## Introduction to the Implicit Function Theorem

Calculus is the study of a continuous rate of change. For this it involves two major branches such as derivative and integrals. There are different laws and rules of calculating derivatives but if the function is implicit, the implicit function theorem is used. Let us learn more about the implicit differentiation theorem with examples.

## Understanding of the Implicit Function Theorem

The implicit function theorem is a fundamental theorem of calculus. It is used to calculate derivative of an implicit function. An implicit function is a polynomial expression which cannot be defined explicitly. Therefore, we cannot calculate derivative of such functions in simple steps. We need to use implicit function theorem. The implicit function theorem states that,

“If $f(x, y)$ is an implicit function, there is a point $c$ in the interval $(x_0, y_0)$ such that $f(x_0, y_0)=c$ and if $f_y(x_0, y_0)\ne 0$ then there exists a unique differentiable function y(x) such that it satisfies $f(x, y(x))$ and $y'(x)= -\frac{d_xf(x, y)}{d_yf(x, y)}$.”

In simple words we can say that the implicit theorem helps to calculate derivative of a function which cannot be written in the form of y = f(x).

## What is Implicit Function Theorem?

The theorem of implicit function is a fundament theorem or law to solve derivative of implicit functions. It states that if a function is defined implicitly over an interval then there will be a point on which its derivative will not be equal to zero, then it will satisfy the following conditions,

- f(x, y) can be represented as f(x, y(x))
- $y'(x) = -\frac{d_xf(x, y)}{d_yf(x, y)}$

For example, the equation of a circle is x2+y2=1. It is clear that this expression is a combination of both independent and dependent variables. Therefore, the derivative of this function can be calculated implicitly which is also known as implicit differentiation.

## Implicit Function Theorem Proof

In this theorem, we will prove that an implicit function can be written in the form of $y = f(x)$. It will help us to develop a formula to calculate derivative of implicit functions. For this, suppose that $F(x, y)$ is continuous at a point in its domain $(x_0, y_0)$ such that it satisfies the following conditions:

- $F(x_0, y_0)=0$
- \frac{df}{dy} ≠ 0$

For some unique value of y, suppose that F(x, y) is continuous and there exists a unique solution y=f(x). Therefore,

$F(x, y) = 0$

Since y = f(x),

$F(x, f(x)) = 0$

Now calculating derivative on the both sides,

$\frac{d}{dx}[F(x,f(x))]=0$

$\frac{\partial F}{\partial x}\frac{\partial}{\partial x}(x)+\frac{\partial F}{\partial y}f'(x)=0$

Now solving for f’(x),

$f'(x)=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

Or,

$f'(x)=-\frac{F_x}{F_y}$

Which is the required formula to calculate derivatives of an implicit function. For more understanding, let us discuss in the following example.

### Implicit function theorem example 1

Consider the equation of a circle whose radius in 1. Let’s calculate the implicit derivative of the equation,

$x^2+y^2=1$

We can write it as,

$F(x,y)=x^2+y^2-1$

Since the implicit function theorem formula is,

$f'(x)=-\frac{F_x}{F_y}$

Calculating partial derivatives,

$\frac{\partial F}{\partail x}=\frac{\partial}{\partail x}(x^2+y^2-1)$

$\frac{\partial F}{\partial x}=2x$

Similarly,

$\frac{\partial F}{\partial y}=\frac{\partial}{\partial y}(x^2+y^2-1)=2y$

Now substituting the values of partial derivatives in the above formula,

$f'(x)=-\frac{2x}{2y}=-\frac{x}{y}$

Hence the implicit derivative of the given function is x/y.

### Implicit function theorem example 2

Find dy/dx of the given function,

$x^4+y^3-3x^2y=0$

We can write it as,

$F(x,y)=x^4+y^3-3x^2y$

Since the implicit function theorem formula is,

$f'(x)=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

Calculating partial derivatives,

$\frac{\partial F}{\partial x}=\frac{\partial}{\partial x}(x^4+y^3-3x^2y)$

By using power rule,

$\frac{\partial F}{\partial x}=4x^3-6xy$

Similarly,

$\frac{\partial F}{\partial y}=\frac{\partial}{\partial y}(x^4+y^3-3x^2y)=3y^2-3x^2$

Now substituting the values of partial derivatives in the above formula, we get

$f'(x)=-\frac{4x^3-6xy}{3y^2-3x^2}=-\frac{2x(2x^2-3y)}{3(y^2-x^2)}$

## How to apply Implicit Function Theorem?

The implicit function differentiation theorem can be applied by finding all partial derivatives of a given expression. It can be applied by using following steps;

- Identify the given function by equating it to zero. If the function is implicit, proceed to the next step. Otherwise find the derivative by using other derivatives rules.
- Find the partial derivative of the given function with respect to all variables involved.
- Now use the implicit function theorem and substitute the values of the partial derivatives in it. Or, divide the partial derivative with respect x with the partial derivative of y and then multiply the answer with the negative sign.
- Simplify the equation if needed.

## Conclusion

In calculus, the implicit function theorem is a fundamental theorem that has a great importance in calculating rate of change. We can easily calculate derivative of an explicit function. But in the case of implicit derivative, it is not possible to use usual derivative formulas and rules such as product rule, quotient rule or power rule. Here, the implicit function theorem helps us to calculate derivative of implicit functions by using partial derivatives. Hence, it is a perfect way to find the rate of change of implicit equations and expressions.