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.