Differential calculus is one of the main types of calculus used to find the tangent line’s slope. It finds the derivative of one, two, and multivariable functions depending upon the types of derivatives. The rules of differential are used to solve the differential calculus problems.

The differential could be calculated by using limits known as the first principle method of derivatives. In this article, we will study the definition, types, rules, and examples of differential calculus.

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## What is differential calculus?

In mathematical analysis, differential calculus is concerned with the problem of determining the rate of change of a function w.r.t the variable on which it depends. This branch of calculus involves evaluating differential and using these terms to solve problems involving non-constant rates of change.

The differential functions could be exponential, constant, linear, logarithmic, trigonometric, polynomial, or quadric.

## Types of differential calculus

There are three well-known types of differential calculus.

- Explicit derivative
- Implicit differentiation
- Partial derivative

**Explicit derivative**

The term explicit differentiation is used to find the differential of a single variable function f(x). the variable of the function could be x, y, z, u, v, w, t, etc. This type of differential calculus is denoted by d/dx or f’(x). for example, the differential of sin(x) is cos(x) with respect to “x”.

**Implicit differentiation **

The implicit differentiation is that type of derivative that is used to find the differential of double variable function or implicit function. In this type of differential calculus, you have to find the derivative of the dependent variable with respect to the independent variable without considering it constant.

Such as dy/dx, the derivative of y^{3} with respect to x must be 3y^{2} dy/dx. It is denoted by dy/dx or y’(x).

**Partial derivative**

The partial derivative deals with the multivariable functions such as f(x, y), f(x, y, z), etc. This type of differential calculus finds the derivative of multivariable functions with respect to one of its independent variables by taking other functions as a constant.

It is denoted by ∂f(x, y, z)/∂x, ∂f(x, y, z)/∂y, and ∂f(x, y, z)/∂z

## Rules of differential calculus

The rules of differential calculus are very helpful for finding the derivative of functions.

- Constant rule
- Power rule
- Sum rule
- Difference rule
- Product rule
- Quotient rule

Use a differentiation calculator to solve the problems of differential calculus according to its rules.

Let us discuss some rules of differential calculus briefly.

**Constant rule of differential calculus**

This rule of differential calculus is applied when the given function is a constant number or variable. According to this rule of differential function, the derivative of the constant function is always zero.

d/dt [L] = 0, where L is any constant

**Example **

Differentiate 8 with respect to “t”.

**Solution **

**Step I:** Apply the differential notation to the given function.

d/dt [8]

**Step II:** Now differentiate the above constant function.

d/dt [8] = 0 (by constant rule of differential function)

**Power rule of differential calculus**

The power rule of the differential calculus is used when the exponential function is given. According to this rule of differential calculus, the power of the variable must be multiplied by the coefficient of the variable and the power must be decreased by 1.

After that take the derivative of the variable of the exponential function. the general formula for the power rule is:

d/dt [h^{n}(t)] = n * h^{n-1}(t) * d/dt (h(t))

**Example of power rule**

Differentiate 8t^{4} with respect to “t”.

**Solution **

**Step I:** Apply the differential notation to the given function

d/dt [8t^{4}]

**Step II:** Now differentiate the above exponential function.

d/dt [8t^{4}] = (8 * 4) t^{4 – 1} d/dt(t) … (by power rule of differential calculus)

= 32 * t^{4 – 1} d/dt(t)

= 32 * t^{3} d/dt(t)

= 32 * t^{3} (1)

= 32t^{3}

**Difference rule of differential calculus**

The difference rule of the differential calculus is used when two or more functions are given along with the subtraction sign among them. According to the difference rule of the differential calculus, the notation of the derivative must be applied to each function separately.

The formula for the 2 and 3 functions are:

- d/dt [h(t) – j(t)] = d/dt [h(t)] – d/dt [j(t)]
- d/dt [h(t) – j(t) – k(t)] = d/dt [h(t)] – d/dt [j(t)] – d/dt [k(t)]

**Example of the difference rule**

Differentiate 6t^{4} – 8t^{3} – 4t with respect to “t”.

**Solution **

**Step I:** Apply the differential notation to the given function

Function = 6t^{4} – 8t^{3} – 4t

d/dt [h(t) – j(t) – k(t)] = d/dt [6t^{4} – 8t^{3} – 4t]

**Step II:** Apply the differential notation to each function separately according to the difference rule of differential calculus.

d/dt [6t^{4} – 8t^{3} – 4t] = d/dt [6t^{4}] – d/dt [8t^{3}] – d/dt [4t]

**Step III:** Now differentiate the above expression.

d/dt [6t^{4} – 8t^{3} – 4t] = d/dt [6t^{4}] – d/dt [8t^{3}] – d/dt [4t]

= 6 d/dt [t^{4}] – 8 d/dt [t^{3}] – 4 d/dt [t]

= 6 [4 t^{4 – 1}] – 8 [3 t^{3 – 1}] – 4 [t^{1 – 1}]

= 6 [4 t^{3}] – 8 [3 t^{2}] – 4 [t^{0}]

= 24 t^{3} – 24 t^{2} – 4 [1]

= 24t^{3} – 24t^{2} – 4

= 4 [6t^{3} – 6t^{2} – 1]

**Product rule of differential calculus**

The product rule of the differential calculus is used when two or more functions are given along with the multiply sign among them. According to the difference rule of the differential calculus, you have to take the derivative of the function while the other function remains unchanged.

Plus, the derivative of the second function while the first function remains unchanged.

- d//dt [h(t) * j(t)] = j(t) d/dt [h(t)] + h(t) d/dt [j(t)]
- d/dt [h(t) * j(t) * k(t)] = [d/dt (h(t)) * j(t) * k(y)] + [h(t) * d/dt (j(t)) * k(t)] + [h(t) * j(t) * d/dt (k(t)]

**Example of the product rule**

Differentiate 3t^{3} * 5t^{2} * 6t with respect to “t”.

**Solution **

**Step I:** Apply the differential notation to the given function

Function = 3t^{3} * 5t^{2} * 6t

d/dt [h(t) * j(t) * k(t)] = d/dt [3t^{3} * 5t^{2} * 6t]

**Step II:** Apply the differential notation to function separately according to the product rule of differential calculus.

d/dt [3t^{3} * 5t^{2} * 6t] = [d/dt (3t^{3}) * 5t^{2} * 6t] + [3t^{3} * d/dt (5t^{2}) * 6t] + [3t^{3} * 5t^{2} * d/dt (6t)]

**Step III:** Now differentiate the above expression.

d/dt [3t^{3} * 5t^{2} * 6t] = [d/dt (3t^{3}) * 5t^{2} * 6t] + [3t^{3} * d/dt (5t^{2}) * 6t] + [3t^{3} * 5t^{2} * d/dt (6t)]

= [(3 x 3 t^{3 – 1}) * 5t^{2} * 6t] + [3t^{3} * (5 x 2 t^{2 – 1}) * 6t] + [3t^{3} * 5t^{2} * (6 x 1 t^{1 – 1})]

= [(9 t^{2}) * 5t^{2} * 6t] + [3t^{3} * (10 t^{1}) * 6t] + [3t^{3} * 5t^{2} * (6 t^{0})]

= [(9 t^{2}) * 5t^{2} * 6t] + [3t^{3} * (10 t) * 6t] + [3t^{3} * 5t^{2} * (6 (1))]

= [(9 t^{2}) * 5t^{2} * 6t] + [3t^{3} * (10 t) * 6t] + [3t^{3} * 5t^{2} * 6]

= [45t^{4} * 6t] + [30t^{4} * 6t] + [15t^{5} * 6]

= [270 t^{5}] + [180 t^{5}] + [90 t^{5}]

= 450 t^{5} + 90 t^{5}

= 540 t^{5}

## Summary

In the above post, we have learned all the basics of differential calculus along with types and rules. Now after reading this article, you can grab all the basics of differential calculus. You can easily solve any problem of differential calculus just by learning its rules.

Last Updated on June 29, 2022 by Admin

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