# 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.

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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 1st, 2nd, & 3rd 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 1st 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 1st order derivative can be evaluated for y= f(x) as:

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

## Types of Derivative Calculus

Here are the types of derivative calculus:

## What are second and third derivatives?

In calculus, the 2nd and 3rd 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.

### 2nd order derivative

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

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

### 3rd order derivative

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

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

d/du [d2/du2 f(u)] = d3/du3 [f(u)]

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

Here we’ll take a function and find its 1st, 2nd, and 3rd-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 u4 – 1] – 6 [3 u3 – 1] + 15 [2 u2 – 1] – 6u3 [2 u2 – 1] – 6u2 [3 u3 – 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 2nd 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 1st 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 u3 – 1] – 18 [2 u2 – 1] + 30 [u1 – 1] – 30 [4 u4 – 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  – 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 3rd 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 2nd 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 2nd 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  – d/du [120u3]

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

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

d/du [60u2 – 36u + 30 – 120u3] = 60 [2 u2 – 1] – 36 [u1 – 1] +  – 120 [3 u3 – 1]

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

d/du [60u2 – 36u + 30 – 120u3] = 60 [2 u] – 36  – 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, 2nd & 3rd order derivatives, and how to evaluate them.

Sharing: Welcome to Mod Education! I'm RK Yadav, an educator passionate about teaching Mathematics and Science. With a teaching journey spanning back to 2012, I've had the privilege of guiding students through the intricate realms of these subjects.