Understanding Fundamentals of First, Second, and Third Derivatives | How to Evaluate Them?

Derivatives: A derivative is one of the main kinds of calculus that plays a crucial role in determining the rate of changes in functions. Calculus is a fundamental kind of mathematics that allows us to study the concept of change and motion.

The derivative is essential for understanding the behaviour of the function and plays a vital role in engineering and mathematical applications. In this article, we will explain the 1^st, 2^nd, & 3^rd order derivatives along with examples.

What is derivative in Calculus?

In calculus, the term derivative is frequently used to find the instantaneous rate of change of functions concerning a variable that is not dependent on the function. The 1^st order derivative is denoted by d/du, f’(u), & D[u].

According to the geometrical definition of the derivative, it is used to represent the slope of the tangent line to the curve of the function at the given point. Mathematically, the 1^st order derivative can be evaluated for y= f(x) as:

dy/du = f’(u) = lim_h→0 [f(u + h) – f(u) / h]

Types of Derivative Calculus

Here are the types of derivative calculus:

Type	Explanation	Example
Explicit differentiation	Simple differentiation of the function is said to be the explicit differentiation. It is used to differentiate the given function with respect to a single variable.In simple words, when a function is given with a single variable that is defined directly in terms of another variable finding the change in the function as the variable changes is said to be the explicit differentiation.	For Example Differentiate y = 3u3 + 4u2 – 5 Solution Step 1: Apply the notation dy/du = d/du [3u3 + 4u2 – 5] Step 2: Apply the notation separately. d/du [3u3 + 4u2 – 5] = d/du [3u3] + d/du [4u2] – d/du [5] Step 3: Take out constant terms and apply power & constant laws. d/du [3u3 + 4u2 – 5] = 3d/du [u3] + 4d/du [u2] – d/du [5] = 3 [3 u3-1] + 4 [2 u2-1] – [0] = 3 [3 u2] + 4 [2 u1] – [0] = 9u2 + 8u
Implicit differentiation	It is the other type of differential calculus that is used to find the derivative of two variable functions. The equation of this type of derivative is not explicit. The way of finding the dependent variable “y” with respect to the independent variable “u” is said to be the implicit differentiation.	For Example Differentiate uy + 2y2 = 12 Solution Step 1: Apply the notation d/du [uy + 2y2] = d/du [12] Step 2: Apply the notation separately. d/du [uy] + d/du [2y2] = d/du [12] Step 3: Take out constant terms and apply the differential. d/du [uy] + 2d/du [y2] = d/du [12]y d/du [u] + u d/du [y] + 2 [2 y2 – 1] = d/du [12]y [u1 – 1] + u d/du [y] + 2 [2 y1] = [0]y + u dy/du + 4y = [0]u dy/du + 5y = 0u dy/du = -5ydy/du = -5y/u
Partial differentiation	It is used to deal with the multivariable calculus functions for finding derivatives of those functions with respect to a particular variable while keeping other variables constant. The function f(u, v, z) can be differentiated with respect to u, v & w but one variable at a time if you want to differentiate the function with respect to “v” the ∂/∂v [f(u, v, z)] will be differentiated and u & z will be considered as constants.	For Example Differentiate 4uv2z + 5v3z – 3 u2v with respect to “v”. Solution Step 1: Apply the notation ∂/∂v [f(u, v, z)] = ∂/∂v [4uv2z + 5v3z – 3 u2v] Step 2: Apply the notation separately. ∂/∂v [4uv2z + 5v3z – 3 u2v] = ∂/∂v [4uv2z] + ∂/∂v [5v3z] – ∂/∂v [3 u2v] Step 3: Take out constant terms and apply the differential. ∂/∂v [4uv2z + 5v3z – 3 u2v] = 4uz ∂/∂v [v2] + 5u2 ∂/∂v [v3] – 3u2 ∂/∂v [v] = 4uz [2 v2 – 1] + 5u2 [3 v3 – 1] – 3u2 [v1 – 1] = 4uz [2 v1] + 5u2 [3 v2] – 3u2 [v0] = 4uz [2 v] + 5u2 [3 v2] – 3u2 [1] = 8uvz + 15u2v2 – 3u2

What are second and third derivatives?

In calculus, the 2^nd and 3^rd order derivatives are considered as higher order differentials that give a piece of detailed information about the given function and how it changes over the domain. 

2^nd order derivative

In calculus, the 2^nd derivative of the given function is the differential of the 1^st derivative. In simple words, the 2^nd-order derivative gives the awareness of the curvature of the graph of a function. 

Mathematically, the 2^nd order differential is represented by f’’(u) & d²(y)/du². The derivative of the first derivative means:

d/du [d/du f(u)] = d²/du² [f(u)]

The second derivative is essential for determining whether the given function is concave or convex:

• If the result of the 2^nd order differential is positive, then the function is convex and the graph of the function makes the upward curves.
• If the result of the 2^nd order differential is negative, then the function is concave and the graph of the function makes the downward curves.

The convex and concave are also known as concave up and concave down respectively. 

3^rd order derivative

In calculus, the 3^rd derivative of the given function is the differential of the 2^nd derivative. In simple words, the 3^rd order derivative gives the awareness of the curvature of the graph of a function. 

Mathematically, the 3^rd order differential is represented by f’’’(u) & d³(y)/du³. The derivative of the 2^nd derivative means:

d/du [d²/du² f(u)] = d³/du³ [f(u)]

How to solve the problems of 1^st, 2^nd, and 3^rd-order derivatives?

Here we’ll take a function and find its 1^st, 2^nd, and 3^rd-order derivatives.

Example 1

Differentiate the given function concerning “u”.

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

Solution

Step 1: First of all, take the given algebraic expression and use the notation of differentiation.  

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

h’(u) = d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)]

Step 2: Now apply the notation of differential separately to each term of the function using sum & difference laws and take out the constant terms.

d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)] = d/du [5u⁴] – d/du [6u³] + d/du [15u²] – d/du [ (2u² x 3u³)]

d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)] = 5d/du [u⁴] – 6d/du [u³] + 15d/du [u²] – d/du [ (2u² x 3u³)]

Step 3: Now apply the product law to the above expression 

d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)] = 5d/du [u⁴] – 6d/du [u³] + 15d/du [u²] – (d/du [2u²] x 3u³ + 2u² d/du [3u³])

d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)] = 5d/du [u⁴] – 6d/du [u³] + 15d/du [u²] – 3u³ d/du [2u²] – 2u² d/du [3u³]

d/du [5u⁴ – 6u³ + 15u² – (2u² x 3u³)] = 5d/du [u⁴] – 6d/du [u³] + 15d/du [u²] – 6u³ d/du [u²] – 6u² d/du [u³]

Step 4: Now use the power rule to differentiate the above expression. 

= 5 [4 u^{4 – 1}] – 6 [3 u^{3 – 1}] + 15 [2 u^{2 – 1}] – 6u³ [2 u^{2 – 1}] – 6u² [3 u^{3 – 1}]

= 5 [4 u³] – 6 [3 u²] + 15 [2 u¹] – 6u³ [2 u¹] – 6u² [3 u²]

= 5 [4 u³] – 6 [3 u²] + 15 [2 u] – 6u³ [2 u] – 6u² [3 u²]

= 20u³ – 18u² + 30u – 12u⁴ – 18u⁴

= 20u³ – 18u² + 30u – 30u⁴

Hence the first derivative of the given function is:

h’(u) = 20u³ – 18u² + 30u – 30u⁴

Example 2:

Find the 2^nd derivative of the given function concerning “u”.

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

Solution

Step 1: First of all, take the given algebraic expression and then write the 1^st derivative of that function.  

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

h’(u) = 20u³ – 18u² + 30u – 30u⁴

Step 2: Now apply the differential notation to the first derivative.

h’’(u) = d/du [20u³ – 18u² + 30u – 30u⁴]

Step 3: Now apply the notation of differential separately to each term of the function using sum & difference laws and take out the constant terms.

d/du [20u³ – 18u² + 30u – 30u⁴] = d/du [20u³] – d/du [18u²] + d/du [30u] – d/du [30u⁴]

d/du [20u³ – 18u² + 30u – 30u⁴] = 20d/du [u³] – 18d/du [u²] + 30d/du [u] – 30d/du [u⁴]

Step 4: Now use the power rule to differentiate the above expression. 

d/du [20u³ – 18u² + 30u – 30u⁴] = 20 [3 u^{3 – 1}] – 18 [2 u^{2 – 1}] + 30 [u^{1 – 1}] – 30 [4 u^{4 – 1}]

d/du [20u³ – 18u² + 30u – 30u⁴] = 20 [3 u²] – 18 [2 u¹] + 30 [u⁰] – 30 [4 u³]

d/du [20u³ – 18u² + 30u – 30u⁴] = 20 [3 u²] – 18 [2 u] + 30 [1] – 30 [4 u³]

d/du [20u³ – 18u² + 30u – 30u⁴] = 60u² – 36u + 30 – 120u³

Hence the second derivative of the given function is:

h’’(u) = 60u² – 36u + 30 – 120u³

Example 3:

Find the 3^rd derivative of the given function concerning “u”.

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

Solution

Step 1: First of all, take the given algebraic expression and then write the 2^nd derivative of that function.  

h(u) = 5u⁴ – 6u³ + 15u² – (2u² x 3u³)

h’’(u) = 60u² – 36u + 30 – 120u³

Step 2: Now apply the differential notation to the 2^nd derivative.

h’’’(u) = d/du [60u² – 36u + 30 – 120u³]

Step 3: Now apply the notation of differential separately to each term of the function using sum & difference laws and take out the constant terms.

d/du [60u² – 36u + 30 – 120u³] = d/du [60u²] – d/du [36u] + d/du [30] – d/du [120u³]

d/du [60u² – 36u + 30 – 120u³] = 60d/du [u²] – 36d/du [u] + d/du [30] – 120d/du [u³]

Step 4: Now use the power rule to differentiate the above expression. 

d/du [60u² – 36u + 30 – 120u³] = 60 [2 u^{2 – 1}] – 36 [u^{1 – 1}] + [0] – 120 [3 u^{3 – 1}]

d/du [60u² – 36u + 30 – 120u³] = 60 [2 u¹] – 36 [u⁰] + [0] – 120 [3 u²]

d/du [60u² – 36u + 30 – 120u³] = 60 [2 u] – 36 [1] – 120 [3 u²]

d/du [60u² – 36u + 30 – 120u³] = 120u – 36 – 360u²

Hence the third derivative of the given function is:

h’’’(u) = 120u – 36 – 360u²

You can also use online platforms like MeraCalculator, which offers a range of calculus tools, including derivative, second-derivative, and third-derivative calculators, to simplify lengthy calculations.

Conclusion

Now you can grab all the basics of differential calculus and higher order derivatives from this post as we have discussed derivative calculus, its types with examples, 2^nd & 3^rd order derivatives, and how to evaluate them.