In mathematics, we may come across with various types of functions and the corresponding properties. In this article, you will learn about two important theorems in calculus, they are inverse function theorem and implicit function theorem. Before getting into detail about inverse and implicit function theorems, let’s recall the meaning and definition of inverse function and implicit function.

Inverse function:

Suppose f is a one-to-one function with domain set A and range set B, then the inverse function f^-1 has domain as B and range as A and is defined as:

f^-1(y) = x if and only if f(x) = y for any y ∈ B.

Learn more about inverse functions here.

Implicit function:

An implicit function is a function of two independent variables, i.e., the functions of the form f(x, y) = c are called implicit functions.

Click here to get more information about implicit functions.

Inverse Function Theorem

In mathematics, particularly differential calculus, the inverse function theorem delivers an acceptable condition for a function to be invertible in the neighbourhood of a point in its domain. Also, the derivative of this function is continuous and non-zero at the point. The theorem also provides a formula to get the derivative of the inverse function.

Let D be an open subset of Rⁿ, and let f: D → Rⁿ be a continuously differentiable function on D. Also, assume a base point x₀ ∈ D and f'(x0): Rⁿ → Rⁿ is invertible. Then, there are open sets U ⊂ D and V ⊂ Rⁿ containing x₀ and f(x₀), respectively (the image of the base point), such that f is a bijection from U to V . Hence, there is an inverse map f⁻¹ : V → U and f⁻¹ is differentiable at y₀, then

(f⁻¹)'(y₀) = 1/(f'(x₀)).

Here, y₀ = f(x₀).

However, if f'(x₀) ≠ 0 for every point x ∈ D, then the above theorem holds for each point of D.

Implicit Function Theorem

Let F be a continuous real-valued function define on some neighbourhood N of the fixed point (a, b) such that F(a, b) = 0 and there exists a continuous differentialble function ∂F/∂y on N, where ∂F/∂y (a, b) ≠ 0. Then there is an unique function g defined on some neighbourhood Na of a so that g(a) = b, F(x, g(x)) = 0 for each x ∈ N_a and g is continuous.

Moreover, if ∂F/∂x exists and is continuous on N, then we can say that g is a continuous differentiable function on N_a. Therefore, g’ is given as:

Applications of Implicit Function Theorem

• In n multivariable calculus, an implicit function theorem acts as a tool that entitles relations to be converted to functions of multiple real variables with the help of a graphical representation of the function.
• The implicit function theorem aims to convey the presence of functions such as g₁(x) and g₂(x), even in cases where we cannot define explicit formulas.
• The implicit function theorem guarantees that the functions g₁(x) and g₂(x) are differentiable.
• The implicit function theorem also works in cases where we do not have a formula for the function f(x, y).

Solved Example

Question:

Show that there exists a continuously differentiable function g defined by the equation F(x,y) = x³ + y³ – 3xy – 4 = 0 in a neighbourhood of x = 2 such that g(2) = 2. Also, find its derivative.

Solution:

Given function is:

F(x, y) = x³ + y³ – 3xy – 4 = 0

And x = 2 and g(2) = 2

Now,

F(2, 2) = (2)³ + (2)³ – 3(2)(2) – 4

= 8 + 8 – 12 – 4

= 0

So, F(2, 2) = 0

∂F/∂x = ∂/∂x (x³ + y³ – 3xy – 4) = 3x² – 3y

∂F/∂y = ∂/∂y (x³ + y³ – 3xy – 4) = 3y² – 3x

Let us calculate the value of ∂F/∂y at (2, 2).

That means, ∂F(2, 2)/∂y = 3(2)2 – 3(2) = 12 – 6 = 6 ≠ 0.

Thus, ∂F/∂y is continuous everywhere.

Hence, by the implicit function theorem, we can say that there exists a unique function g defined in the neighbourhood of x = 2 by g(x) = y, where F(x, y) = 0 such that g(2) = 2.

Also, we know that ∂F/∂x is continuous.

Now, by implicit function theorem, we get;

g’(x) = -[∂F(x, y)/∂x]/ [∂F(x, y)/ ∂y]

= -(3x² – 3y)/(3y² – 3x)

= -3(x² – y)/ 3(y² – x)

= -(x² – y)/(y² – x)