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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.

In vacuum, magnetization is zero and magnetic flux density obeys $\mathbf{B}={\mu }_0\mathbf{H}$. The applied quantities ${\mathbf{H}}_{\mathrm{app}}$ and ${\mathbf{B}}_0={\mu }_0{\mathbf{H}}_{\mathrm{app}}$ mean the field of the external sources at the sample position with the sample absent. The internal magnetic field intensity includes the sample's demagnetizing field; its internal magnetic flux density additionally includes ${\mu }_0\mathbf{M}$. A field quoted in the tesla is a $B$ quantity, even when an instrument informally calls it “$H$”.

For a uniformly magnetized ellipsoid, along a principal axis with demagnetizing factor $N_d$ in the International System of Units,For a sphere, $N_d=1/3$. The axial value approaches zero in the long slender ellipsoid limit and the normal value approaches one in the thin flat ellipsoid limit. Thus a stipulated intrinsic volume magnetic susceptibility ${\chi }_v=3$ would yield apparent values $1.5$ in a sphere and approximately $3$ in a long axial sample. These are calculated geometry effects, not measured material values. For fixed $N_d$, implicit differentiation gives the same conversion for local differential magnetic susceptibility, $dM/d{H}_{\mathrm{app}}={\chi }_{\mathrm{diff}}/(1+N_d{\chi }_{\mathrm{diff}})$.

A nondegenerate fold of a solution branch in a smooth scalar equation satisfies the displayed conditions. The implicit function theorem still gives $\lambda =\lambda (x)$ because $F_{\lambda }\ne 0$. Implicit differentiation gives ${\lambda }^{\prime }(x_{\ast })=0$ and ${\lambda }^{\prime \prime }(x_{\ast })=-F_{xx}/F_{\lambda }\ne 0$: the parameter has a genuine local extremum on the curve.

In the cubic polynomial, $F_x=0$ gives $x_{\ast }=\pm 1/\sqrt{3}$, and substituting into $F=0$ gives ${\lambda }_{\ast }=2x_{\ast }/3$. Furthermore, $F_{xx}=6x_{\ast }\ne 0$. Yet $\nabla F=(3x^2-1,1)$ never vanishes, so the full solution branch remains a regular level set everywhere. The obstruction belongs to the projection onto $\lambda$, not to the curve itself.

For systems, the cited continuation account uses a one-dimensional kernel of $F_u$ and $F_{\lambda }$ outside its image of a linear map to define a simple fold (its Definition 2.2). These conditions make $[F_u{F}_{\lambda }]$ have full rank and support a transverse correction. A nonzero second-order term along the kernel is additionally needed to describe a nondegenerate quadratic turning point; in the scalar example this is $F_{xx}\ne 0$.^[ref]

An implicit equation specifies a relation between variables without isolating a designated output as an explicit function of the inputs. For example, $x^2+y^2=1$ specifies a circle. On $-1<x<1$, it determines the two real branches $y=\pm \sqrt{1-x^2}$, each an implicit function when one branch is selected. The implicit function theorem gives local conditions for a unique differentiable branch; implicit differentiation computes its derivative without first solving explicitly.

An implicit function is a function specified through a relation between its input and output, rather than an explicit output formula. Locally, an equation $F(x,y)=0$ may specify $y=g(x)$ if one output $y$ is selected for each input $x$ in a suitable domain. Different branches can define different functions; the relation itself need not give a unique output globally. For example, $x^2+y^2=1$ gives the two branches $y=\pm \sqrt{1-x^2}$ on $-1<x<1$.

The implicit function theorem gives sufficient conditions for a unique local differentiable branch, while implicit differentiation computes its derivative by the chain rule. If $F$ is continuously differentiable and $F_y\ne 0$, that derivative is $g^{\prime }(x)=-F_x/F_y$ along the branch.

An equation can define a function without giving an explicit formula for it. The implicit function theorem tells us when the constraint $F(x,y)=0$ determines $y$ uniquely from nearby $x$, and makes the resulting function continuously differentiable. It is useful when solving an equation explicitly is cumbersome or impossible: the partial derivatives already reveal whether local recovery is possible.

Let $F$ be a continuously differentiable map from an open subset of ${\mathbb{R}}^m\times {\mathbb{R}}^n$ to ${\mathbb{R}}^n$. Suppose $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map. There are open neighbourhoods $U$ of $x_0$ and $V$ of $y_0$ and a unique continuously differentiable function $g:U\to V$ such that, for $(x,y)\in U\times V$,Thus the nearby level set is a graph of a function. The derivative of this graph isThe matrix inverse acts on the left; the order matters in several dimensions.

For example, take the unit circle, written as the level set $F(x,y)=x^2+y^2-1=0$. At $(x_0,y_0)=(3/5,4/5)$, $F_y=2y_0=8/5\ne 0$, so the local upper branch is $y=g(x)=\sqrt{1-x^2}$. Implicit differentiation givesThis tangent line slope is available directly from the constraint. The negative square root branch is elsewhere on the same circle; it does not destroy uniqueness near the selected upper point.

At $(1,0)$, however, $F_y=0$ and the circle cannot be represented locally as one function $y=g(x)$: just to the left of $x=1$ there are two nearby values of $y$. The circle is still smooth there. Since $F_x=2\ne 0$, use the other coordinate instead: $x=h(y)=\sqrt{1-y^2}$, with $h^{\prime }(0)=0$. A failed coordinate choice is different from a geometric singularity.

The derivative condition is sufficient, not necessary for every possible implicit graph. For $F(x,y)=y^3$, the entire zero set is the smooth graph $y=0$, even though $F_y$ vanishes there. In this case the defining equation itself has a degenerate derivative; the theorem's test gives no conclusion.

The connection to the inverse function theorem is particularly elegant. Augment the constraint to the map $H(x,y)=(x,F(x,y))$, whose Jacobian matrix isIts determinant is $\det D_{y}F$. When this is nonzero, locally invert $H$ and set its second output to zero. The recovered second input is $g(x)$. In this way an implicit constraint becomes an ordinary inverse function problem. ^[ref]

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

For a monic polynomial $p(x)={\prod }_i(x-a_i)$ with distinct ordered real roots of a polynomial, each critical point $\xi$ lies between two consecutive roots by Rolle's theorem. Its logarithmic derivative gives ${\sum }_i(\xi -a_i{)}^{-1}=0$. Implicit differentiation gives the displayed formula, so moving any root to the right moves every ordered critical point to the right, while the roots remain distinct. Iterating this statement for derivatives preserves componentwise ordering of their roots. Weak inequalities extend by continuity to repeated roots.

Use implicit differentiation while holding $x$ fixed. Writing $y_z=(\partial y/\partial z{)}_x$, the constraint givesThe chain rule applied to $v=x{y}^3$ now yieldsThis local formula requires $2xy+z^2\ne 0$, as in the implicit function theorem; that denominator equals $3$ at the specified point.

Implicit differentiation gives $2y^{\prime }=15x^2+(\cos y)y^{\prime }$. ThereforeFor real $y$, the denominator is at least one, so division introduces no singularity. In fact the function $2y-\sin y$ is strictly increasing, consistent with a unique real branch $y(x)$.

Let $D=y(3y-2)$. Implicit differentiation of the two constraints gives the linear systemAdding them, then substituting back, gives at regular pointsThe relevant Jacobian determinant is $2(1+a)yz(3y-2)$. Thus the special parameter value is $\boxed{a=-1}$, assuming the other factors are nonzero. At that parameter, adding the original constraints gives $(x+1{)}^2=0$: the intersection curve lies wholly in the plane $x=-1$ and cannot be parametrized by $x$ on an interval. Calling its derivatives “infinite” describes a vertical projection, not a finite-valued differentiable pair $y(x),z(x)$ at $a=-1$; the displayed ratios then have simultaneous zero numerators on the intersection and are undefined.

There are also point-dependent failures when $y=0$, $y=2/3$ or $z=0$. These can produce vertical tangents or require a different local parameter, and simultaneous numerator zeros need separate analysis. The implicit function theorem guarantees finite $y^{\prime },z^{\prime }$ only where the stated Jacobian determinant is nonzero. This distinction is the geometry of a singular projection of an implicit curve.

The projective completion has identity $O=[0:1:0]$. For this Weierstrass equation of an elliptic curve, negation isFor distinct affine points $P,Q$, draw their chord; for $P=Q$, use the tangent with its intersection multiplicity. If the third intersection is $R$, define $P\oplus Q=-R$. A vertical chord pairs $P$ with $-P$ and gives their sum $O$; also $P\oplus O=P$. A vertical tangent at a point satisfying $2y+1=0$ gives $2P=O$.

For a nonvertical line $y=mx+b$, substitution into the cubic givesThus if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, the sum has coordinatesHere $m=(y_2-y_1)/(x_2-x_1)$ for a chord, and implicit differentiation givesfor a nonvertical tangent. These formulas include the repeated-root case. Associativity can be seen without a lengthy coordinate calculation: a line cuts the cubic in a divisor equivalent to $3(O)$, and $P\mapsto [(P)-(O)]$ identifies the curve with its degree-zero divisor class group. The chord-and-tangent operation becomes addition of divisor classes, hence is associative.

Work locally where the needed derivatives of the level set can be represented by the implicit function theorem. Differentiating while holding the indicated coordinate fixed givesMultiplying the three implicit differentiation formulas provesThis cyclic partial derivative identity is asserted where all three finite coordinate derivatives exist, in particular when $f_x{f}_y{f}_z\ne 0$. It is not a globally defined product at points where one of the necessary coordinate parametrizations is singular.

The partial derivatives of $f$ are $f_x=2x$, $f_y=4y(a-y^2)$ and $f_z=2z$. Therefore the three implicit differentiation formulas areTheir product is $-1$ wherever $xyz(y^2-a)\ne 0$. The excluded factors identify exactly where these quotient formulas cannot all represent finite partial derivatives.

Assume that $\mathbf{v}$ is constant. SetImplicit differentiation of $t=\tau +R(\tau )/c$ givesSince $d\mathbf{R}/d\tau =-\mathbf{v}$, direct differentiation of $D$ yieldsThereforeThe Lienard-Wiechert potentials for uniform motion have $\varphi =q/(4\pi {ϵ}_{0}D)$ and $\mathbf{A}=\mathbf{v}\varphi /c^2$. The preceding identity consequently impliesAs $\mathbf{v}$ is constant,Thus the explicit uniformly moving potentials satisfy the Lorenz gauge.

Solve the two algebraic constraints for the squared coordinates:On any nonsingular real branch, implicit differentiation at fixed $u$ gives $(\partial x/\partial z{)}_u=z/(3x)$ and $(\partial y/\partial z{)}_u=z/(3y)$. Differentiating $z^2=2x^2+y^2$ with the other original coordinate fixed gives $(\partial z/\partial x{)}_y=2x/z$ and $(\partial z/\partial y{)}_x=y/z$. ThereforeThis is the multivariable chain rule applied to $z(x(z,u),y(z,u))=z$. These formulas assume the relevant denominators are nonzero and the chosen variables are valid local coordinates.

For $F(x,y)=e^{xy}+y\ln x-x^2$, the real domain has $x>0$. Its partial derivatives are $F_x=y{e}^{xy}+y/x-2x$ and $F_y=x{e}^{xy}+\ln x$. At $(1,0)$ they equal $-2$ and $1$, respectively. Thus implicit differentiation gives $y^{\prime }=2$, and the tangent line is

Use implicit differentiation on $F(x,y)=x{y}^3-y{x}^3-6=0$. Along the curve,At $(1,2)$, $F_x=2$ and $F_y=11\ne 0$, so the slope is $-2/11$. The tangent line is

The point on the parabola is $(1,2)$. Implicit differentiation gives $2ydy/dx=4$, so the tangent line has slope one there. Its perpendicular normal line to a plane curve therefore has slope $-1$. The required line is

Use implicit differentiation, treating $y$ as a function of $x$:Where $y\ne 0$, this is $\boxed{dy/dx=-2x/(3y)}$. At the two points where $y=0$, the ellipse has vertical tangent lines rather than a finite slope.