# How to perform implicit differentiation on the different types of functions?

In calculus, differentiation is used on a larger scale to calculate the change in the given function with respect to its variable. There are two types of differentiation in calculus. One is explicit differentiation and the other is implicit differentiation. To calculate the derivative of simple variables explicit differentiation is used, while on the other hand, implicit differentiation is used to calculate the derivative of an equation.

## Implicit differentiation

Implicit differentiation is a type of differentiation in calculus. When we have to calculate the differentiation of an equation, we use implicit differentiation in which we find dy/dx of the given equation.

The equation has dependent variables as well as independent variables such as

x^{2} + y = 3y + x

in the above equation, we cannot apply the explicit differentiation, so we have to use the explicit differentiation for the calculation of the given problem. Simply apply the differentiation on both sides of the given function, and the differentiation of y must be dy/dx, and the differentiation of y^{2} must be 2y dy/dx, and so on.

A method to calculate the derivative of an implicit function or an equation is said to be the implicit differentiation. While on the other hand, explicit is applicable on single-sided values such as y = f(x).

## How to perform implicit differentiation?

For the calculation of implicit function, there is no specific formula. In order to calculate the implicit function simply take the derivative on both sides of the function. In implicit differentiation, we have to calculate the dy/dx of the given function.

The differentiation of y must be dy/dx, and the differentiation of y^{2} must be 2y dy/dx, and so on. While for the calculation of x terms, simply apply explicit differentiation on them.

For the calculation of the implicit function, all the basic rules of derivatives must be used. For accurate and step-by-step solutions to your problem, the implicit differentiation calculator is very useful for this purpose. Letâ€™s go through some examples of implicit to understand more accurately.

**Example 1**

Find the implicit differentiation of the given function, 3x^{4} + 2y^{2} = 9x + 18

**Solution **

**Step 1:** write the given implicit function.

3x^{4} + 2y^{2} = 9x + 18

**Step 2:** Apply d/dx on both sides of the given implicit function.

d/dx (3x^{4} + 2y^{2}) = d/dx (9x + 18)

**Step 3:** Apply the rule of the differentiation.

d/dx (3x^{4}) +d/dx (2y^{2}) = d/dx (9x) + d/dx (18)

**Step 4:** Solve the derivative.

12x^{3 }+ 4y dy/dx = 9 + 0

12x^{3 }+ 4y dy/dx = 9

**Step 5:** place dy/dx on one side of the equation.

12x^{3 }+ 4y dy/dx = 9

4y dy/dx = 9 – 12x^{3 }

dy/dx = 9 – 12x^{3} / 4y

Hence, the implicit differentiation of the given function is dy/dx = 9 – 12x^{3} / 4y.

**Example 2**

Find the implicit differentiation of the given function, 3x^{2} + 2y^{2} = 6y^{3} â€“ 9x + 34.

**Solution **

**Step 1:** write the given implicit function.

3x^{2} + 2y^{2} = 6y^{3} â€“ 9x + 34

**Step 2:** Apply d/dx on both sides of the given implicit function.

Dy/dx (3x^{2} + 2y^{2}) = dy/dx (6y^{3} â€“ 9x + 34)

**Step 3:** Apply the rules of the differentiation.

d/dx (3x^{2}) + d/dx (2y^{2}) = d/dx (6y^{3}) â€“ dy/dx (9x) + dy/dx (34)

**Step 4:** Solve the derivative.

6x + 4y dy/dx = 18y^{2} dy/dx â€“ 9 + 0

6x + 4y dy/dx = 18y^{2} dy/dx â€“ 9

**Step 5:** place dy/dx on one side of the equation.

6x + 4y dy/dx = 18y^{2} dy/dx â€“ 9

4y dy/dx â€“ 18y^{2} dy/dx = â€“ 9 â€“ 6x

(4y â€“ 18y^{2}) dy/dx = â€“ (9 + 6x)

dy/dx = â€“ (9 + 6x) / (4y â€“ 18y^{2})

Hence, the implicit differentiation of the given function is dy/dx = â€“ (9 + 6x) / (4y â€“ 18y^{2}).

**Example 3**

Find the implicit differentiation of the given function, x^{3} â€“ x^{2} + 3y^{2} = 7y^{2} + 7y + sin(x)

**Solution **

**Step 1:** write the given implicit function.

x^{3} â€“ x^{2} + 3y^{2} = 7y^{2} + 7y + sin(x)

**Step 2:** Apply d/dx on both sides of the given implicit function.

d/dx (x^{3} â€“ x^{2} + 3y^{2}) = dy/dx (7y^{2} + 7y + sin(x))

**Step 3:** Apply the rules of the differentiation.

d/dx (x^{3}) â€“ d/dx (x^{2}) +d/dx (y^{2}) = d/dx (7y^{2}) + dy/dx (7y) + d/dx (sin(x))

**Step 4:** Solve the derivative.

3x^{2} â€“ 2x + 2y dy/dx = 14y dy/dx +7 dy/dx + cos(x)

**Step 5:** Arrange the above equation to get dy/dx.

3x^{2} â€“ 2x + 2y dy/dx = 14y dy/dx +7 dy/dx + cos(x)

+ 2y dy/dx = 14y dy/dx + 7 dy/dx + cos(x) â€“ 3x^{2} + 2x

2y dy/dx â€“ 14y dy/dx â€“ 7 dy/dx = cos(x) â€“ 3x^{2} + 2x

(2y â€“ 14y â€“ 7) dy/dx = cos(x) â€“ 3x^{2} + 2x

(â€“ 12y â€“ 7) dy/dx = cos(x) â€“ 3x^{2} + 2x

dy/dx = (cos(x) â€“ 3x^{2} + 2x) / (â€“ 12y â€“ 7)

Hence, the implicit differentiation of the given function is dy/dx = (cos(x) â€“ 3x^{2} + 2x) / (â€“ 12y â€“ 7)

**Alternate method**

There is another method to calculate the implicit differentiation of the given function known as the chain rule.

**Example **

Find the implicit differentiation of the given function, 2x^{3}Â + 3y^{2}Â = 15Â with respect toÂ x.

**Solution**

**Step 1: **Â write the given implicit function.

2x^{3}Â + 3y^{2}Â = 15

**Step 2:** Apply d/dx on both sides of the given implicit function.

d/dx (2×3+3y2) = d/dx (15)

**Step 3:Â **Apply the rules of the differentiation.

d/dx (2×3) + d/dx (3y2) = d/dx (15)

**Step 4:Â **use the power rule.

d/dx (3y2) + 6×2 = d/dx (15)

**Step 4:**Â Use the chain rule, where u = y and

d/du (u2) = 2u

6×2 + 6y d/dx (y) = d/dx (15)

**Step 5:**Â Using the chain rule, where u = x.

d/du (y(u)) = yâ€²(u)

dx + (d/dx(x)) yâ€²(x)2y = d/dx (15)

The derivative of x is 1.

(6×2+1)6yyâ€²(x) = d/dx (5)

**Step 6:Â **Use the coefficient rule.

6×2 + 6yyâ€²(x) = 0

6yyâ€²(x) = âˆ’6×2

yâ€²(x) = -6x/6y

yâ€²(x) = -x/y

Hence, the implicit differentiation of the given function is dy/dx = -x/y

## Summary

Implicit differentiation is a type of differentiation in calculus. It is a process of finding dy/dx of the given implicit function. For the calculation of implicit function, the differentiation notation must be applied on both sides of the function.