# Implicit Differentiation Formula

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## Implicit Differentiation

Implicit differentiation is the process of finding the derivative of an implicit function. There are two types of functions: explicit function and implicit function. An explicit function is of the form y = f(x) with the dependent variable “y” is on one of the sides of the equation. But it is not necessary always to have ‘y’ on one side of the equation. For example, consider the following functions:

• x2 + y = 2
• xy + sin (xy) = 0

In the first case, though ‘y’ is not one of the sides of the equation, we can still solve it to write it like y = 2 – x2 and it is an explicit function. But in the second case, we cannot solve the equation easily for ‘y’, and this type of function is called an implicit function and in this page, we are going to see how to find the derivative of an implicit function by using the process of implicit differentiation.

## What is Implicit Differentiation?

Implicit differentiation is the process of differentiating an implicit function. An implicit function is a function that can be expressed as f(x, y) = 0. i.e., it cannot be easily solved for ‘y’ (or) it cannot be easily got into the form of y = f(x). Let us consider an example of finding dy/dx given the function xy = 5. Let us find dy/dx in two methods: (i) Solving it for y (ii) Without solving it for y.

• Method – 1:
xy = 5
y = 5/x
y = 5x-1
Differentiating both sides with respect to x:
dy/dx = 5(-1x-2) = -5/x2
• Method – 2:
xy = 5
Differntiating both sides with respect to x:
d/dx (xy) = d/dx(5)
Using product rule on the left side,
x d/dx(y) + y d/dx(x) = d/dx(5)
x (dy/dx) + y (1) = 0
x(dy/dx) = -y
dy/dx = -y/x
From xy = 5, we can write y = 5/x.
dy/dx = -(5/x)/x = -5/x2

In Method -1, we have converted the implicit function into the explicit function and found the derivative using the power rule. But in method-2, we differentiated both sides with respect to x by considering y as a function of x, and this type of differentiation is called implicit differentiation. But for some functions like xy + sin (xy) = 0, writing it as an explicit function (Method – 1) is not possible. In such cases, only implicit differentiation (Method – 2) is the way to find the derivative.

### Implicit Derivative

The derivative that is found by using the process of implicit differentiation is called the implicit derivative. For example, the derivative dy/dx found in Method-2 (in the above example) at first was dy/dx = -y/x and it is called the implicit derivative. An implicit derivative usually is in terms of both x and y.

## Implicit Differentiation and Chain Rule

The chain rule of differentiation plays an important role while finding the derivative of implicit function. The chain rule says d/dx (f(g(x)) = (f’ (g(x)) · g'(x). Whenever we come across the derivative of y terms with respect to x, the chain rule comes into the scene and because of the chain rule, we multiply the actual derivative (by derivative formulas) by dy/dx. Here is an example.

Here are more examples to understand the chain rule in implicit differentiation.

• d/dx (y2) = 2y dy/dx
• d/dx (sin y) = cos y dy/dx
• d/dx (ln y) = 1/y · dy/dx
• d/dx (tan-1y) = 1/(1 + y2) · dy/dx

In other words, wherever y is being differentiated, write dy/dx also there. It is suggested to go through these examples again and again as these are very helpful in doing implicit differentiation.

## How to Do Implicit Differentiation?

In the process of implicit differentiation, we cannot directly start with dy/dx as an implicit function is not of the form y = f(x), instead, it is of the form f(x, y) = 0. Note that we should be aware of the derivative rules such as the power rule, product rule, quotient rule, chain rule, etc before learning the process of implicit differentiation. Here is the flowchart of the steps for performing implicit differentiation.

Now, these steps are explained by an example where are going to find the implicit derivative dy/dx if the function is y + sin y = sin x.

• Step – 1: Differentiate every term on both sides with respect to x.
Then we get d/dx(y) + d/dx(sin y) = d/dx(sin x).
• Step – 2: Apply the derivative formulas to find the derivatives and also apply the chain rule.
(All x terms should be directly differentiated using the derivative formulas; but while differentiating the y terms, multiply the actual derivative by dy/dx)

In this example, d/dx (sin x) = cos x whereas d/dx (sin y) = cos y (dy/dx).
Then the above step becomes:
(dy/dx) + (cos y) (dy/dx) = cos x
• Step – 3: Solve it for dy/dx.
Taking dy/dx as common factor:
(dy/dx) (1 + cos y) = cos x
dy/dx = (cos x)/(1 + cos y)
This is the implicit derivative.

### Implicit Differentiation Formula

We have seen the steps to perform implicit differentiation. Did we come across any particular formula along the way? No!! There is no particular formula to do implicit differentiation, rather we perform the steps that are explained in the above flow chart to find the implicit derivative.

Important Notes on Implicit Differentiation:

• Implicit differentiation is the process of finding dy/dx when the function is of the form f(x, y) = 0.
• To find the implicit derivative dy/dx, just differentiate on both sides and solve for dy/dx. But in this process, write dy/dx wherever we are differentiating y.
• All derivative formulas and techniques are to be used in the process of implicit differentiation as well.

## Implicit Differentiation Examples

Example 2: Find the implicit derivative y’ if the function is defined as x + ay2 = sin y, where ‘a’ is a constant.

Solution:

The given equation is:

x + ay2 = sin y

We find the derivative by using implicit differentiation.

Taking derivative of each term on both sides with respect to x:

d/dx (x) + a d/dx (y2) = d/dx (sin y)

(keep in mind that ‘a’ is a constant here and hence

By chain rule,

1 + a (2y · dy/dx) = cos y · dy/dx

Replace dy/dx by y’ and solve for y’.

1 + 2ay y’ = cos y · y’

cos y · y’ – 2ay · y’ = 1

y’ (cos y – 2ay) = 1

y’ = 1/(cos y – 2ay).

Answer: y’ = 1/(cos y – 2ay).

Example 3: Find the second implicit derivative if x2 + y2 = 4.

Solution:

The given equation is,

x2 + y2 = 4

Finding first implicit derivative:

Differentiating both sides with respect to x:

2x + 2y dy/dx = 0

2y dy/dx = -2x

dy/dx = (-2x)/(2y)

dy/dx = -x/y (or)

y’ = -x/y

Finding second implicit derivative:

Differentiating both sides of dy/dx again with respect to x:

d/dx (y’) = d/dx (-x/y) (or)

y” = d/dx (-x/y)

By quotient rule,

y” = [ y d/dx (-x) – (-x) d/dx (y) ] / y2

= [-y + xy’]/y2

Substitute y’ = -x/y,

y” = [-y + x (-x/y) ] / y2

= [-y2 – x2]/y3

Answer:The second implicit derivative is, y” = [-y2 – x2]/y3.

## FAQs on Implicit Differentiation

### What is Implicit Differentiation?

Implicit differentiation is the process of differentiating an implicit function which is of the form f(x, y) = 0 and finding dy/dx. To find it,

• Differentiate both sides of f(x, y) = 0 with respect to x
• Apply usual derivative formulas to differentiate the x terms
• Apply usual derivative formulas to differentiate the y terms along with multiplying the derivative by dy/dx
• Solve the resultant equation for dy/dx (by isolating dy/dx).

### How to Find Implicit Derivative?

To find the implicit derivative of an equation, for example, say, x2 + sin (y) = 0:

• Take the derivative with respect to x on both sides.
Then we get d/dx(x2) + d/dx (sin y) = 0.
• Multiply by dy/dx wherever we are differentiating something with y.
2x + cos y dy/dx = 0.
• Solve it for dy/dx.
cos y dy/dx = -2x
dy/dx = -2x/cos y

### How to Do Implicit Differentiation With Trig Functions?

When we do the implicit differentiation of trig functions, then just apply the normal trig derivatives such as d/dx(sin x) = cos x, d/dx(cos x) = – sin x, etc and then apply chain rule. It means we should multiply the actual derivative by the derivative of the inside function. For example, d/dx (sin y2) = cos y2 d/dx (y2) = 2y cos y dy/dx.

### What are Implicit Differentiation Rules?

While finding an implicit derivative, we just differentiate an equation in terms of x and y on both sides with respect to x, use dy/dx also whenever we are differentiating something with y, and solve the resultant equation for dy/dx.

### What is Implicit Differentiation Meaning?

The meaning of implicit differentiation, as its name suggests, is the process of differentiating an implicit function f(x, y) = 0 and finding the derivative dy/dx. To know how to do implicit differentiation

### How to Find Second Implicit Derivative?

When an implicit function f(x, y) = 0 is given, use the process of implicit differentiation to find the first derivative dy/dx (or) y’. We then differentiate the first derivative y’ with respect to x on both sides to find the second implicit derivative. In this process, we may have to use the answer of y’ also.

### What is Implicit Differentiation Formula?

There is no specific formula for doing implicit differentiation. Instead, we just differentiate the function on both sides without ignoring the chain rule and solving the resultant equation for dy/dx. Wherever we differentiate something with y, just multiply the derivative by dy/dx also.

## Implicit vs Explicit

A function can be explicit or implicit:

Explicit: “y = some function of x”. When we know x we can calculate y directly.

Implicit: “some function of y and x equals something else”. Knowing x does not lead directly to y.

## How to do Implicit Differentiation

• Differentiate with respect to x

Again, all we did was differentiate with respect to y and multiply by dy/dx

## Explicit

Let’s also find the derivative using the explicit form of the equation.

• To solve this explicitly, we can solve the equation for y
• Then differentiate
• Then substitute the equation for y again

### Chain Rule Again!

Yes, we used the Chain Rule again. Like this (note different letters, but same rule):

## Using The Derivative

OK, so why find the derivative y’ = −x/y ?

Well, for example, we can find the slope of a tangent line.

## Another Example

Sometimes the implicit way works where the explicit way is hard or impossible.

## Inverse Functions

Implicit differentiation can help us solve inverse functions.

The general pattern is:

• Start with the inverse equation in explicit form. Example: y = sin−1(x)
• Rewrite it in non-inverse mode: Example: x = sin(y)
• Differentiate this function with respect to x on both sides.
• Solve for dy/dx

As a final step we can try to simplify more by substituting the original equation.

An example will help:

## Summary

• To Implicitly derive a function (useful when a function can’t easily be solved for y)
• Differentiate with respect to x
• Collect all the dy/dx on one side
• Solve for dy/dx
• To derive an inverse function, restate it without the inverse then use Implicit differentiation

## Implicit Vs Explicit Functions

But to really understand this concept, we first need to distinguish between explicit functions and implicit functions.

An explicit function is an equation written in terms of the independent variable, whereas an implicit function is written in terms of both dependent and independent variables.

Explicit Vs. Implicit

Notice how all the explicit functions are solved for one variable (i.e., one variable on the left-hand-side and every other term is on the right) while the implicit functions have the variables all intermixed on both sides of the equation. This does not leave an easy way for us to solve for one variable.

## How To Do Implicit Differentiation

In all of our previous derivative lessons, we have dealt with explicit functions only, as they are already solved for one variable in terms of another. But now it’s time to learn how to find the derivative, or rate of change, of equations that contain one or more variables and when x and y are intermixed.

1. Take the derivative of every variable.
2. Whenever you take the derivative of “y” you multiply by dy/dx.
3. Solve the resulting equation for dy/dx.

### Example

Let’s use this procedure to solve the implicit derivative of the following circle of radius 6 centered at the origin.

Implicit Differentiation Example – Circle

And that’s it!

The trick to using implicit differentiation is remembering that every time you take a derivative of y, you must multiply by dy/dx.

Furthermore, you’ll often find this method is much easier than having to rearrange an equation into explicit form if it’s even possible.

### Example

Let’s look a harder problem with trig where x’s and y’s are intermixed.

Implicit Derivative – Trig And Exponential Functions

### Example

And sometimes, we will experience implicit functions with more than one y-variable. All this means is that we will have multiple dy/dx terms that we will need to collect in order to simplify, as the following example nicely highlights.

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