Second Order Derivative: Definition, Representation and Sample Questions

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Jasmine Grover

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The derivative of a function of a real variable tells us the rate of change of the function with respect to its argument. It is represented as dy/dx. The term dy/dx tells us the rate of change in y with respect to x. The simplest definition of the Second-Order Derivative is the derivative of the first-order derivative of the given function.

Read More: Continuity and Differentiability

Key Takeaways: Derivative of a function, rate of change, velocity, first-order derivative, second-order derivative, derivatives, function, tangent line


Derivative Introduction

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The derivative of a function of a real variable tells us the rate of change of the function with respect to its argument. It is represented as dy/dx. The term dy/dx tells us the rate of change of y with respect to x.

Read More: Trigonometry Table

The derivative can also be defined as the slope of the tangent line to the graph of the function. The derivative can be of the first order, second order, third order, and so on. To deduce the higher derivative, the previous derivatives need to be deduced first. 

An example of the derivative is velocity. Velocity is the rate of change of an object’s position. It is a first-order derivative that can be calculated by the division of displacement and time. The second-order derivative in this situation will give Acceleration. 

Discover about the Chapter video:

Continuity and Differentiability Detailed Video Explanation:

Read More: Cosine Formula


Second-Order Derivative

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The simplest definition of the Second-Order Derivative is the derivative of the first-order derivative of the given function. For example, acceleration is the second-order derivative of the distance covered with respect to time and tells us the rate of change of velocity. 

Just like the first-order derivative tells us about the slope of the tangent line to the graph of the given function, the second-order derivative tells us about the shape of the graph and its concavity. 

The second-order derivative is represented using f’’(x) or d2y/dx2

The video below explains this:

Second-Order Derivative Detailed Video Explanation:

Read More: Differentiation and Integration Formula


Representation of Second-Order Derivative 

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Let there be a function: 

y = f(x) 

The first-order derivative of the function can be written as: 

dy/dx = f’(x)

As mentioned above, Second-Order Derivative is the derivative of the first-order derivative of the given function. Therefore, 

d/dx (dy/dx) = d2y/dx= f’’(x)

gives us the Second-Order Derivative. 

Similarly, the higher derivatives can be represented as d3y/dx3, d4y/dx4, and so on. 

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Graphical Representation of Second-Order Derivative 

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Let there be a function y = x4 + x3 + 1. The first-order derivative of this function will come out to be: 

f’(x) = 4x3 + 3x2 - (1)

The graphical representation of the equation (1) is:

The red line represents the slope of the given function. 

To find the Second-Order Derivative, we will differentiate the equation (1) with respect to x. We will get: 

f’’(x) = 12x2 + 6x - (2)

The graphical representation of equation (2) is: 

The second-order derivative tells us about the concavity of the graph. The upward and downward concavity of the graph can be determined by the positive and negative values of the second-order derivative respectively. 

Read More: Integers As Exponents


Concavity and Point of Inflection

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Concavity refers to whether the graph will be open upwards or downwards. 

  • Concave Up: If the value of the second-order derivative comes out to be positive, it is said to be Concave Up. This also means that the tangent line will lie below the graph of the function. 
  • Concave Down: If the value of the second-order derivative comes out to be negative, it is said to be Concave Down. This also means that the tangent line will lie above the graph of the function. 
  • Point of Inflection: Point of Inflection refers to the point where the second derivative of a function changes its sign and so does its concavity, it switches from concave up to concave down and vice-versa. The value of the second derivative at the point of inflection comes out to be zero.

Read More: Differential Equation


Second-Order Derivative Test

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Second-Order Derivative Test is used to confirm the presence of local maxima and minima. The test is as follows: 

  • If f’’(x) < 0, the function f(x) has a local maxima at x. 
  • If f’’(x) > 0, the function f(x) has a local minima at x.
  • If f’’(x) = 0, then x is a point of inflection. In that case, the second-order derivative test fails and we need to move to the higher derivatives.

Read More: First Order Differential Equation


Things to Remember 

  • The derivative of a function of a real variable tells us the rate of change of the function with respect to its argument.
  • The derivative is represented as dy/dx. The term dy/dx tells us the rate of change in y with respect to x.
  • The derivative can also be defined as the slope of the tangent line to the graph of the function.
  • Second-Order Derivative is the derivative of the first-order derivative of the given function. It tells us about the shape of the graph and its concavity.
  • The second-order derivative is represented using f’’(x) or d2y/dx2
  • If the value of the second-order derivative comes out to be positive, it is said to be Concave Up.
  • If the value of the second-order derivative comes out to be negative, it is said to be Concave Down.
  • Point of Inflection refers to the point where the second derivative of a function changes its sign and so does its concavity, it switches from concave up to concave down and vice-versa.
  • Second-Order Derivative Test is used to confirm the presence of local maxima and minima.

Read More: Inverse Trigonometric Formulas


Sample Questions 

Ques: What is Second-Order Derivative? (1 mark)

Ans: Second-Order Derivative is the derivative of the first-order derivative of the given function. For example, acceleration is the second-order derivative of the distance covered with respect to time and tells us the rate of change of velocity. It tells us about the shape of the graph and its concavity. The second-order derivative is represented using f’’(x) or d²y/dx

Ques: Describe the Second-Order Derivative Test. (1 mark)

Ans: Second-Order Derivative Test is used to confirm the presence of local maxima and minima. The test is as follows: 

  • If f’’(x) < 0, the function f(x) has a local maxima at x. 
  • If f’’(x) > 0, the function f(x) has a local minima at x.
  • If f’’(x) = 0, then x is a point of inflection. In that case, the second-order derivative test fails and we need to move to the higher derivatives. 

Ques: Find d2y/dx2 of y = 4x+ 3x + 1. (3 mark)

Ans: y = 4x2 + 3x + 1

Differentiating with respect to x, we get 

dy/dx = 4 d/dx (x2) + 3 d/dx (x) + d/dx (1)

dy/dx = 4 * 2x + 3 * 1 + 0

dy/dx = 8x + 3

Differentiating with respect to x once again, we get

d2y/dx= 8 d/dx (x) + d/dx (3)

d2y/dx2 = 8 * 1 + 0

d2y/dx2 = 8

Therefore, the second-order derivative of y = 4x2 + 3x + 1 comes out to be 8. 

Ques: Find d2y/dx2 of y = 4sinx + 3cosx. (3 mark)

Ans: y = 4sinx + 3cosx

Differentiating with respect to x, we get 

dy/dx = 4 d/dx (sinx) + 3 d/dx(cosx)

dy/dx = 4 * cosx + 3 * (-sinx)

dy/dx = 4cosx - 3sinx

Differentiating with respect to x once again, we get

d2y/dx2 = 4 d/dx (cosx) - 3 d/dx (sinx)

d2y/dx2 = 4 * (-sinx) - 3 * cosx

d2y/dx2 = -4sinx - 3cosx

d2y/dx2 = -(4sinx + 3cosx)

d2y/dx2 = -y

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CBSE CLASS XII Related Questions

  • 1.

    Find:
    Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

      • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
      • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
      • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
      • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)

    • 2.
      Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).


        • 3.
          Find:

          If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

            • \(0\)
            • \(-2\)
            • \(-1\)
            • \(2\)

          • 4.
            Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


              • 5.
                Find:

                If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

                  • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
                  • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
                  • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
                  • \(p = 0, \, q = 0\)

                • 6.
                  Find:

                  The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

                    CBSE CLASS XII Previous Year Papers

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