b)=0 \]and
\[ \boxed{ F_y(a,b)\neq0 }, \]then under standard smoothness assumptions the equation
\[ F(x,y)=0 \]defines \(y\) locally as a differentiable function of \(x\).
Local, Not Necessarily Global
The implicit function theorem is a local result.
A relationship may define \(y\) as a function of \(x\) near one point while failing to do so over the entire curve.
The circle is a standard example.
Circle as a Local Graph
Near the top of the unit circle,
\[ (0,1), \]we have
\[ F_y=2. \]Therefore, the circle can locally be written as
\[ y=\sqrt{1-x^2}. \]Near the bottom point,
\[ (0,-1), \]it can locally be written as
\[ y=-\sqrt{1-x^2}. \]Failure at a Vertical Tangent
At
\[ (1,0), \]the circle has
\[ F_y=0. \]It cannot be represented locally as a differentiable graph
\[ y=f(x) \]through that point.
It can, however, be represented locally as
\[ x=g(y) \]because
\[ F_x=2\neq0. \]Three-Variable Implicit Function Theorem
Suppose
\[ F(a,b,c)=0 \]and
\[ \boxed{ F_z(a,b,c)\neq0 }. \]Under standard smoothness assumptions, the equation
\[ F(x,y,z)=0 \]defines \(z\) locally as a differentiable function
\[ z=z(x,y). \]Regular Points
A point on
\[ F(x,y)=0 \]or
\[ F(x,y,z)=0 \]is called a regular point when
\[ \boxed{ \nabla F\neq\mathbf{0} }. \]Regular points have a well-defined local normal direction.
Singular Points
A point where
\[ \nabla F=\mathbf{0} \]is a singular point of the level set.
At such points, the usual tangent geometry may fail or become ambiguous.
Example: Singular Curve
Consider
\[ F(x,y)=y^2-x^3. \]The curve
\[ y^2=x^3 \]has
\[ \nabla F = \langle-3x^2,2y\rangle. \]At the origin,
\[ \nabla F(0,0)=\mathbf{0}. \]The origin is a singular point.
Implicit Differentiation with More Complicated Expressions
Consider
\[ e^{xy}+x^2+y^2=5. \]Differentiate with respect to \(x\):
\[ e^{xy} \left( y+x\frac{dy}{dx} \right) + 2x + 2y\frac{dy}{dx} = 0. \]Solve for the Implicit Derivative
Collect the terms containing
\[ \frac{dy}{dx}. \]We obtain
\[ \left( xe^{xy}+2y \right) \frac{dy}{dx} = - \left( ye^{xy}+2x \right). \]Therefore,
\[ \boxed{ \frac{dy}{dx} = - \frac{ye^{xy}+2x} {xe^{xy}+2y} }. \]Higher Implicit Derivatives
Once
\[ \frac{dy}{dx} \]has been found implicitly, we can differentiate again to obtain
\[ \frac{d^2y}{dx^2}. \]The calculation is usually more involved because the first derivative is itself a function of both \(x\) and \(y\).
Second Derivative of an Implicit Curve
Starting from
\[ F_x+F_y y'=0, \]differentiate again with respect to \(x\):
\[ F_{xx} + 2F_{xy}y' + F_{yy}(y')^2 + F_y y'' = 0 \]when the relevant second partials are sufficiently regular.
Therefore,
\[ \boxed{ y'' = - \frac{ F_{xx} + 2F_{xy}y' + F_{yy}(y')^2 }{F_y} }. \]Implicit Relationships and Constraints
The previous lesson used constraint equations such as
\[ g(x,y)=c. \]These are implicit relationships.
The gradient
\[ \nabla g \]is normal to the feasible curve or surface, which is exactly why it enters the Lagrange multiplier condition.
Implicit Relationships and Level Sets
Every level set
\[ f(x,y)=c \]can be rewritten as
\[ F(x,y)=f(x,y)-c=0. \]Thus level curves and level surfaces are special cases of implicit relationships.
Implicit Relationships and Differential Forms
For
\[ F(x,y)=0, \]the total differential along the relationship satisfies
\[ dF=0. \]Therefore,
\[ \boxed{ F_x\,dx+F_y\,dy=0 }. \]Solving for \(dy/dx\) recovers the implicit derivative formula.
Three-Variable Differential Relation
For
\[ F(x,y,z)=0, \]infinitesimal changes along the surface satisfy
\[ \boxed{ F_x\,dx + F_y\,dy + F_z\,dz = 0 }. \]This is the differential expression of tangent-plane motion.
Tangent Space Viewpoint
For a regular level surface
\[ F(x,y,z)=0, \]tangent vectors
\[ \mathbf{v} \]satisfy
\[ \boxed{ \nabla F\cdot\mathbf{v}=0 }. \]The tangent plane is therefore the set of directions orthogonal to the gradient.
Systems of Implicit Equations
Several equations may define several variables implicitly.
For example,
\[ F(x,y,u,v)=0 \]and
\[ G(x,y,u,v)=0 \]may determine \(u\) and \(v\) locally as functions of \(x\) and \(y\).
The relevant derivative information is organized using a Jacobian matrix.
Jacobian Condition for a System
To solve locally for
\[ (u,v) \]in terms of
\[ (x,y), \]a key condition is that
\[ \boxed{ \frac{\partial(F,G)} {\partial(u,v)} \neq0 }. \]This is the multivariable system version of requiring a nonzero derivative in the one-equation case.
Connection to Coordinate Changes
The previous lesson used nonzero Jacobians to guarantee local invertibility of coordinate transformations.
The implicit function theorem uses the same local linear algebra: a nonsingular derivative block allows one group of variables to be solved locally in terms of another.
Implicit Sensitivity
Suppose a relationship
\[ F(x,y)=0 \]links two quantities.
Then
\[ \frac{dy}{dx} = -\frac{F_x}{F_y} \]measures the local sensitivity of \(y\) to changes in \(x\) while remaining on the relationship.
Economic Equilibrium Interpretation
Suppose an equilibrium condition is written
\[ F(p,q)=0, \]where \(p\) is price and \(q\) is quantity.
Then
\[ \frac{dq}{dp} = -\frac{F_p}{F_q} \]describes the local equilibrium response of quantity to a change in price, provided \(F_q\neq0\).
Engineering Constraint Interpretation
If design variables satisfy
\[ F(x,y,z)=0, \]then their allowable infinitesimal changes must satisfy
\[ F_x\,dx+F_y\,dy+F_z\,dz=0. \]This equation describes how one variable must adjust when the others change while maintaining the design constraint.
Common Error: Treating \(y\) as Constant
In implicit differentiation,
\[ y=y(x). \]Therefore,
\[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx}, \]not merely \(0\) and not merely \(2y\).
Common Error: Forgetting the Chain Rule
If
\[ \sin y \]appears in an implicit equation, then
\[ \frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx}. \]The factor \(dy/dx\) is essential.
Common Error: Dividing by Zero
The formula
\[ \frac{dy}{dx} = -\frac{F_x}{F_y} \]cannot be used when
\[ F_y=0. \]Such points may correspond to vertical tangents, singular points, or a different local graph orientation.
Common Error: Assuming a Global Function Exists
An implicit relationship may define several branches.
The circle cannot be written globally as one real-valued function
\[ y=f(x) \]over its entire extent.
Local solvability does not imply global single-valuedness.
Common Error: Ignoring Singular Points
If
\[ \nabla F=\mathbf{0}, \]the standard tangent normal is unavailable.
Singular points require separate geometric or algebraic analysis.
A Reliable Implicit-Differentiation Procedure
- Write the relationship as an equation involving the dependent and independent variables.
- Differentiate both sides with respect to the independent variable.
- Use the chain rule on every occurrence of the dependent variable.
- Collect all terms containing the desired derivative.
- Factor out that derivative.
- Solve for it.
- Check the denominator and the point of evaluation.
A Reliable Multivariable Implicit Procedure
- Write the relationship as \(F=0\).
- Decide which variable is being treated as dependent.
- Differentiate with respect to each independent variable as needed.
- Apply the chain rule through the dependent variable.
- Solve for the desired partial derivatives.
- Verify that the relevant partial derivative of \(F\) is nonzero.
- Use the gradient for tangent and normal geometry when useful.
Example: Full Two-Variable Workflow
Consider
\[ x^2+xy+y^2=7. \]Differentiate with respect to \(x\):
\[ 2x + \left( y+x\frac{dy}{dx} \right) + 2y\frac{dy}{dx} = 0. \]Collect derivative terms:
\[ (x+2y)\frac{dy}{dx} = -(2x+y). \]Complete the Two-Variable Workflow
Therefore,
\[ \boxed{ \frac{dy}{dx} = -\frac{2x+y}{x+2y} }. \]This is valid wherever
\[ x+2y\neq0. \]Example: Full Three-Variable Workflow
Suppose
\[ x^2+y^2+z^2+xyz=10 \]defines \(z\) locally as a function of \(x\) and \(y\).
Define
\[ F(x,y,z) = x^2+y^2+z^2+xyz-10. \]Then
\[ F_x=2x+yz, \] \[ F_y=2y+xz, \] \[ F_z=2z+xy. \]Complete the Three-Variable Workflow
Whenever
\[ 2z+xy\neq0, \]we have
\[ \boxed{ z_x = -\frac{2x+yz}{2z+xy} } \]and
\[ \boxed{ z_y = -\frac{2y+xz}{2z+xy} }. \]Practice 1: Basic Implicit Derivative
Find
\[ \frac{dy}{dx} \]for
\[ x^2+y^2=25. \]Practice 2: Product Term
Find
\[ \frac{dy}{dx} \]for
\[ x^2+xy+y^2=3. \]Practice 3: Exponential Relationship
Find
\[ \frac{dy}{dx} \]for
\[ e^{x+y}+xy=4. \]Practice 4: Tangent Slope
Find the tangent slope to
\[ x^2+y^2=25 \]at
\[ (3,4). \]Practice 5: Gradient Normal
Find a normal vector to
\[ x^2+2y^2=6 \]at
\[ (2,1). \]Practice 6: Implicit Partial Derivatives
Suppose
\[ x^2+y^2+z^2=16 \]defines \(z=z(x,y)\).
Find \(z_x\) and \(z_y\).
Practice 7: Tangent Plane
Find the tangent plane to
\[ x^2+y^2+z^2=14 \]at
\[ (1,2,3). \]Practice 8: Local Solvability
For
\[ F(x,y)=x^2+y^2-1, \]determine whether the implicit function theorem guarantees that \(y\) can be solved locally as a differentiable function of \(x\) near
\[ (0,1). \]Practice 9: Singular Point
For
\[ F(x,y)=y^2-x^3, \]determine whether the origin is a regular or singular point.
Practice 10: Identify the Error
A student differentiates
\[ x^2+y^2=1 \]and writes
\[ 2x+2y=0. \]Explain the error.
Practice Answers
Practice 1
\[ 2x+2y\frac{dy}{dx}=0. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = -\frac{x}{y} }. \]Practice 2
Differentiate:
\[ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = -\frac{2x+y}{x+2y} }. \]Practice 3
Differentiate:
\[ e^{x+y} \left( 1+\frac{dy}{dx} \right) + y + x\frac{dy}{dx} = 0. \]Collect derivative terms:
\[ \left( e^{x+y}+x \right) \frac{dy}{dx} = - \left( e^{x+y}+y \right). \]Thus,
\[ \boxed{ \frac{dy}{dx} = - \frac{e^{x+y}+y} {e^{x+y}+x} }. \]Practice 4
\[ \frac{dy}{dx} = -\frac{x}{y}. \]At \((3,4)\),
\[ \boxed{ \frac{dy}{dx} = -\frac34 }. \]Practice 5
Let
\[ F(x,y)=x^2+2y^2. \]Then
\[ \nabla F = \langle2x,4y\rangle. \]At \((2,1)\),
\[ \boxed{ \nabla F(2,1)=\langle4,4\rangle }. \]Practice 6
Let
\[ F=x^2+y^2+z^2-16. \]Since
\[ F_x=2x, \quad F_y=2y, \quad F_z=2z, \]we have
\[ \boxed{ z_x=-\frac{x}{z} } \]and
\[ \boxed{ z_y=-\frac{y}{z} }. \]Practice 7
Define
\[ F=x^2+y^2+z^2. \]Then
\[ \nabla F(1,2,3) = \langle2,4,6\rangle. \]Therefore,
\[ \boxed{ 2(x-1)+4(y-2)+6(z-3)=0 }. \]Practice 8
We have
\[ F_y=2y. \]At \((0,1)\),
\[ F_y=2\neq0. \]Therefore, under the usual smoothness assumptions, the implicit function theorem guarantees that \(y\) can be solved locally as a differentiable function of \(x\).
Practice 9
We have
\[ \nabla F = \langle-3x^2,2y\rangle. \]At the origin,
\[ \nabla F(0,0)=\mathbf{0}. \]Therefore,
\[ \boxed{ (0,0)\text{ is singular} }. \]Practice 10
The variable \(y\) depends on \(x\), so the chain rule is required:
\[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx}. \]The correct differentiation is
\[ \boxed{ 2x+2y\frac{dy}{dx}=0 }. \]What to Remember
Implicit relationships describe dependence without requiring one variable to be solved explicitly in advance.
- Write an implicit curve or surface as \(F=0\).
- If \(F_y\neq0\), then locally \(dy/dx=-F_x/F_y\).
- If \(F_z\neq0\), then a surface can locally define \(z=z(x,y)\) with \(z_x=-F_x/F_z\) and \(z_y=-F_y/F_z\).
- The gradient is normal to regular level curves and surfaces.
- The implicit function theorem provides local solvability under nondegeneracy and smoothness conditions.
- Local solvability does not guarantee one global explicit branch.
- Singular points occur where the gradient vanishes and require special attention.
The central relationship is
\[ \boxed{ F=0 \Longrightarrow \text{local dependence} + \text{gradient geometry} + \text{implicit sensitivity} }. \]Where We Go Next
Implicit relationships describe how scalar variables are linked by equations. The next step is to study functions whose outputs are vectors rather than single scalar values.
The next lesson, Vector-Valued Functions, develops parameterized curves and trajectories, componentwise differentiation and integration, velocity, acceleration, and the geometry of motion in several dimensions.