# Implicit Function

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 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.

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

Example 1:Find $\frac {dy}{dx}$ if y = $5x^2 – 9y$

Solution 1: The given functiony = $5x^2 – 9y$   can be rewritten as

⇒ 10y = 5 $x^2$

⇒ y = $\frac {1}{2} x^2$

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

$\frac {dy}{dx}$ = 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.

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. Visit byjus.com to learn the concepts in a way you have never learnt before.