Question

What is the Implicit function theorem and how do you prove it?

Answer 1

Hint: A function tells us the link between unknown variable quantities. A function that is expressed in terms of one variable is known as an explicit function, it can be expressed as \[f(x) = {x^2} - x + 1\] , whereas in an implicit function, the function is expressed in terms of two variables, it can be expressed as $ f(x,y) = 0 $ and is of the form $ {x^2} + {y^2} + 3 = 0 $ . In differentiation, one variable is differentiated with respect to the other variable so to differentiate implicit functions, we use the implicit function theorem.

Complete step-by-step answer:
Differentiation is a process of finding a very small change in a given quantity with respect to another quantity, it is written as $ \dfrac{{dy}}{{dx}} $ where $ dy $ represents a very small change in y and $ dx $ represents a very small change in x.
To differentiate implicit functions, we use partial derivatives.
Suppose there is an equation in terms of x and y, such that one variable cannot be expressed in terms of the other variable, let the function be –
 $ F(x,y) = 0 $
Differentiating both sides with respect to x using partial derivatives, we get –
\[
  \dfrac{{\partial F}}{{\partial x}}\dfrac{{dx}}{{dx}} + \dfrac{{\partial F}}{{\partial y}}\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\dfrac{{\partial F}}{{\partial x}}}}{{\dfrac{{\partial F}}{{\partial y}}}} \;
 \]
This is the implicit function theorem.
For example, let the equation of a circle be –
 $ {x^2} + {y^2} = 1 $
This function can be written in the form of $ F(x,y) = 0 $ as $ {x^2} + {y^2} - 1 = 0 $
 $ \dfrac{{\partial F}}{{\partial x}} = 2x $ and $ \dfrac{{\partial F}}{{\partial y}} = 2y $
Now, according to the implicit function theorem,
 $
  \dfrac{{dy}}{{dx}} = - \dfrac{{\dfrac{{\partial F}}{{\partial x}}}}{{\dfrac{{\partial F}}{{\partial y}}}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{2y}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{y} \;
  $

Note: We can check if the above answer obtained is correct or not by differentiating the function implicitly –
 $
  \dfrac{d}{{dx}}({x^2} + {y^2}) = \dfrac{{d(1)}}{{dx}} \\
   \Rightarrow \dfrac{{d{x^2}}}{{dx}} + \dfrac{{d{y^2}}}{{dx}} = 0 \\
   \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{y} \;
  $
While doing partial derivatives of a function, we treat the other variable as constant.