How implicit differentiation is applied to functions and equations?

The differential is one of the calculus main topics. There is a wide number of topics in calculus used for different purposes in mathematics, biology, physics, chemistry, engineering, and many other forms of study.

For linear and polynomial equations of algebra have two variables, one is dependent and the other is independent we use differential to calculate the slope and area by using implicit differentiation. These kinds of problems are solved by the use of this type of derivative.

What is differentiation?

Differentiation is a process to calculate the rate of change of a function according to its desired variable. The main purpose of differentiation is to boost the performance of all the students, the students are those who fall behind or ahead of year-level hopes. The benefits of differentiation are very high for the students.

In simple words, the derivative of the function is also named differentiation. To determine the instant rate of change of one amount with respect to another amount. Sometimes we also use a word for differentiation in calculus such as differential.

Differentiation is usually stated as a process by which the rate of change of a few amounts with respect to another amount is measured is termed as the differential. After applying the differentiation, the required outputs are said to be derivatives.

y = f(x) is a general term used in derivatives, such as y is a dependent variable and x is an independent variable.

What is Implicit Differentiation?

Implicit differentiation is widely used in derivatives to calculate the differential of an equation or functions having equations. Implicit differentiation is a state-forward topic in which we have to determine the dy/dx of the equation.

In simple words, implicit differentiation is used to calculate the rate of change of the dependent variable with respect to the independent variable. When we have to calculate the differential of the linear equation or polynomial equation, we use implicit differentiation.

How differentiation is applied to calculate the derivative of an equation?

The differential is applied on both sides of the equation. When the differential is applied to both sides of the equation, apply all the rules and formulas to simplify the left-hand side and the right-hand side of the equation.

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The derivative of y in the equation with respect to x is not considered in implicit differentiation. Although the derivative of the dependent variable y is dy/dx. Every equation has a target to calculate the dy/dx of the given equation. To solve the derivatives with implicit differentiation, you can use implicit derivative calculator. Let’s go through few examples to understand the basic concept of implicit derivatives.

Example 1

Evaluate the derivative of the given equation by using implicit differentiation with respect to x.

y2 = (2x – 5y) * (14x + 3x2)

Solution

Step 1: Write the given polynomial function.

y2 = (2x – 5y) * (14x + 3x2)

Step 2: Take the general form.

y = f(x)

Step 3: Apply the d/dx notation on both sides of the given polynomial equation.

d/dx (y2) = d/dx ((2x – 5y) * (14x + 3x2))

Step 4: Apply product law of the function on right side of the equation.

d/dx (y2) = (2x – 5y) d/dx (14x + 3x2) + (14x + 3x2) d/dx (2x – 5y)

Step 5: Apply the laws of sum, difference, power, and constant to simplify the above equation.

2y dy/dx = (2x – 5y) d/dx (14x + 3x2) + (14x + 3x2) d/dx (2x – 5y)

2y dy/dx = (2x – 5y) (d/dx (14x) + d/dx (3x2)) + (14x + 3x2) (d/dx (2x) – d/dx (5y))

2y dy/dx = (2x – 5y) ((14x1-1) + (3×2 x2-1)) + (14x + 3x2) ((2x1-1) – 5dy/dx)

2y dy/dx = (2x – 5y) (14 + 6x) + (14x + 3x2) (2 – 5dy/dx)

2y dy/dx = (2x – 5y) (14 + 6x) + (14x + 3x2) (2 – 5dy/dx)

2y dy/dx = 28x + 12x2 – 70y – 30xy + 28x – 6x2 – 70x dy/dx – 5×3 x2 dy/dx

2y dy/dx = 28x + 12x2 – 70y – 30xy + 28x – 6x2 – 70x dy/dx – 15x2 dy/dx

2y dy/dx + 70x dy/dx + 15x2 dy/dx = 56x + 6x2 – 70y – 30xy

(2y – 70x + 15x2) dy/dx = 56x + 6x2 – 70y – 30xy

dy/dx = (56x + 6x2 – 70y – 30xy) / (2y – 70x + 15x2)

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Hence, the required dy/dx of the equation by using of implicit differentiation.

Example 2

Evaluate the derivative of the given equation by using implicit differentiation with respect to x.

9y2 + 2x = (3x2 – 5y)

Solution

Step 1: Write the given polynomial function.

9y2 + 2x = (3x2 – 5y)

Step 2: Take the general form.

y = f(x)

Step 3: Apply the d/dx notation on both sides of the given polynomial equation.

d/dx (9y2 + 2x) = d/dx (3x2 – 5y)

Step 4: Apply the sum and difference law of the function on both side of the equation.

d/dx (9y2 + 2x) = d/dx (3x2 – 5y)

d/dx (9y2) + d/dx (2x) = d/dx (3x2) – d/dx (5y)

Step 5: Evaluate the differential by using laws of differentiation.

(9 x 2) y2-1 dy/dx + 2x1-1 = (3 x 2) x2-1 – 5dy/dx

18y dy/dx + 2 = 6x – 5dy/dx

18y dy/dx + 5dy/dx = 6x – 2

(18y + 5) dy/dx = (6x – 2)

dy/dx = (6x – 2) / (18y + 5)

Example 2

Evaluate the derivative of the given equation by using implicit differentiation with respect to x.

9y2 = (6x3 – 5y2)

Solution

Step 1: Write the given polynomial function.

9y2 = (6x3 – 5y2)

Step 2: Take the general form.

y = f(x)

Step 3: Apply the d/dx notation on both sides of the given polynomial equation.

d/dx (9y2) = d/dx (6x3 – 5y2)

Step 4: Apply the sum and difference law of the function on both side of the equation.

d/dx (9y2) = d/dx (6x3 – 5y2)

d/dx (9y2) = d/dx (6x3) – d/dx (5y2)

Step 5: Evaluate the differential by using laws of differentiation.

(9 x 2) y2-1 = (6 x 3) x3-1 – (5 x 2) y2-1 dy/dx

18y dy/dx = 18x2 – 10y dy/dx

18y dy/dx + 10y dy/dx = 18x2

(18y + 10y) dy/dx = 18x2

dy/dx = 18x2 / (18y + 10y)

dy/dx = 18x2 / (28y)

dy/dx = 9x2 / 14y

Summary

Differentiation of the equation is done by using implicit differentiation. In which we calculate the dy/dx of the given equation. Dy/dx is the general form of the equation. The equation could be linear or polynomial.

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