In multivariable calculus, the implicit function theorem is a theorem that provides sufficient conditions under which a planar curve specified by F ( x , y ) = 0 {\displaystyle F(x,y)=0} can also be specified as the graph of a function f {\displaystyle f} , so that for each point ( x , y ) {\displaystyle (x,y)} on part of the curve, one has y = f ( x ) {\displaystyle y=f(x)} . An example is the unit circle, whose points ( x , y ) {\displaystyle (x,y)} satisfy x 2 + y 2 − 1 = 0 {\displaystyle x^{2}+y^{2}-1=0} , which can locally be solved (if y > 0 {\displaystyle y>0} ) by y = 1 − x 2 {\displaystyle y={\sqrt {1-x^{2}}}} , expressing the top semicircle as a graph. It is not always possible to solve the equation F ( x , y ) = 0 {\displaystyle F(x,y)=0} for y {\displaystyle y} algebraically, and the implicit function theorem gives analytic conditions under which there exists a function f {\displaystyle f} whose graph belongs to the given curve, and, in some formulations, also gives a way of constructing approximations to f {\displaystyle f} . More generally, given a system of m equations fi (x1, ..., xn, y1, ..., ym) = 0, i = 1, ..., m (often abbreviated into F(x, y) = 0), the theorem states that, under a mild condition on the partial derivatives (with respect to each yi ) at a point, the m variables yi are differentiable functions of the xj in some neighbourhood of the point. As these functions generally cannot be expressed in closed form, they are implicitly defined by the equations, and this motivated the name of the theorem. In other words, under a mild condition on the partial derivatives, the set of zeros of a system of equations is locally the graph of a function.
History Augustin-Louis Cauchy (1789–1857) is credited with the first rigorous form of the implicit function theorem. Ulisse Dini (1845–1918) generalized the real-variable version of the implicit function theorem to the context of functions of any number of real variables.
Two variables case Let f : R 2 → R {\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} } be a continuously differentiable function defining the implicit equation of a curve f ( x , y ) = 0 {\displaystyle f(x,y)=0} . Let ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} be a point on the curve, that is, a point such that f ( x 0 , y 0 ) = 0 {\displaystyle f(x_{0},y_{0})=0} . In this simple case, the implicit function theorem can be stated as follows:
Proof. By differentiating the equation f ( x , φ ( x ) ) = 0 {\displaystyle f(x,\varphi (x))=0} , one gets
∂ f ∂ x ( x , φ ( x ) ) + φ ′ ( x ) ∂ f ∂ y ( x , φ ( x ) ) = 0. {\displaystyle {\frac {\partial f}{\partial x}}(x,\varphi (x))+\varphi '(x)\,{\frac {\partial f}{\partial y}}(x,\varphi (x))=0.}
and thus
φ ′ ( x ) = − ∂ f ∂ x ( x , φ ( x ) ) ∂ f ∂ y ( x , φ ( x ) ) . {\displaystyle \varphi '(x)=-{\frac {{\frac {\partial f}{\partial x}}(x,\varphi (x))}{{\frac {\partial f}{\partial y}}(x,\varphi (x))}}.}
This gives an ordinary differential equation for φ {\displaystyle \varphi } , with the initial condition φ ( x 0 ) = y 0 {\displaystyle \varphi (x_{0})=y_{0}} . Since ∂ f ∂ y ( x 0 , y 0 ) ≠ 0 , {\textstyle {\frac {\partial f}{\partial y}}(x_{0},y_{0})\neq 0,} the right-hand side of the differential equation is continuous. Hence, the Peano existence theorem applies so there is a (possibly non-unique) solution. To see why φ {\textstyle \varphi } is unique, note that the function g x ( y ) = f ( x , y ) {\textstyle g_{x}(y)=f(x,y)} is strictly monotone in a neighbourhood of x 0 , y 0 {\textstyle x_{0},y_{0}} (as ∂ f ∂ y ( x 0 , y 0 ) ≠ 0 {\textstyle {\frac {\partial f}{\partial y}}(x_{0},y_{0})\neq 0} ), thus it is injective. If φ , ϕ {\textstyle \varphi ,\phi } are solutions to the differential equation, then g x ( φ ( x ) ) = g x ( ϕ ( x ) ) = 0 {\textstyle g_{x}(\varphi (x))=g_{x}(\phi (x))=0} and by injectivity we get, φ ( x ) = ϕ ( x ) {\textstyle \varphi (x)=\phi (x)} .
First example
If we define the function f(x, y) = x2 + y2, then the equation f(x, y) = 1 cuts out the unit circle as the level set {(x, y) | f(x, y) = 1}. There is no way to represent the unit circle as the graph of a function of one variable y = g(x) because for each choice of x ∈ (−1, 1), there are two choices of y, namely ± 1 − x 2 {\displaystyle \pm {\sqrt {1-x^{2}}}} . However, it is possible to represent part of the circle as the graph of a function of one variable. If we let g 1 ( x ) = 1 − x 2 {\displaystyle g_{1}(x)={\sqrt {1-x^{2}}}} for −1 ≤ x ≤ 1, then the graph of y = g1(x) provides the upper half of the circle. Similarly, if g 2 ( x ) = − 1 − x 2 {\displaystyle g_{2}(x)=-{\sqrt {1-x^{2}}}} , then the graph of y = g2(x) gives the lower half of the circle. The implicit function theorem says that under some mild assumptions, functions like g1(x) and g2(x) always exist, even in situations where they cannot be written down by explicit formulas. It guarantees that g1(x) and g2(x) are differentiable, and it even works in situations where we do not have a formula for f(x, y).
General case Let f : R n + m → R m {\displaystyle f:\mathbb {R} ^{n+m}\to \mathbb {R} ^{m}} be a continuously differentiable function. We think of R n + m {\displaystyle \mathbb {R} ^{n+m}} as the Cartesian product R n × R m , {\displaystyle \mathbb {R} ^{n}\times \mathbb {R} ^{m},} and we write a point of this product as ( x , y ) = ( x 1 , … , x n , y 1 , … y m ) . {\displaystyle (\mathbf {x} ,\mathbf {y} )=(x_{1},\ldots ,x_{n},y_{1},\ldots y_{m}).} Starting from the given function f {\displaystyle f} , our goal is to construct a function g : R n → R m {\displaystyle g:\mathbb {R} ^{n}\to \mathbb {R} ^{m}} whose graph ( x , g ( x ) ) {\displaystyle ({\textbf {x}},g({\textbf {x}}))} is precisely the set of all ( x , y ) {\displaystyle ({\textbf {x}},{\textbf {y}})} such that f ( x , y ) = 0 {\displaystyle f({\textbf {x}},{\textbf {y}})={\textbf {0}}} . As noted above, this may not always be possible. We will therefore fix a point ( a , b ) = ( a 1 , … , a n , b 1 , … , b m ) {\displaystyle ({\textbf {a}},{\textbf {b}})=(a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})} which satisfies f ( a , b ) = 0 {\displaystyle f({\textbf {a}},{\textbf {b}})={\textbf {0}}} , and we will ask for a g {\displaystyle g} that works near the point ( a , b ) {\displaystyle ({\textbf {a}},{\textbf {b}})} . In other words, we want an open set U ⊂ R n {\displaystyle U\subset \mathbb {R} ^{n}} containing a {\displaystyle {\textbf {a}}} , an open set V ⊂ R m {\displaystyle V\subset \mathbb {R} ^{m}} containing b {\displaystyle {\textbf {b}}} , and a function g : U → V {\displaystyle g:U\to V} such that the graph of g {\displaystyle g} satisfies the relation f = 0 {\displaystyle f={\textbf {0}}} on U × V {\displaystyle U\times V} , and that no other points within U × V {\displaystyle U\times V} do so. In symbols,
{ ( x , g ( x ) ) ∣ x ∈ U } = { ( x , y ) ∈ U × V ∣ f ( x , y ) = 0 } . {\displaystyle \{(\mathbf {x} ,g(\mathbf {x} ))\mid \mathbf {x} \in U\}=\{(\mathbf {x} ,\mathbf {y} )\in U\times V\mid f(\mathbf {x} ,\mathbf {y} )=\mathbf {0} \}.}
To state the implicit function theorem, we need the Jacobian matrix of f {\displaystyle f} , which is the matrix of the partial derivatives of f {\displaystyle f} . Abbreviating ( a 1 , … , a n , b 1 , … , b m ) {\displaystyle (a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})} to ( a , b ) {\displaystyle ({\textbf {a}},{\textbf {b}})} , the Jacobian matrix is
( D f ) ( a , b ) = [ ∂ f 1 ∂ x 1 ( a , b ) ⋯ ∂ f 1 ∂ x n ( a , b ) ∂ f 1 ∂ y 1 ( a , b ) ⋯ ∂ f 1 ∂ y m ( a , b ) ⋮ ⋱ ⋮ ⋮ ⋱ ⋮ ∂ f m ∂ x 1 ( a , b ) ⋯ ∂ f m ∂ x n ( a , b ) ∂ f m ∂ y 1 ( a , b ) ⋯ ∂ f m ∂ y m ( a , b ) ] = [ X Y ] {\displaystyle (Df)(\mathbf {a} ,\mathbf {b} )=\left[{\begin{array}{ccc|ccc}{\frac {\partial f_{1}}{\partial x_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1}}{\partial x_{n}}}(\mathbf {a} ,\mathbf {b} )&{\frac {\partial f_{1}}{\partial y_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1}}{\partial y_{m}}}(\mathbf {a} ,\mathbf {b} )\\\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\{\frac {\partial f_{m}}{\partial x_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m}}{\partial x_{n}}}(\mathbf {a} ,\mathbf {b} )&{\frac {\partial f_{m}}{\partial y_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m}}{\partial y_{m}}}(\mathbf {a} ,\mathbf {b} )\end{array}}\right]=\left[{\begin{array}{c|c}X&Y\end{array}}\right]}
where X {\displaystyle X} is the matrix of partial derivatives in the variables x i {\displaystyle x_{i}} and Y {\displaystyle Y} is the matrix of partial derivatives in the variables y j {\displaystyle y_{j}} . The implicit function theorem says that if Y {\displaystyle Y} is an invertible matrix, then there are U {\displaystyle U} , V {\displaystyle V} , and g {\displaystyle g} as desired. Writing all the hypotheses together gives the following statement.
Statement of the theorem Let f : R n + m → R m {\displaystyle f:\mathbb {R} ^{n+m}\to \mathbb {R} ^{m}} be a continuously differentiable function, and let R n + m {\displaystyle \mathbb {R} ^{n+m}} have coordinates ( x , y ) {\displaystyle ({\textbf {x}},{\textbf {y}})} . Fix a point ( a , b ) = ( a 1 , … , a n , b 1 , … , b m ) {\displaystyle ({\textbf {a}},{\textbf {b}})=(a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})} with f ( a , b ) = 0 {\displaystyle f({\textbf {a}},{\textbf {b}})=\mathbf {0} } , where 0 ∈ R m {\displaystyle \mathbf {0} \in \mathbb {R} ^{m}} is the zero vector. If the Jacobian matrix (this is the right-hand panel of the Jacobian matrix shown in the previous section):
J f , y ( a , b ) = [ ∂ f i ∂ y j ( a , b ) ] {\displaystyle J_{f,\mathbf {y} }(\mathbf {a} ,\mathbf {b} )=\left[{\frac {\partial f_{i}}{\partial y_{j}}}(\mathbf {a} ,\mathbf {b} )\right]}
is invertible, then there exists an open set U ⊂ R n {\displaystyle U\subset \mathbb {R} ^{n}} containing a {\displaystyle {\textbf {a}}} such that there exists a unique function g : U → R m {\displaystyle g:U\to \mathbb {R} ^{m}} such that g ( a ) = b {\displaystyle g(\mathbf {a} )=\mathbf {b} } , and f ( x , g ( x ) ) = 0 for all x ∈ U {\displaystyle f(\mathbf {x} ,g(\mathbf {x} ))=\mathbf {0} ~{\text{for all}}~\mathbf {x} \in U} . Moreover, g {\displaystyle g} is continuously differentiable and, denoting the left-hand panel of the Jacobian matrix shown in the previous section as:
J f , x ( a , b ) = [ ∂ f i ∂ x j ( a , b ) ] , {\displaystyle J_{f,\mathbf {x} }(\mathbf {a} ,\mathbf {b} )=\left[{\frac {\partial f_{i}}{\partial x_{j}}}(\mathbf {a} ,\mathbf {b} )\right],}
the Jacobian matrix of partial derivatives of g {\displaystyle g} in U {\displaystyle U} is given by the matrix product:
[ ∂ g i ∂ x j ( x ) ] m × n = − [ J f , y ( x , g ( x ) ) ] m × m − 1 [ J f , x ( x , g ( x ) ) ] m × n {\displaystyle \left[{\frac {\partial g_{i}}{\partial x_{j}}}(\mathbf {x} )\right]_{m\times n}=-\left[J_{f,\mathbf {y} }(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times m}^{-1}\,\left[J_{f,\mathbf {x} }(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times n}}
A proof may be found in the inverse function theorem article. Here, the two-dimensional case is detailed.
Higher derivatives If, moreover, f {\displaystyle f} is analytic or continuously differentiable k {\displaystyle k} times in a neighbourhood of ( a , b ) {\displaystyle ({\textbf {a}},{\textbf {b}})} , then one may choose U {\displaystyle U} in order that the same holds true for g {\displaystyle g} inside U {\displaystyle U} . In the analytic case, this is called the analytic implicit function theorem.
The circle example Let us go back to the example of the unit circle. In this case n = m = 1 and f ( x , y ) = x 2 + y 2 − 1 {\displaystyle f(x,y)=x^{2}+y^{2}-1} . The matrix of partial derivatives is just a 1 × 2 matrix, given by
( D f ) ( a , b ) = [ ∂ f ∂ x ( a , b ) ∂ f ∂ y ( a , b ) ] = [ 2 a 2 b ] {\displaystyle (Df)(a,b)={\begin{bmatrix}{\dfrac {\partial f}{\partial x}}(a,b)&{\dfrac {\partial f}{\partial y}}(a,b)\end{bmatrix}}={\begin{bmatrix}2a&2b\end{bmatrix}}}
Thus, here, the Y in the statement of the theorem is just the number 2b; the linear map defined by it is invertible if and only if b ≠ 0. By the implicit function theorem we see that we can locally write the circle in the form y = g(x) for all points where y ≠ 0. For (±1, 0) we run into trouble, as noted before. The implicit function theorem may still be applied to these two points, by writing x as a function of y, that is, x = h ( y ) {\displaystyle x=h(y)} ; now the graph of the function will be ( h ( y ) , y ) {\displaystyle \left(h(y),y\right)} , since where b = 0 we have a = 1, and the conditions to locally express the function in this form are satisfied. The implicit derivative of y with respect to x, and that of x with respect to y, can be found by totally differentiating the implicit function x 2 + y 2 − 1 {\displaystyle x^{2}+y^{2}-1} and equating to 0:
2 x d x + 2 y d y = 0 , {\displaystyle 2x\,dx+2y\,dy=0,}
giving
d y d x = − x y {\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}}
and
d x d y = − y x . {\displaystyle {\frac {dx}{dy}}=