Implicit Function

An implicit function is a function, written in terms of both dependent and independent variables, like y-3x2+2x+5 = 0. Whereas an explicit function is a function which is represented in terms of an independent variable. For example, y = 3x+1 is explicit where y is a dependent variable and is dependent on the independent variable x. In the case of differentiation, an implicit function can be easily differentiated without rearranging the function and differentiating each term instead. As y is a function of x, therefore we will apply chain rule as well as product and quotient rule.

What is an Implicit function?

When in a function the dependent variable is not explicitly isolated on either side of the equation then the function becomes an implicit function.

It is very easy to solve when the equations take the form y = f(x). When a function is expressed in such a form it represents the explicit function. But it is possible to express y implicitly in terms of f(x). In such a case we use the concept of implicit function differentiation.

The unit circle can be specified implicitly as the set of points (x,y) fulfilling the equation, x+ y2=1.

To make our point more clear let us take some implicit functions and see how they are differentiated.

Implicit Function Theorem

In mathematics, especially in multivariable calculus, the implicit function theorem is a mechanism that enables relations to be transformed to functions of various real variables. It is possible by representing the relation as the graph of a function. An individual function graph may not represent the complete relation, but there could be such a function on a constraint of the domain of the relation. The implicit function theorem gives a satisfactory condition to assure that there is such a function.

Suppose a function with n equations is given, such that, fi (x1 , …, xn, y1, …, yn) = 0, where i = 1, …, n or we can also represent as F(xi, yi) = 0, then the implicit theorem states that, under a fair condition on the partial derivatives at a point, the m variables yi  are differentiable functions of the xj  in some section of the point. Since, we cannot express these functions in closed form, therefore they are implicitly defined by the equations.

Implicit Function Examples

Example 1:Find dy/dx if y = 5x2 – 9y 

Solution 1: The given function, y = 5x2 – 9y  can be rewritten as:

⇒ 10y = 5x2

⇒ y = 1/2 x2

Since this equation can explicitly be represented in terms of y, therefore, it is an explicit function.

Now, as it is an explicit function, we can directly differentiate it w.r.t. x,

Since, \( \frac {d(x^n)}{dx}\) =\( nx^{n-1} \)

⇒ dy/dx = x

Example 2:Find,  \( \frac {dy}{dx} \) if y = \( 5x^2 – 9e^y \) .

Solution:The given function  y = \( 5x^2 – 9e^y \) can be rewritten as \( y + 9e^y = 5x^2 \) . But it is not possible to completely isolate  and represent it as a function of. This type of function is known as an implicit function.

To differentiate an implicit function, we consider  y as a function of x  and then we use the chain rule to differentiate any term consisting of y.

Now to differentiate the given function, we differentiate directly w.r.t. x the entire function. This step basically indicates the use of chain rule.

⇒ \( \frac {dy}{dx} + \frac {d(9e^y)}{dx}\) = \(\frac {d(5x^2)}{dx} \)

⇒ \( \frac {dy}{dx} + 9e^y \frac {dy}{dx}\) = 10x

⇒ \( \frac {dy}{dx} (1 + 9e^y) \) = 10x

⇒ \( \frac {dy}{dx} \) = \( \frac {10x}{1+9y^y}\)

Example 3: Find \( \frac {dy}{dx}\) . if \( x^4 + y^3 – 3x^2 y \) = 0 .

Solution 3: The given function \( x^4 + y^3 – 3x^2y = 0 \)  can be differentiated using the concept of implicit function differentiation.

Therefore differentiating both the sides w.r.t. x, we get,

\( 4x^3 + 3y^2 \frac {dy}{dx} – 3 \left( 2xy + x^2 \frac {dy}{dx} \right) \) = 0

\( \frac {dy}{dx} (3x^2 – 3y^2) \) = \( 4x^3 – 6xy \)

⇒ \( \frac {dy}{dx} \) = \( \frac {4x^3 – 6xy}{3x^2 – 3y^2}\)

Example 4: Find the slope of the tangent to the curve y = \( x^2 + 3y^2 + xy \) .

Solution 4: In this example, we are asked to find a tangent to the given curve. To find a tangent we find \( \frac {dy}{dx} \)  which represents the slope of the given curve. Since it is an implicit function, on differentiating both the sides w.r.t. x we get,

\( \frac {dy}{dx} \)= 2x + 6y\( \frac {dy}{dx} + y + x \frac {dy}{dx} \)

⇒ \( \frac {dy}{dx} (1 – x – 6y)\) = 2x + y

⇒ \( \frac {dx}{dy} \) = \( \frac {2x+y}{1-x-6y}\)

This represents the slope of the given curve.

Example 5: Differentiate x+ y2=25 implicitly.

Solution: Differentiating x+ y2=25 with respect to x we get;

2x + 2y dy/dx = 0

2y dy/dx = -2x

dy/dx = -2x/2y

dy/dx = -x/y

Example 6: Differentiate x+ y2=16 implicitly.

Solution: Differentiating x+ y2=16 with respect to x we get;

3x+ 2y dy/dx =0

2y dy/dx = -3x2

dy/dx = -3x2/2y

Now it might be very clear to you what exactly the difference between an implicit and an explicit function. The method of finding derivatives of implicit function would also be very clear by now.