Practical Uses of Calculus

Introduction to practical uses of calculus:-

In this article we are going to discuss about the solve calculus problems step by step concept. The process of abbreviation the calculus comparable to over screening the important concept and problem in calculus are referred as review calculus. This article helps to improve the practical knowledge for calculus and below the problems are helping toll for the practical. Overtake the practical get review to this article. Calculus practical use solutions also show below. I like to share this Calculus Chain Rule with you all through my article.

Level One Example Practical Uses of Calculus Problems:-
Practical uses of calculus problem 1:-

Integrate the following equation

`int (2x^4)+(10x^5)+(28x^3) dx`

Solution:-

Step 1:-

`int (2x^4)+(10x^5)+(28x^3) dx`

Step 2:-

`= int(2x^4)dx + int(10x^5)dx + int (28 x^3) dx`

Integrating the above equation

We get

Step 3:-

`= (2x^5)/(5) + (10x^6)/(6) + (28x^4)/(4)`

Step 4:-

`= (2x^5)/ (5) + (10x^6)/ (6) + (28x^4)/ (4)`

Step 5:-

`= (2x^5)/ (5) + (10x^6) + (7x^4)`

Practical uses of calculus problem 2:-

Solve by differentiating the following equation and get the first derivative second derivative and third derivative

`y = (3x^3)+(2x^2)+(2x^5)+(5x)`

Solution:-

Given equation is` y = (2x^3) +(2x^2) + (2x^5) + (5x)`

Step 1:-

To get the 1st derivative differentiate the above given equation

`(dy)/(dx) = (6x^2)+(4x)+(10x)+(5)`

Step 2:-

To get the 2nd derivative differentiate the ist derivative

`(d^2y)/(dx^2)= 12x+14`

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Second One Example Practical Uses of Calculus Problems:-

Practical uses of calculus problem 1:-

Integrat the following equation

`int 2x^4+8x^5+16x^3 dx`

Solution:-

`int (2x^4)+(8x^5)+(16x^3) dx`

Step 1:-

`= int (2x^4) dx + int(8x^5) dx + int (16 x^3) dx`

Integrating the above equation

We get

Step 2:-

`= (2x^5)/(5) + 8 (x^6)/(6) + (16x^4)/(4)`

Step 3:-

`= (2x^5)/(5) + (8x^6)/(6)+ (4x^4)`

Practical uses of calculus problem 2:-

Differentiate the following equation and get the 1st, 2nd and 3rd derived

`y = 3x^3+2x^2+x^1 + 10`

Solution:-

Step 1:-

By differentiate the given equation with respect to x to get the 1st derivative

`y= dy/dx =9x^2+4x^1+1.`

Step 2:-

To get the 2nd derived differentiate the 1st derivative of the given equation.

`y =(d^2x)/(dy^2)= 18x + 4`

Step 3:-

To get the 3rd derived differentiate the 2nd derivative of the given equation.

`y=(d^3x)/(dy^3) = 18.`