Open In App

Implicit Differentiation

Last Updated : 21 May, 2024
Like Article

Implicit Differentiation is a useful tool in the arsenal of tools to tackle problems in calculus and beyond which helps us differentiate the function without converting it into the explicit function of the independent variable. Suppose we don’t know the method of implicit differentiation. In that case, we have to convert each implicit function into an explicit function, which is sometimes very hard and sometimes it is not even possible.

Implicit differentiation makes these problems very easy to solve. In this article, we will learn all the necessary basics we need to know about implicit differentiation formula, chain rule, implicit differentiation of inverse trigonometric functions, etc.

Implicit Differentiation vs Explicit Differentiation

What is Implicit Differentiation?

Implicit Differentiation is the process of differentiation in which we differentiate the implicit function without converting it into an explicit function. For example, we need to find the slope of a circle with an origin at 0 and a radius r. Its equation is given as x2 + y2 = r2

Now, to find the slope we need to find the dy/dx of the given function, so without implicit differentiation, we have to convert this function into an explicit function i.e., y = ∓√(r2 – x2) . The explicit function of this is comparatively hard to differentiate. Thus, we need to learn the implicit differentiation by which this can be very easily differentiated.

Read More: Calculus in Maths

Prerequisite for Implicit Differentiation

There are some prerequisite concepts that we need to know before learning Implicit Differentiation, these prerequisites are as follows:

Chain Rule 

The chain rule is a formula to compute the derivative of a composite function. That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g as f(g(x))’ = f'(g(x)) × g'(x)

Implicit Function

When a function is not defined explicitly in terms of a single independent variable. Implicit Function is represented as f(x, y) = k, where k is a real number, then the function is called the implicit function. For example, y + x2 = 5,  x2 + y2 = r2, etc.

Explicit function

When a function is defined in terms of a single independent variable explicitly such as y = f(x), then the function is called the explicit function. For example, y = x2, y = 3x+7, y = sin x, etc.

 f(x, y) = 0 e.g. y + x2 = 5

Note: Here we took only 2 variables x and y to define the implicit function. But you can have any number of variables.

Chain Rule in Implicit Differentiation

The general chain rule that we follow in differentiation is also followed here. We represent the derivative of two functions in product form. This can be understood from the following example:

For example, we have to differentiate 3xy2 implicitly with respect to x.

d/dx(3xy) = 3y2+3x.2ydy/dx.

Implicit Differentiation Formula

In Implicit differentiation, we do the differentiation of functions expressed in terms of more than one variable. The formulas for derivatives of the functions are the same as normal differentiation but the derivative of another variable is done following chain rule. The general representation of implicit differentiation can be d/dx{f(x,y)}.

df/dx = (df/dy).(dy/dx)

How to do Implicit Differentiation

The following steps need to be followed to differentiate any implicit function.

Step 1: Follow the rules of differentiation to differentiate both sides of the equation with respect to x.

Step 2: Use the chain rule to differentiate expressions involving y.

Step 3: Solve the equation for dy/dx.

Example: Differentiate x2 + y2 = r2.


Given equation: 

x2 + y2 = r2

Step 1: Differentiate both sides wrt to x and follow the rules of differentiation.

d/dx{x2 + y2} = d/dx(r2)

Step 2: Using the chain rule

2x + 2y(dy/dx) = 0

Step 3: Simplify the equation

2y(dy/dx) = -2x

dy/dx = -x/y

Thus, dy/dx = -x/y

Implicit Differentiation of Inverse Trigonometric Functions

Implicit differentiation is very useful in finding derivatives of Inverse trigonometric Functions.

Let us consider y = sin-1(x) and we need to find its derivative, 

Take sin both sides of the equation,

sin y = sin(sin-1(x)

sin y = x

Differentiating the above equation w.r.t x, we get

d/dx(sin y) = d/dx (x)

cos y(dy/dx) = 1

dy/dx = 1/(cos y)

Now, sin(y) = x

⇒ x2 = sin2(y

⇒ x2 + cos2(y) = 1

⇒ cos2(y) = 1 – x2

⇒ cos(y) = ​​



Substituting the value, we get

dy/dx = 1/(cos y)

dy/dx = 1/√(1 – x2)

Difference between Implicit and Explicit Differentiation

Implicit Differentiation and Explicit Differentiation are the two methods of differentiation that are used in calculus. The differences between them are explained in the table below,

Implicit Differentiation

Explicit Differentiation

Used to find the derivative of a function that  cannot be easily solved for y

Used to find the derivative of a function that can be easily solved for y

Does not require the function to be expressed  in terms of y

Requires the function to be expressed in terms of y

Treats y as an implicit function of x

Treats y as an explicit function of x

Involves taking the derivative of both sides of the equation with respect to x

Involves directly differentiating the function with respect to x

Involves the chain rule and product rule to differentiate the function

Involves the power rule, product rule, quotient rule, and chain rule to differentiate the function

Often used when solving equations that involve multiple variables or when finding higher-order derivatives

Often used when finding the slope of a tangent line or the instantaneous rate of change of a function

Read More,

Implicit Differentiation Examples

Example 1: Find the derivative of y + x + 5 = 0.


Using Explicit Differentiation

y + x + 5 = 0

⇒ y = -(x + 5)

\begin{aligned}\\ \Rightarrow \frac{dy}{dx}&=-(1+0) \\\Rightarrow \frac{dy}{dx}&=-1 \end{aligned}


Using Implicit Differentiation

y + x + 5 = 0

Differentiating both sides wrt x



Isolate dy/dx



Example 2: Find the derivative of y5 – y = x.


y5 – y = x

Differentiating the above equation with respect to x, we get

\begin{aligned} & \Rightarrow 5y^4\frac{dy}{dx}-\frac{dy}{dx}&=1 \\ & \Rightarrow (5y^4-1)\frac{dy}{dx}&=1 \\ & \Rightarrow \frac{dy}{dx}=\frac{1}{5y^4-1} \end{aligned}


Example 3: Find the derivative of 10x4 – 18xy2 + 10y3 = 48.



10x4 – 18xy2 + 10y3 = 48

Differentiating both sides w.r.t x

10(4x^3) − 18(x.2y\frac{dy}{dx} + y^2) + 10(3y^2 \frac{dy}{dx}) = 0


40x^3 − 36xy \frac{dy}{dx} − 18y^2 + 30y^2 \frac{dy}{dx} = 0


Keeping all the terms involving dy/dx on left and rest terms on right side of equation

−36xy \frac{dy}{dx} + 30y^2 \frac{dy}{dx} = −40x^3+ 18y^2


(30y^2−36xy)\frac{dy}{dx}= 18y^2 − 40x^3


Dividing both sides by 2

(15y^2−18xy) \frac{dy}{dx}=9y^2 − 20x^3


Finally Isolate dy/dx

\bold{\frac{dy}{dx}=\frac{9y^2 − 20x^3}{15y^2−18xy}}


For the term xy2 we used the Product Rule: (f.g)’ = f.g’ + f’.g

\begin{aligned} (xy^2)’  &= x(y^2)’ + (x)’y^2 \\ &=x({2y\frac{dy}{dx}})+y^2 \\ &=2xy\frac{dy}{dx}+y^2 \end{aligned}


Example 4: Find the derivative of x4 + 2y2 = 8.



x4 + 2y2 = 8

\begin{aligned}  4x^3+4y\frac{dy}{dx}&=0 \\ \frac{dy}{dx}&=\frac{-x^3}{y} \end{aligned}


Practice Questions on Implicit Differentiation

Q1: Differentiate 3x2 + y3 = 6x

Q2: Differentiate sin x = cos y

Q3: Find dy/dx if y2 + 2xy + x3 = 21

Q4: Find dy/dx if x2/p2 + y2/q2 = 2

Implicit Differentiation-FAQs

1. What is Implicit Differentiation?

Implicit differentiation is the process of differentiation of implictit function, where we differentiate the given implicit function without converting it into explicit fuction.

2. Why do we need Implicit Differentiation?

We need implicit differentiation, as not all implicit functions can be converted into explicit function and some which are possible to covert into explicit function are sometimes becomes very complicated and hard to calculate.

3. What is Implicit Differentiation Method?

Implicit differentiation is a method in calculus that is used to solve various problems of calculus. Implicit Differntiation is an application of Chain Rule.

4. What is the chain rule in Implicit Differentiation?

To differentiate a function f(g(x)), we use chain rule which states

d/dx(f(g(x))) = f'(g(x))×g'(x)

Where ‘ represents the first-order derivative of the function.

5. Can Implicit Differentiation be used to find Higher-Order Derivatives?

Yes, implicit differentiation can be used to find higher-order derivatives. To find higher-order derivatives just differentiate the equation as many times as needed.

6. What is the Implicit Differentiation Formula?

The Implicit Differentiation Formula is, du/dx = du/dy.dy/dx.

7. When Can Implicit Differntiation be Used?

Implicit functions are the function that have both dependent and independent variable multiplied with each other. These types of functions are differentiated using Implicit Differntiation method.

Previous Article
Next Article

Similar Reads

Implicit differentiation - Advanced Examples
In the previous article, we have discussed the introduction part and some basic examples of Implicit differentiation. So in this article, we will discuss some advanced examples of implicit differentiation. Table of Content Implicit DifferentiationMethod to solveImplicit differentiation Formula Solved Example Implicit DifferentiationImplicit differe
5 min read
Derivatives of Implicit Functions - Continuity and Differentiability | Class 12 Maths
Implicit functions are functions where a specific variable cannot be expressed as a function of the other variable. A function that depends on more than one variable. Implicit Differentiation helps us compute the derivative of y with respect to x without solving the given equation for y, this can be achieved by using the chain rule which helps us e
4 min read
Proofs for the derivatives of eˣ and ln(x) - Advanced differentiation
In this article, we are going to cover the proofs of the derivative of the functions ln(x) and ex. Before proceeding there are two things that we need to revise: The first principle of derivative Finding the derivative of a function by computing this limit is known as differentiation from first principles. Derivative by the first principle refers t
3 min read
Disguised Derivatives - Advanced differentiation | Class 12 Maths
The dictionary meaning of “disguise” is “unrecognizable”. Disguised derivative means “unrecognized derivative”. In this type of problems, the definition of derivative is hidden in the form of limit. At a glance, the problem seems to be solvable using limit properties but it is much easier to solve it using the first principle of derivative. So, thi
4 min read
Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.6
Question 1. If [Tex]y=\sqrt{x+\sqrt{x+\sqrt{\ \infin}}}[/Tex], prove that [Tex]\frac{dy}{dx}=\frac{1}{2y-1}[/Tex] Solution: We have, [Tex]y=\sqrt{x+\sqrt{x+\sqrt{\ \infin}}}[/Tex] ⇒ [Tex]y=\sqrt{x+y}[/Tex] Squaring both sides, we get, y2 = x + y [Tex]⇒2y\frac{dy}{dx}=1+\frac{dy}{dx}\\ ⇒\frac{dy}{dx}(2y-1)=1\\ ⇒\frac{dy}{dx}=\frac{1}
3 min read
Class 12 RD Sharma Solutions- Chapter 11 Differentiation - Exercise 11.8 | Set 1
Question 1: Differentiate x2 with respect to x3. Solution: Let u = x2, and let v = x3 Differentiating u with respect to x, du/dx = 2x -----(i) Differentiating v with respect to x, dv/dx = 3x2 ------(ii) Dividing equation (i) by (ii) (du/dx) / (dv/dx) = 2x/3x2 (du/dv) = 2/3x (Ans) Question 2: Differentiate log (1+x2) with respect to tan-1x. Solution
10 min read
Class 12 RD Sharma Solutions- Chapter 11 Differentiation - Exercise 11.8 | Set 2
Question 11: Differentiate sin-1(2x√(1-x2)) with respect to tan-1(x/√(1-x2)) if -1/√2<x <1/√2. Solutions: Let u = sin-1(2x√(1-x2)) Substitute x = sin θ ⇒ θ = sin-1x u = sin-1(2 sin θ √(1 - sin2θ)) u = sin-1(2 sin θ cos θ) [sin2θ + cos2θ = 1] u = sin-1(sin 2θ) -----(i) [sin 2θ = 2 sin θ cos θ] Let, v = tan-1(x/√(1-x2)) v = tan-1(sin θ/ √(1 - s
8 min read
Class 12 RD Sharma Solutions- Chapter 11 Differentiation - Exercise 11.4 | Set 1
Find dy/dx in each of the following:Question 1. xy = c2 Solution: We have xy=c2 Differentiating both sides with respect to x. d(xy)/dx = d(c2)/dx By product rule, y+x*dy/dx=0 Therefore the answer is. dy/dx=-y/x Question 2. y3 -3xy2=x3+3x2y Solution: We have y3-3xy2=x3+3x2y Differentiating both sides with respect to x, d(y3-3xy2)/dx=d(x3+3x2y)/dx By
3 min read
Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.4 | Set 2
Find dy/dx in each of the following.Question 16. If x√(1+y) +y√(1+x) =0 then prove that (1+x)2dy/dx +1=0 Solution: We have, x√(1+y) +y√(1+x) =0 =>x√(1+y)=-y√(1+x) On squaring both sides, we have x2(1+y)=y2(1+x) =>x2+ x2y-y2-y2x=0 =>(x+y)(x-y)+xy(x-y)=0 =>(x-y)(x+y+xy)=0 So, either (x-y)=0 or, x+y+xy=0 =>x+y(1+x)=0 =>y=-x/(1+x) On
3 min read
Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.1
Question 1. Differentiate the following functions from first principles e-x Solution: We have, Let, f(x)=e-x f(x+h)=e-(x+h) [Tex]\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/Tex] [Tex]=\lim_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}[/Tex] [Tex]=\lim_{h\to0}\frac{e^{-x}.e^{-h}-e^{-x}}{h}[/Tex] [Tex]=\lim_{h\to0}[e^{-x}(\frac{e^{-h}-1}{-h})](-1)[/Tex]
2 min read