The notion of explicit and implicit functions is of utmost importance while solving real-life problems. Also, you must have read that the differential equations are used to represent the dynamics of the real-world phenomenon. Therefore, we must learn to differentiate the explicit and implicit functions.

Likewise, to arrive at the required differential equations and gain a better understanding of the physical process under consideration. We’ll learn how to find the derivatives of implicit functions in this section. Before jumping into the topic, let us first clarify ourÂ concepts.

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## What are Explicit and Implicit Functions?

I’ll make you understand the difference between the two on the basis of how much we have learned so far. Up until now, we have only learned how to differentiate functions of the form y = f(x). This is actually the explicit form of the dependent variable y in terms of the independent variable x.

Basically, an explicit form is one in which yourÂ equation is simplified to such an extent that the variables get separated out and can be expressed in terms of each other.Â For eg, consider the equation of a circle of unit radius: $${ x^2 + y^2 = 1 } \text{…………….Â The implicit form of the function}$$

The alternate form is as follows: $$ { y = \pm{\sqrt{1 – x^2}}} \text{…………… The explicit form of the function} $$ As you may notice, we have conveniently expressed y here, as a function of x. Thus, this representation of the equation is explicit in nature with respect to y (as well as x, since you can follow the same procedure for x).

However, if you consider the equation of a two-dimensional curve, given as: $$ { x^2 + xyÂ + y^2 = 1 } $$ You may note that it is not possible to separate out the variables in this case. Such an equation is completely implicit in nature with respect to the variables x and y.

To get the slope of such a curve at any point, we cannot follow the direct procedure of differentiating the function y = f(x) and putting the value of the x-coordinate of the point in \(\frac{dy}{dx}\) to get theÂ slope. Instead, we will have to follow the process of implicit differentiation and solve forÂ \(\frac{dy}{dx}\) here, as will now be demonstrated.

## Implicit Differentiation

Given an implicit function with the dependent variable y and the independent variable x (or the other way around), our aim is to solve forÂ \(\frac{dy}{dx}\) or higher order derivatives, in terms of the variables x and y or any lower order derivatives. For achieving this, we’ll adopt the following procedure:

- Differentiate the entire equation with respect to the independent variable (it could be x or y).
- Apply the Chain Rule of Differentiation.
- Write the derivatives of y as y’ and y” and so on, instead ofÂ \(\frac{dy}{dx}\) andÂ \(\frac{d^2y}{dx^2}\), to make our expressions seem less cumbersome. Similarly, the derivatives of x with respect to y can be written as x’ and x” and so on.
- Solve the resultant equation for y’ (or x’ likewise) or differentiate again if the higher order derivatives are needed.

Now let us solve a couple of questions to get a better idea of how to proceed with this new style of differentiation.

**Browse more Topics under Continuity And Differentiability**

- Continuity
- Algebra of Continuous Functions
- Algebra of Derivatives
- Derivatives of Composite Functions
- Derivatives of Functions in Parametric Forms
- DerivativesÂ of Inverse Trigonometric Functions
- Exponential and Logarithmic Functions
- Logarithmic Differentiation
- Mean Value Theorem
- Second Order Derivatives

## Important Points to Remember

The method of implicit differentiation used here is a general technique. It is used to find the derivatives of unknown quantities. Don’t interpret it as useful only in calculating \(\frac{dy}{dx}\) or the other derivatives we have solved for.

For example, Let the position x and the velocity v of a body be given as functions of time t i.e. position: x(t) and velocity: v(t). If it is given that the body follows a path given by the equation: 7v^{2}t^{2} – 5x = 2

Then you can easily get a relation between the acceleration (v'(t)) and the velocity (x'(t)) of the body by implicitly differentiating the given expression with respect to time and proceeding in the same manner as discussed above (i.e. apply the chain rule on each term).

## Solved Examples for You

**Question 1:** Find the expression for the first derivative of the function y(x) given implicitly by the equation: x^{2}y^{3} – 4yÂ + 3x^{3} = 2.

Solution: We begin by first differentiating the given equation with respect to x.

$$ { x^2y^3 – 4y + 3x^3 = 2 } $$

$$ {\frac{d}{dx}{(x^2y^3 – 4y + 3x^3)} = \frac{d}{dx}{(2)}} $$

Now let us understand how to take the different derivatives here. Starting from the left hand side, we first encounter the term x^{2}y^{3}. Its derivative using the chain rule is:

$$ { \frac{d}{dx}{(x^2y^3)} = (\frac{d}{dx}x^2).y^3 + (\frac{d}{dx}y^3).x^2 } $$

\( {= (2x).(y^3) + (3y^2).(y’).(x^2) } \)

$$ {= 2xy^3 + 3x^2y^2y’ } $$

Similarly, the derivative of the subsequent terms can be calculated.

$$ { \frac{d}{dx}(-4y) = -4y’} $$

$$ { \frac{d}{dx}(3x^3) = 9x^2 } $$

The right-hand side’s derivative will simply be 0, since it is a constant.

Thus, we have the resultant expression as:

$$ {Â 2xy^3 + 3x^2y^2y’ – 4y’ + 9x^2 = 0 } $$

Collect the terms involving y’ on one side and take the remaining terms on the other side to get:

$$ { y’.(3x^2y^2 – 4) = -9x^2 – 2xy^3 } $$

$$ { y’ = -\frac{9x^2 + 2xy^3}{3x^2y^2 – 4} } $$

This is the required expression for the first derivative at any point on the curve.

### Question 2

Find the second derivative of y, given implicitly as: y – tan^{-1}yÂ = x.

Solution: Differentiating the equation with respect to x:

$$ { \frac{d}{dx}(y – tan^{-1}y) = \frac{d}{dx}(x) } $$

$$ { y’ – \frac{1}{1 + y^2}.y’ = 1 } $$

Solve for y’:

$$ { y'(1 –Â \frac{1}{1 + y^2}) = 1 } $$

\( { y’ = \frac{1}{1 –Â \frac{1}{1 + y^2}} } \)

Simplify the R.H.S. to get:

$$ { y’ = \frac{1}{y^2} + 1 } $$

Differentiate with respect to x again to get the expression for y”:

$$ { \frac{d}{dx}y’ = \frac{d}{dx}(\frac{1}{y^2} + 1) } $$

$$ { y” = \frac{-2y}{y^4}.y’ } $$

$$ {Â y” = -\frac{2y’}{y^3} } $$

Note that this expression involves y’ in it. This, we can substitute from what we had calculated above:

$${ y” = -\frac{2}{y^3}.(\frac{1}{y^2} + 1) } $$

$${ y” = -( \frac{2}{y^5} + \frac{2}{y^3}) } $$

$${ y” = -\frac{2(1 + y^2)}{y^5}} $$

This is the required expression for the second derivative of y.

This concludes our discussion on the topic derivatives of implicit functions.

**Solved Questions for You**

**Question 1: What is implicit and explicit?**

**Answer:** There is no room for doubt left as everything is communicated in quite a clear and direct manner. This is what majorly splits these two words. We label something as implicit when something is implied however, not stated in a direct manner. On the other hand, explicit means when something is stated in a direct manner leaving no room for any doubts.

**Question 2: What is the point of implicit differentiation?**

**Answer:** The system of implicit differentiation lets us find the derivative of y with regards to x without solving the given equation for y. Thus, we must make use of the chain rule whenever the function y is being distinctive because of our supposition that we may express y as a function of x.

**Question 3: Who invented implicit differentiation?**

**Answer:** Isaac Newton invented implicit differentiation. He was a popular physicist and mathematician who made use of it in several physics problems he encountered. On the other hand, another mathematician from Germany, Gottfried W. Leibniz has been said to also develop the technique in parallel to Newton around the same time.

**Question 4: What happens when the derivative is zero?**

**Answer:** A positive derivative is when the function increases. A negative derivative is when the function decreases. Finally, a zero derivative is when the function has got some special behaviour at the specified point. It may consist of a local maximum, a local minimum.

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