# Blog

### Derivative in calculus: An introduction with examples

A derivative is one of the main branches of calculus used to find the differential of the functions. Calculus is a main type of mathematics that deals with properties and finding the complex problems of differential and integration.

Integration is the other kind of calculus that is used to find the area under the curve with or without using the boundary values. The limit of calculus plays a very essential role to find both types of calculus.

Let’s discuss the definition, formula, & rules of differentiation with solved examples.

## What does derivative mean in calculus?

In calculus, a derivative is the instantaneous rate of change of a function with respect to its variable or finding the slope of the tangent line. Derivatives reverse the process of integration, that is why it is also known as differentiation.

The functions can be the linear or polynomial algebraic expression, constant, logarithmic, or exponential for finding the derivative. All the simple, hyperbolic, & inverse trigonometric functions can be differentiated by using the chain rule of derivative.

Derivative in calculus used different notations and techniques for different kinds of functions, for example, to solve one variable function, you have to find d/dx or f’(x) with respect to a variable. This process is said to be the explicit differentiation.

To solve the equation or implicit functions, you have to calculate dy/dx or y’(x) without taking “y” as a constant, and this process is known as implicit differentiation. If the functions have several variables in the function, you have to solve it partially (∂/∂x, ∂/∂y, or ∂/∂z) & known as a partial derivative.

### Rules of differentiation

There are several rules of derivatives used to solve its problems.

 Rules name Rules Quotient rule d/dx [a(x) / b(x)] = 1/[b(x)]2 [b(x) d/dx [a(x)] - a(x) d/dx [b(x)]] Product rule d/dx [a(x) * b(x)] = b(x) d/dx [a(x)] + a(x) d/dx [b(x)] Constant rule d/dx [C] = 0 Difference rule d/dx [a(x) - b(x)] = d/dx [a(x)] - d/dx [b(x)] Sum rule d/dx [a(x) + b(x)] = d/dx [a(x)] + d/dx [b(x)] Exponential rule d/dx [ex] = ex Constant function rule d/dx [Ca(x)] = C d/dx [a(x)] Power rule d/dx [a(x)]n = n [a(x)]n-1 * d/dx [a(x)]

## How to solve the problems of derivatives?

All the problems of derivatives in calculus can be solved easily by using its rules. Let us take some examples to learn how to calculate derivatives.

Example 1

Evaluate 15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y with respect to “x”.

Solution

Step 1: Write the given function equal to a(x) and apply the notation of derivative.

a(x) = 15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y

d/dx [a(x)] = d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y]

Step 2: Apply d/dx to each function separately by using the sum & difference rules of differential and write the constant coefficients outside the differential notation.

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = d/dx [15x3] + d/dx [6u2] – d/dx [9x3] + d/dx [14sin(x)] + d/dx [2x] + d/dx [19y]

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 15d/dx [x3] + 6d/dx [u2] – 9d/dx [x3] + 14d/dx [sin(x)] + 2d/dx [x] + 19d/dx [y]

Step 4: Now differentiate the above expression by using the constant, trigonometric, & power rules of differentiation.

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 15 [3x3-1] + 6 [0] – 9 [3x3-1] + 14 [cos(x)] + 2 [1] + 19 [0]

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 15 [3x2] + 6 [0] – 9 [3x2] + 14 [cos(x)] + 2 [1] + 19 [0]

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 15 [3x2] + 0 – 9 [3x2] + 14 [cos(x)] + 2 [1] + 0

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 45x2 + 0 – 27x2 + 14cos(x) + 2 + 0

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = (45 – 27) x2 + 14cos(x) + 2

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 18x2 + 14cos(x) + 2

d/dx [15x3 + 6u2 – 9x3 + 14sin(x) + 2x + 19y] = 2(9x2 + 7cos(x) + 1)

This example of differentiation can also be solved by using a derivative calculator to get the result in a fraction of seconds. To learn how to calculate derivative using calculator, follow the below steps.

Step 1: Enter the function.

Step 2: Select the variable & write the order of differentiation.

Step 3: Press the calculate button. The result will show in a fraction of a second.

Example 2

Evaluate 5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19 with respect to “x”.

Solution

Step 1: Write the given function equal to a(x) and apply the notation of derivative.

a(x) = 5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19

d/dx [a(x)] = d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19]

Step 2: Apply d/dx to each function separately by using the sum & difference rules of differential and write the constant coefficients outside the differential notation.

d/dx [a(x)] = d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = d/dx [5xsin(x)] + d/dx [6x2] – d/dx [27z3] + d/dx [15x2] + d/dx [9x] + d/dx [19]

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5d/dx [xsin(x)] + 6d/dx [x2] – 27d/dx [z3] + 15d/dx [x2] + 9d/dx [x] + d/dx [19]

Step 4: Now differentiate the above expression by using the constant, trigonometric, product, & power rules of differentiation.

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5[x d/dx sin(x) + sin(x) d/dx (x)] + 6d/dx [x2] – 27d/dx [z3] + 15d/dx [x2] + 9d/dx [x] + d/dx [19]

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5[x (cos(x)) + sin(x) (1)] + 6 [2x2-1] – 27 [0] + 15 [2x2-1] + 9 [1] + [0]

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5[x (cos(x)) + sin(x) (1)] + 6 [2x] – 0 + 15 [2x] + 9 [1] + [0]

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5[xcos(x) + sin(x)] + 6 [2x] + 15 [2x] + 9 [1]

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5xcos(x) + 5sin(x) + 12x + 30x + 9

d/dx [5xsin(x) + 6x2 – 27z3 + 15x2 + 9x + 19] = 5xcos(x) + 5sin(x) + 42x + 9

## Summary

Now you are witnessed that the derivative is not a difficult topic. You can solve any problem of derivatives in calculus after reading the above articles. We’ve discussed almost all the basics of derivatives along with solved examples.

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Last updated: Sunday, April 19th, 2020 - 12:42PM