In calculus, there are various kinds of functions that are to be evaluated on the basis of the order of the derivatives. The order of derivatives plays a vital role in the calculation of complex calculus functions.

In this section, we will define derivative calculus, its orders, and how to evaluate the problems of differential calculus along with solved examples.

## What is derivative calculus?

Derivative calculus is a fascinating branch of mathematics that deals with the rates of change of functions. It can be used to find out how a function changes as its input changes, and it can also be used to find local minimums and maximums, points of inflection, and extreme points.

The process of calculating the differential of a function is known as differentiation. Differentiation can be defined as the process of transforming one function into another so that it can be more easily analyzed or manipulated.

There are several different types of derivatives that can be calculated. These include linear derivatives (such as dy/dx), quadratic derivatives (such as y2-x2), cubic derivatives (such as x3-y3), quartic derivatives (such as y4-x4), quantic derivatives (such as x5-y5), and sextic derivatives (such as x6-y6).

It is important to note that not all functions have every type of derivative possible; for example, linear functions only have linear derivatives while nonlinear functions may have both linear and quadratic derivatives.

## Orders of derivatives in calculus

There are three main orders of derivatives:

• First order

• Second order

• Third order

Each has its own specific uses and properties.

### 1. First order derivatives

First-order derivatives are often used when looking for points of inflection or extreme points. It can be further divided into two categories:

• Directional derivatives

• Scalar (linear) derivative

**Directional derivatives**refer to the changes in direction of a function;

**Scalar (linear) derivative**refers to the magnitude only of a function's change.

For example, if you have a function that increases by 2 units along one axis but decreases by 1 unit on another axis, then the directional derivative would be (+2/-1), while the scalar derivative would simply be 1.

Another common type of first-order derivative is

**D'Alembert's equation**, which is used to calculate tangents at points on curves### 2. Second order derivatives

Second order derivatives are often used when trying to find local minimums or maximums. It can be subdivided into two categories as well:

• Functional (vector) derivatives

• Additive (time-dependent) derivative

**Functional (vector) derivatives**refer to the changes in direction and magnitude at different locations within a domain;

**Additive (time-dependent) derivatives**refer to how their sum affects the final result over time.

Second-order partial differentiation also exists – this allows us to differentiate between several functions with respect to certain variables while leaving others constant, resulting in potentially simplified formulas compared to multiple individual implementations.

Another common type of second-order derivative is Laplace's equation, which calculates potential energy at points on surfaces under various circumstances.

### 3. Third order derivatives

Third-order derivatives are more general and can be used in a variety of situations.

Third-order differential equations exist too – these are equations that involve higher orders than second order and cannot always be solved using simpler methods like those mentioned above.

Some examples include Euler's equation governing fluid flow or Neumann’s inequality determining whether two objects collide or not.

## How to calculate the differential calculus problems?

The problems of differential calculus can be evaluated easily with the help of its rules. The process of calculating the derivative of a function is simple, just a little effort is required. You can follow the steps in the following examples to learn how to evaluate the problems of derivatives.

**Example 1**

Calculate the derivative of f(t) = 2t4 + t6 – 12t4 + 11t + cos(t) with respect to “t”.

**Solution**

**Step 1**: First of all, take the given algebraic function and write the differential notation according to the independent variable and apply it to the given expression.

f(t) = 2t4 + t6 – 12t4 + 11t + cos(t)

d/dt [f(t)] = d/dt [2t4 + t6 – 12t4 + 11t + cos(t)]

**Step 2:**Now apply the notation of differential calculus separately with each term of the function with the help of the sum and difference rule of differentiation. And take out the constant coefficients.

d/dt [2t4 + t6 – 12t4 + 11t + cos(t)] = d/dt [2t4] + d/dt [t6] – d/dt [12t4] + d/dt [11t] + d/dt [cos(t)]

d/dt [2t4 + t6 – 12t4 + 11t + cos(t)] = 2d/dt [t4] + d/dt [t6] – 12d/dt [t4] + 11d/dt [t] + d/dt [cos(t)]

**Step 3**: Now evaluate the differential of the above expression with the help of the power and trigonometry rules of differentiation.

d/dt [2t4 + t6 – 12t4 + 11t + cos(t)] = 2 [4 t4-1] + [6 t6-1] – 12 [4 t4-1] + 11 [t1-1] + [-sin(t)]

= 2 [4 t3] + [6 t5] – 12 [4 t3] + 11 [t0] + [-sin(t)]

= 2 [4 t3] + [6 t5] – 12 [4 t3] + 11 [1] + [-sin(t)]

= 8 t3 + 6 t5 – 48 t3 + 11 + [-sin(t)]

= 8 t3 + 6 t5 – 48 t3 + 11 – sin(t)

= 6t5 – 40t3 + 11 – sin(t)

You can also try a differentiation calculator to get rid of these lengthy and time-consuming calculations.

**Example 2**

Find the derivative of f(w) = 2sin(w) + w cos(w) – 2wz + 5w3 – tan(w) with respect to “w”.

**Solution**

**Step 1**: First of all, take the given algebraic function and write the differential notation according to the independent variable and apply it to the given expression.

f(w) = 2sin(w) + w cos(w) – 2wz + 5w3 – tan(w)

d/dw [f(w)] = d/dw [2sin(w) + w cos(w) – 2wz + 5w3 – tan(w)]

**Step 2**: Now apply the notation of differential calculus separately with each term of the function with the help of the sum and difference rule of differentiation. And take out the constant coefficients.

d/dw [2sin(w) + w cos(w) – 2wz + 5w3 – tan(w)] = d/dw [2sin(w)] + d/dw [w cos(w)] – d/dw [2wz] + d/dw [5w3] – d/dw [tan(w)]

d/dw [2sin(w) + w cos(w) – 2wz + 5w3 – tan(w)] = 2d/dw [sin(w)] + d/dw [w cos(w)] – 2zd/dw [w] + 5d/dw [w3] – d/dw [tan(w)]

**Step 3**: Now evaluate the differential of the above expression with the help of the power and trigonometry rules of differentiation.

d/dw [2sin(w) + w cos(w) – 2wz + 5w3 – tan(w)] = 2 [cos(w)] + [w d/dw [cos(w)] + cos(w) d/dw [w]] – 2z [w1-1] + 5 [3w3-1] – [sec2(w)]

= 2 [cos(w)] + [w [-sin(w)] + cos(w) [w1-1]] – 2z [w0] + 5 [3w2] – [sec2(w)]

= 2 [cos(w)] + [w [-sin(w)] + cos(w) [w0]] – 2z [1] + 5 [3w2] – [sec2(w)]

= 2 [cos(w)] + [w [-sin(w)] + cos(w) [1]] – 2z [1] + 5 [3w2] – [sec2(w)]

= 2cos(w) + [-w sin(w) + cos(w)] – 2z + 15w2 – sec2(w)

= 2cos(w) – w sin(w) + cos(w) – 2z + 15w2 – sec2(w)

= 3cos(w) – w sin(w) – 2z + 15w2 – sec2(w)

## Sum Up

Now you can grab all the basics of the orders of derivatives and find the differential of complex calculus functions from this section. As we have discussed the definition of derivatives, the order of derivatives, and solved examples of derivatives in this post.