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Implicit Differentiation
Implicit differentiation is the process of finding the derivative of an implicit function. i.e., this process is used to find the implicit derivative. 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:
 x^{2} + 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  x^{2} 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/x^{2}  Method  2:
xy = 5
Differentiating 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/x^{2}
In Method 1, we have converted the implicit function into the explicit function and found the derivative using the power rule. But in method2, 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 Method2 (in the above example) at first was dy/dx = y/x and it is called the implicit derivative. This is because we have differentiated the implicit function xy = 5 directly without solving it for y. 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 (y^{2}) = 2y dy/dx
 d/dx (sin y) = cos y dy/dx
 d/dx (ln y) = 1/y · dy/dx
 d/dx (tan^{1}y) = 1/(1 + y^{2}) · 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.
Implicit Differentiation of Inverse Trigonometric Functions
The process of implicit differentiation is helpful in finding the derivatives of inverse trig functions. Let us find the derivative of y = tan^{1}x using implicit differentiation. From the definition of arctan, y = tan^{1} x ⇒ tan y = x. Differentiating this equation both sides with respect to x,
sec^{2}y × dy/dx = 1 ( because the derivative of tan x is sec^{2}x)
dy/dx = 1/sec^{2}y
dy/dx = 1 / (1 + tan^{2}y) ( by one of the trigonometric identities)
dy/dx = 1 / (1 + x^{2}) (because tan y = x)
In this way, the implicit differentiation process can be used to find the derivatives of any inverse function.
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.
☛Related Topics:
Implicit Differentiation Examples

Example 1: Find dy/dx by implicit differentiation: 3x + 2y = cos y.
Solution:
The given equation is,
3x + 2y = cos y
Differentiating both sides with respect to x:
3 d/dx (x) + 2 d/dx(y) = d/dx (cos y)
3(1) + 2 (dy/dx) = sin y dy/dx
2 (dy/dx)  sin y dy/dx = 3
dy/dx (2 + sin y) = 3
dy/dx = 3/(2 + sin y)
Answer:The implicit derivative, dy/dx = 3/(2 + sin y).

Example 2: Find the implicit derivative y' if the function is defined as x + ay^{2} = sin y, where 'a' is a constant.
Solution:
The given equation is:
x + ay^{2} = 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 (y^{2}) = d/dx (sin y)
(keep in mind that 'a' is a constant here and hence its derivative is 0)
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 x^{2} + y^{2} = 4.
Solution:
The given equation is,
x^{2} + y^{2} = 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) ] / y^{2}
= [y + xy']/y^{2}
Substitute y' = x/y,
y'' = [y + x (x/y) ] / y^{2}
= [y^{2}  x^{2}]/y^{3}
Answer:The second implicit derivative is, y'' = [y^{2}  x^{2}]/y^{3}.
FAQs on Implicit Differentiation
What is the Definition of Implicit Differentiation in Calculus?
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 the implicit derivative,
 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, x^{2} + sin (y) = 0:
 Take the derivative with respect to x on both sides.
Then we get d/dx(x^{2}) + 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 y^{2}) = cos y^{2} d/dx (y^{2}) = 2y cos y^{2} 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, click here.
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 the 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.
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