Implicit differentiation is the main type of differential calculus. It is widely used to find the derivative of one variable with respect to another variable. The derivative is a major topic of calculus used to calculate the rate of change in the given function.

The other type of derivative is explicit differentiation. Both types are used to solve the problems of differential calculus. In this post, we will discuss the definition, methods, and examples of implicit differentiation.

Index

**What is implicit differentiation?**

Implicit differentiation is a branch of differentiation in which you can calculate the derivative of an equation. In this type of derivative, two variables are used like x and y. These variables behave as one is the function of the other and you have to calculate dy/dx of the given function.

In implicit differentiation, the term y with respect to x is not considered as constant. The derivative of **y**^{2}** **in the implicit differentiation must be **2y dy/dx**. All the formulas and rules remain the same in this type of differentiation.

The implicit differentiation can be defined as calculating the derivative of **y** with respect to **x **without solving the given equation for **y**. The main purpose of this type of differentiation is to calculate the dy/dx term of the given equation. More simply, we can say that implicit differentiation is used to calculate the differential of the inverse function.

**Rules of implicit differentiation**

Rules of differentiation remains unchanged for implicit differentiation.

**Sum rule **

d/dx (u + v) = d/dx (u) + d/dx (v)

**Difference rule **

d/dx (u – v) = d/dx (u) – d/dx (v)

**Product rule**

d/dx (u * v) = u d/dx (v) + v d/dx (u)

**Quotient rule **

d/dx (u / v) = 1/v^{2} [v d/dx (u) – u d/dx (v)]

**Power rule**

d/dx (u^{n}) = n (u)^{n-1} d/dx (u)

By using rules and formulas of differentiation, we can easily find the implicit differentiation of the given equation.

**How to calculate the implicit differentiation?**

Implicit differentiation of the given function can be calculated by treating x and y as variables and applying the rules of differentiation. You can also use an implicit differentiation calculator for getting the step-by-step solution of the given problems. In just a single click, you can calculate the result. You have to put the question into the input box like:

Then click the “**calculate” **button the result will show below the calculate button in a couple of seconds.

By pressing show steps, you can see the step-by-step solution of the given problem.

**Example 1**

Calculate the derivative of the given function with respect to **x**, 9x^{2}y + 17xy^{2} – 19xy^{3} = 3x^{5} / 3x^{2} + 17x + 2y + 3?

**Solution**

**Step 1:** First of all, write the given equation.

9x^{2}y + 17xy^{2} – 2y^{3} = 3x^{5} / 3x^{2} + 17x + 2y + 3

**Step 2:** Now apply the differential operator on both side in the given equation.

d/dx (9x^{2}y + 17xy^{2} – 2y^{3}) = d/dx (3x^{5} / 3x^{2} + 17x + 2y + 3)

**Step 3:** Apply the difference, product, sum, and quotient rules on the above equation.

d/dx (9x^{2}y) + d/dx (17xy^{2}) – d/dx (2y^{3}) = d/dx (3x^{5} / 3x^{2}) + d/dx (17x) + d/dx (2y) + d/dx (3)

9x^{2} d/dx (y) + y d/dx (9x^{2}) + x d/dx (17y^{2}) + y^{2} d/dx 17x – d/dx (2y^{3}) = 1/(3x^{2})^{2} [3x^{2} * d/dx (3x^{5}) – 3x^{5} d/dx (3x^{2})] + d/dx (17x) + d/dx (2y) + d/dx (3)

**Step 4:** Now apply the constant and power rules on the above equation.

9x^{2} d/dx (y) + 9y d/dx (x^{2}) + 17x d/dx (y^{2}) + 17y^{2} d/dx (x) – 2d/dx (y^{3}) = 1/(3x^{2})^{2} [3x^{2} * 3d/dx (x^{5}) – 3x^{5} 3d/dx (x^{2})] + 17d/dx (x) + 2d/dx (y) + 0

9x^{2} dy/dx + 9y (2x^{2-1}) + 17x (2y^{2-1}) dy/dx + 17y^{2} (x^{1-1}) – 2 (3y^{3-1}) = 1/(9x^{4}) [9x^{2} * (5x^{5-1}) – 9x^{5} (2x^{2-1})] + 17 (x^{1-1}) + 2dy/dx

9x^{2} dy/dx + 18xy + 34xy dy/dx + 17y^{2} – 6y^{2} = 1/(9x^{4}) [9x^{2} * (5x^{4}) – 9x^{5} (2x)] + 17 + 2dy/dx

9x^{2} dy/dx + 18xy + 34xy dy/dx + 17y^{2} – 6y^{2} dy/dx = 1/(9x^{4}) [45x^{6} – 18x^{6}] + 17 + 2dy/dx

9x^{2} dy/dx + 18xy + 34xy dy/dx + 17y^{2} – 6y^{2} dy/dx = 1/(9x^{4}) [27x^{6}] + 17 + 2dy/dx

9x^{2} dy/dx + 18xy + 34xy dy/dx + 17y^{2} – 6y^{2} dy/dx = 3x^{2} + 17 + 2dy/dx

**Step 5:** Take the dy/dx term on one side and take it as common.

9x^{2} dy/dx + 34xy dy/dx – 6y^{2} dy/dx – 2dy/dx = 3x^{2} + 17 – 18xy – 17y^{2}

(9x^{2} + 34xy – 6y^{2}– 2) dy/dx = 3x^{2} + 17 – 18xy – 17y^{2}

dy/dx = (3x^{2} + 17 – 18xy – 17y^{2}) / (9x^{2} + 34xy – 6y^{2}– 2)

**Example 2**

Calculate the derivative of the given function with respect to x, 3xy^{2} + 4x^{2}y – 13y = 3x^{5} * 19y^{2} + 34x + 2?

**Solution**

**Step 1:** First of all, write the given equation.

3xy^{2} + 4x^{2}y – 13y = 3x^{5} * 19y^{2} + 34x + 2

**Step 2:** Now apply the differential operator on both side in the given equation.

d/dx (3xy^{2} + 4x^{2}y – 13y) = d/dx (3x^{5} * 19y^{2} + 34x + 2)

**Step 3:** Apply the difference, product, sum, and quotient rules on the above equation.

d/dx (3xy^{2}) + d/dx (4x^{2}y) – d/dx (13y) = d/dx (3x^{5} * 19y^{2}) + d/dx (34x) + d/dx (2)

3x d/dx (y^{2}) + y^{2} d/dx (3x) + 4x^{2 }d/dx (y) + y d/dx (4x^{2}) – d/dx (13y) = 19y^{2} d/dx (3x^{5}) + 3x^{5} d/dx (19y^{2}) + d/dx (34x) + d/dx (2)

**Step 4:** Now apply the constant and power rules on the above equation.

3x (2y^{2-1}) dy/dx + y^{2} (3(1)) + 4x^{2 }dy/dx + y ((4 * 2) x^{2-1}) – 13 dy /dx = 19y^{2} ((3 * 5) x^{5-1}) + 3x^{5} ((19 * 2) y^{2-1}) dy/dx + (34(1)) + 0

3x (2y dy/dx) + y^{2} (3) + 4x^{2 }dy/dx + y (8x) – 13 dy /dx = 19y^{2} (15x^{4}) + 3x^{5} (38y) dy/dx + (34(1))

6xy dy/dx + 3y^{2} + 4x^{2 }dy/dx + 8xy – 13 dy /dx = 285x^{4}y^{2} + 114x^{5}y dy/dx + 34

**Step 5:** Take the dy/dx term on one side and take it as common.

6xy dy/dx + 4x^{2 }dy/dx– 13 dy /dx – 114x^{5}y dy/dx = 285x^{4}y^{2} + 34 – 8xy – 3y^{2}

(6xy + 4x^{2} – 13 – 114x^{5}y) dy/dx = 285x^{4}y^{2} + 34 – 8xy – 3y^{2}

dy/dx = (285x^{4}y^{2} + 34 – 8xy – 3y^{2}) / (6xy + 4x^{2} – 13 – 114x^{5}y)

**Conclusion **

Implicit differentiation is used to find the values of dy/dx by applying rules to the given equation without solving the terms of ** y**. You can easily solve the problems of implicit differentiation, once you grab all the basics of implicit differentiation.