Implicit differentiation is a type of derivative in which we have to find the derivative of the equation. It is widely used in calculus to find the derivatives of the algebraic expression. In this post, we will learn the definition of Implicit differentiation and how to find them with a lot of examples.
What is Implicit differentiation?
The process of finding the derivative of a dependent variable in an implicit function by differentiating each term separately by expressing the derivative of the dependent variable as a symbol, and solving a resulting expression for a symbol. In implicit differentiation, we differentiate each side of an equation with two variables usually x and y by treating one of the variables as a function of the other. The technique of implicit differential allows you to find the derivative of y with respect to x without having to solve the given equation for y.
The process of finding dy/dx is known as implicit differentiation. In simple words to find the derivative of an equation we use implicit differentiation. In this process, we apply the derivative on both sides. In the equation, we also find the derivative of y with respect to x, the derivative of y with respect to x is dy/dx and the derivative of y2 is 2y*dy/dx.

Example 1
Find the derivative of x2 + y2 = 10
Solution
Step 1: First of all, take derivative on both sides
ddx(x2 + y2) = ddx10
Step 2: Apply the sum rule
ddx(x2) +ddx (y2) = ddx10
Step 3: Now apply the power and constant rule
2×2-1+2ydydx = 0
2×1+2ydydx = 0
2ydydx = -2x
dydx = -2x/2y
= -x/y
Example 2
Find the derivative of 2xy + y2 = 2y
Solution
Step 1: First of all, take derivative on both sides
ddx(2xy + y2) = ddx2y
Step 2: Apply the sum rule
ddx(2xy) +ddx (y2) = ddx2y
Step 3: Now apply the product and power rule
2xdydx+2ydxdx+2ydydx = 2dydx
2xdydx+2y+2ydydx = 2dydx
2xdydx+2ydydx-2dydx = -2y
(2x+2y-2)dydx = -2y
dydx = -2y/ (2x+2y-2)
dydx = -y/ (x+y-1)
If you are looking for a tool to perform implicit differentiation without wasting your time, implicit Differentiation calculator will be very helpful because it shows all steps of the calculation along with plot.
Example 3
Find the derivative of 4x3 + y2 = 100
Solution
Step 1: First of all, take derivative on both sides
ddx(4x3 + y2) = ddx100
Step 2: Apply the sum rule
ddx(4x3) +ddx (y2) = ddx100
Step 3: Now apply the power and constant rule
4*3×3-1+2ydydx = 0
12×2+2ydydx = 0
2ydydx = -12x2
dydx = -12x2/2y
dydx = -6x2/y
Example 4
Find the derivative of 2xy + y2 = 2y2
Solution
Step 1: First of all, take derivative on both sides
ddx(2xy + y2) = ddx2y2
Step 2: Apply the sum rule
ddx(2xy) +ddx (y2) = ddx2y2
Step 3: Now apply the product and power rule
2xdydx+2ydxdx+2ydydx = 4ydydx
2xdydx+2y+2ydydx = 4ydydx
2xdydx+2ydydx-4ydydx = -2y
(2x+2y-4y)dydx = -2y
(2x-2y)dydx = -2y
dydx = -2y/ (2x-2y)
dydx = -y/ (x-y)
Implicit differentiation can also be done by using the chain rule. This rule is applicable for complicated problems such as you are stuck on a problem and feel difficult to solve the problem use this rule in which you can suppose the functions.
Example
Find the derivative of x3 + y2 = 5 with respect to x.
Solution
It is important to choose a differentiation notation. The above calculator uses dy/dx. We will also use the same notation in this example.
d/dx(x3+y2) = d/dx (5)
Step 1: Differentiate the expression using linearity. i.e., differentiate each term separately.
d/dx (x3) + d/dx (y2) = d/dx (5)
Step 2: use the power rule.
d/dx (y2) + 3x2 = d/dx (5)
Step 3: Use the chain rule, where u = y and
d/du (u2) = 2u
3x2+2y d/dx (y) = d/dx (5)
Step 4: Using the chain rule, where u = x and
d/du (y(u)) = y′(u)
dx+(d/dx(x)) y′(x)2y = d/dx (5)
The derivative of x is 1.
(3x2+1)2yy′(x) = d/dx (5)
Step 5: Use the coefficient rule.
3x2 + 2yy′(x) = 0
2yy′(x) = −3x2
y′(x) = -3x/2y
Difference between Implicit and Explicit Differentiation
The process of finding dy/dx is known as implicit differentiation. In simple words to find the derivative of an equation we use implicit differentiation. While on the other hand, explicit differentiation is simply taking the derivative of a function. In an explicit function, one variable is defined completely in terms of the other. This usually means that the independent variable x is written explicitly in terms of the dependent variable y.
In implicit equations are involved while in explicit only a function is involved to be calculated.
Let us take some examples in order to show the difference between implicit differentiation and explicit differentiation.
Examples of Implicit differentiation:
Example 1
Find the derivative of x2y + y2 = y2
Solution
Step 1: First of all, take derivative on both sides
ddx(x2y + y2) = ddxy2
Step 2: Apply the sum rule
ddx(x2y) +ddx (y2) = ddxy2
Step 3: Now apply the product and power rule
x2dydx+yddxx2+2ydydx = 2ydydx
x2dydx+y(2x)+2ydydx = 2ydydx
x2dydx+2ydydx-2ydydx = -2xy
(x2+2y-2y)dydx = -2xy
(x2)dydx = -2xy
dydx = -2xy/ x2
dydx = -2y/ x
Example 2
Find the derivative of x3 + y2 = 10
Solution
Step 1: First of all, take derivative on both sides
ddx(x3 + y2) = ddx10
Step 2: Apply the sum rule
ddx(x3) +ddx (y2) = ddx10
Step 3: Now apply the power and constant rule
3×3-1+2ydydx = 0
3×2+2ydydx = 0
2ydydx = -3x2
dydx = -3x2/2y
Examples of Explicit Differentiation:
Example 1
Find the Explicit differentiation of x2+2x?
Solution
Step 1: write the derivative notation.
d/dx (x2 + 2x)
Step 2: Apply sum rule.
d/dx (x2 + 2x) = d/dx (x2) + d/dx (2x)
Step 2: Apply power rule and solve.
d/dx (x2 + 2x) = 2x2-1 + 2x1-1
= 2x1 + 2x0
= 2x + 2 = 2(x + 1)
Example 2
Find the derivative of x2 – 3x?
Solution
Step 1: write the derivative notation.
d/dx (x2 – 3x)
Step 2: Apply difference rule.
d/dx (x2 – 3x) = d/dx (x2) – d/dx (3x)
Step 2: Apply power rule and solve.
d/dx (x2 – 3x) = d/dx (x2) – d/dx (3x)
= 2x2-1 – 3x1-1
= 2x1 – 3x0
= 2x – 3
Summary
In this post we have learnt about the types of derivatives one is implicit and the other is explicit. We discussed the definitions and the basics of implicit and explicit along with examples. Now with a little practice, you will be the master in this topic. This topic is not much difficult but if you did not grab the basic knowledge about this topic or you did not practice this topic you will find this topic rather difficult.