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# Difference Between Derivative and Antiderivative The derivative and antiderivative are the two main types of calculus used to calculate the slope of the tangent line and area under the curve respectively. Both subtypes of calculus can be defined and evaluated with the help of limit calculus.

The main difference between a derivative and an antiderivative is that the antiderivative reverses the process that a derivative does. In this post, we will learn all the basics of derivatives and antiderivatives along with their definitions, formulas, and examples.

## What is the Derivative in Calculus?

In calculus, the term that is used to evaluate the slope of the tangent line is known as the derivative. It is also known as differential and is defined as the instantaneous rate of change of the functions corresponding to their independent variables.

The process of finding the differential of a function (differentiating the function with respect to an independent variable) is known as differentiation. The derivative of the functions is to be evaluated with the help of the first principle method in which limits are involved. It is denoted by d/dr where r is the independent variable.

### Formula of derivative by the first principle

The first principle method is widely used to calculate the differential of the function. Here is the general expression of this method.

d/dr [f(r)] = limh→0 [f(r + h) – f(r)] / h

### Types of derivative

There are different types of derivatives in calculus for solving various kinds of complex problems. These subtypes of the derivative are:

1. Explicit derivative
2. Implicit derivative
3. Partial derivative
4. Directional derivative

These types of derivatives are frequently used to differentiate the single, double, and multivariable functions. These types will cover all the basics of differentiation by solving the linear, algebraic, trigonometric, geometric, logarithmic, exponential, and many other functions and expressions.

### Rules of derivative in calculus

Rules of differentiation are helpful in differentiating the functions easily. Here are some well-known rules of differentiation that are very helpful in calculating the derivative of the functions.

### How to calculate the problems of the derivative?

The problems of derivatives in calculus can be solved easily either by using its rules or a derivative calculator. Here is an example of a differential to solve manually.

Example

Evaluate the derivative of the given function with respect to “r”.

p(r) = 2r2 + 4r3 – 2r / 5r2 + 20r

Solution

Step 1: First of all, take the given function p(r) and apply the notation d/dr to it.

p(r) = 2r2 + 4r3 – 2r / 5r2 + 20r

d/dr p(r) = d/dr [2r2 + 4r3 – 2r / 5r2 + 20r]

Step 2: Now apply the differential notation [d/dr] to each term of the above expression separately by using the sum and difference law of differentiation.

d/dr [2r2 + 4r3 – 2r / 5r2 + 20r] = d/dr [2r2] + d/dr [4r3] – d/dr [2r / 5r2] + d/dr [20r]

Step 3: Now use the quotient rule of the derivative.

d/dr [2r2 + 4r3 – 2r / 5r2 + 20r] = d/dr [2r2] + d/dr [4r3] – [1/(2r)2 (2r d/dr [5r2] – 5r2 d/dr [2r])] + d/dr [20r]

d/dr [2r2 + 4r3 – 2r / 5r2 + 20r] = d/dr [2r2] + d/dr [4r3] – 1/4r2 (2r d/dr [5r2]) + 1/4r2 (5r2 d/dr [2r]) + d/dr [20r]

Step 4: Now take the constant coefficients outside the differential notation.

d/dr [2r2 + 4r3 – 2r / 5r2 + 20r] = 2d/dr [r2] + 4d/dr [r3] – 1/4r2 (2r d/dr [5r2]) + 1/4r2 (5r2 d/dr [2r]) + 20d/dr [r]

Step 5: Now use the power rule to differentiate the above function with respect to “r”.

= 2[2r2-1] + 4 [3r3-1] – 1/4r2 (2r [10r2-1]) + 1/4r2 (5r2 [2r1-1]) + 20 [r1-1]

= 2[2r1] + 4 [3r2] – 1/4r2 (2r [10r1]) + 1/4r2 (5r2 [2r0]) + 20 [r0]

= 2[2r] + 4 [3r2] – 1/4r2 (2r [10r]) + 1/4r2 (5r2 [2(1)]) + 20 

= 2[2r] + 4 [3r2] – 1/4r2 (20r2) + 1/4r2 (10r2) + 20

= 4r + 12r2 – 20r2/4r2 + 10r2/4r2 + 20

= 4r + 12r2 – 5 + 5/2 + 20

= 4r + 12r2 + 5/2 + 15

= 4r + 12r2 + 35/2

## What is the Antiderivative?

In calculus, the reverse process of the differential is known as the antiderivative. It is widely used to calculate the area under the curve. It calculated the new function whose original function is derivative and calculate the numerical value of the function.

Limit calculus is also used in this subtype of calculus which is helpful in calculating the numerical value of the function by applying the upper and lower limit value. The fundamental theorem of calculus is a help full technique to apply the upper and lower limit to the integrated function.

### Types of antiderivative (integral)

There are two further sub-types of antiderivatives,

1. Definite integral
2. Indefinite integral

The definite and indefinite integrals are widely used to calculate the numerical value of the function and the new function whose original function is derivative respectively. The limit values are used in the definite integral.

### Formulas of antiderivative

The formula for finding the numerical value of the function is:

p(r) dr = P[r]ts = P(t) – P(s) = N

The formula for calculating the new function is:

ʃ p(r) dr = P(r) + C

An antiderivative calculator can be used to solve the integral problems according to the above formulas.

### Rules of antiderivative

Here are some basic rules of antiderivative used to integrate the functions.

### How to calculate the problems of antiderivative?

Here is an example to understand how to calculate the problem of the antiderivative.

Example

Evaluate the antiderivative of the function with respect to “r”.

p(r) = 2r3 + 3r5 – 14r6 + 2cos(r) + 12r – 120

Solution

Step 1: First of all, take the given function p(r) and apply the notation “ʃ” to it.

p(r) = 2r3 + 3r5 – 14r6 + 2cos(r) + 12r – 120

ʃ p(r) dr = ʃ [2r3 + 3r5 – 14r6 + 2cos(r) + 12r – 120] dr

Step 2: Now apply the integral notation (ʃ) to each term of the above expression separately by using the sum and difference laws of integration.

ʃ [2r3 + 3r5 – 14r6 + 2cos(r) + 12r – 120] dr = ʃ [2r3] dr+ ʃ [3r5] dr – ʃ [14r6] dr + ʃ [2cos(r)] dr + ʃ [12r] dr – ʃ  dr

Step 3: Now take out the constant coefficients outside the integral notation.

ʃ [2r3 + 3r5 – 14r6 + 2cos(r) + 12r – 120] dr = 2ʃ [r3] dr+ 3ʃ [r5] dr – 14ʃ [r6] dr + 2ʃ [cos(r)] dr + 12ʃ [r] dr – ʃ  dr

Step 4: Now integrate the above expression with the help of power and trigonometry laws of integration.

= 2 [r3+1 / 3 + 1]+ 3 [r5+1 / 5 + 1] – 14 [r6+1 / 6 + 1] + 2 [sin(r)] + 12 [r1+1 / 1 + 1] – [120r] + C

= 2 [r4 / 4]+ 3 [r6 / 6] – 14 [r7 / 7] + 2 [sin(r)] + 12 [r2 / 2] – [120r] + C

= 2/4 [r4]+ 3/6 [r6] – 14/7 [r7] + 2 [sin(r)] + 12/2 [r2] – [120r] + C

= 1/2 [r4]+ 1/2 [r6] – 2 [r7] + 2 [sin(r)] + 6 [r2] – [120r] + C

= r4/2+ r6/2 – 2r7 + 2sin(r) + 6r2 – 120r + C

## Final words

Now you are witnessed that the derivative and antiderivative are not difficult topics. In this post, we have covered all the basics of the derivative and antiderivative. Now you can solve any problem related to these types of calculus easily.

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