2.3 The Second Derivative
Differentiating a function \(y = f(x)\) with respect to \(x\) produces the first derivative, \(\frac{dy}{dx}\) or \(f'(x)\), which is itself a new function of \(x\). When this first derivative is differentiated again with respect to \(x\), the result is called the second derivative of the function, denoted as \(\frac{d^2y}{dx^2}\) or \(f''(x)\). The second derivative plays an essential role in further mathematical analyses, such as finding rates of change of gradients and identifying characteristics of coordinates.
Learning Objectives
- Define the second derivative of an algebraic function.
- Determine the first derivative \(\frac{dy}{dx}\) (or \(f'(x)\)) and the second derivative \(\frac{d^2y}{dx^2}\) (or \(f''(x)\)) for algebraic functions.
- Evaluate the value of the second derivative at specific values of \(x\).
- Solve equations involving the first and second derivatives, such as finding values of \(x\) when \(f'(x) = f''(x)\) or \(\frac{dy}{dx} = 0\).
Flowchart of Successive Differentiation
See how a function transforms from its original form to its first derivative, then to its second derivative.
Short Explanation
The second derivative is obtained by differentiating a function twice consecutively.
Definition and Notation
Given \(y = f(x)\), the first derivative is \(\frac{dy}{dx}\) or \(f'(x)\). Differentiating this first derivative again with respect to \(x\) gives the second derivative: \[\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \quad \text{or} \quad f''(x) = \frac{d}{dx}[f'(x)]\]
Worked Example 11
For \(y = x^3 + \frac{4}{x^2}\), rewrite as \(y = x^3 + 4x^{-2}\). First derivative: \(\frac{dy}{dx} = 3x^2 - 8x^{-3} = 3x^2 - \frac{8}{x^3}\). Second derivative: \(\frac{d^2y}{dx^2} = 6x + 24x^{-4} = 6x + \frac{24}{x^4}\).
Worked Example 12
Given \(f(x) = x^3 + 2x^2 + 3x + 4\), find \(x\) such that \(f'(x) = f''(x)\). \(f'(x) = 3x^2 + 4x + 3\), \(f''(x) = 6x + 4\). Equating: \(3x^2 + 4x + 3 = 6x + 4\). \(3x^2 - 2x - 1 = 0\). \((3x + 1)(x - 1) = 0\). Thus, \(x = -\frac{1}{3}\) or \(x = 1\).
Critical Distinction
If \(y = 5x - 3\), then \(\frac{dy}{dx} = 5\), so \(\left(\frac{dy}{dx}\right)^2 = 25\). However, \(\frac{d^2y}{dx^2} = \frac{d}{dx}(5) = 0\). Therefore, \(\left(\frac{dy}{dx}\right)^2 \neq \frac{d^2y}{dx^2}\).
General definition of the second derivative
\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \quad \text{or} \quad f''(x) = \frac{d}{dx}[f'(x)]\)Differentiate the first derivative once more with respect to \(x\).
Try to Answer First
Answer in your mind, then press "Check Answer".
Find \(\frac{d^2y}{dx^2}\) for \(y = 4x^2 - \frac{2}{x}\).
Given \(g(x) = hx^3 - 4x^2 + 5x\). If \(g''(1) = 4\), determine the value of the constant \(h\).
For the curve \(y = x^3 + 3x^2 - 9x + 2\), find the value of \(\frac{d^2y}{dx^2}\) at the point where \(x = 1\).
Common Mistakes
- Confusing the second derivative notation \(\frac{d^2y}{dx^2}\) with the square of the first derivative \(\left(\frac{dy}{dx}\right)^2\).
- Incorrectly handling negative signs when differentiating negative powers, for example differentiating \(-8x^{-3}\) and writing \(-24x^{-4}\) instead of \(+24x^{-4}\).
- Forgetting to multiply by the power when taking the derivative a second time.
- Substituting a numerical value into the original function or the first derivative instead of the second derivative expression.
Concept Misunderstandings
Assuming that \(\frac{d^2y}{dx^2}\) means \(\left(\frac{dy}{dx}\right)^2\).
\(\frac{d^2y}{dx^2}\) is an operation where differentiation is performed twice consecutively: \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\). In contrast, \(\left(\frac{dy}{dx}\right)^2\) is simply the numerical or algebraic square of the first derivative function. For instance, if \(y = 5x - 3\), \(\left(\frac{dy}{dx}\right)^2 = 25\), whereas \(\frac{d^2y}{dx^2} = 0\).
Thinking that the second derivative of any quadratic term is zero.
The first derivative of a quadratic term like \(ax^2\) is \(2ax\), and its second derivative is the non-zero constant \(2a\). Only linear terms (\(bx\)) or constants (\(c\)) have a second derivative equal to zero.
Summary
- The first derivative of \(y = f(x)\) with respect to \(x\) is denoted as \(\frac{dy}{dx}\) or \(f'(x)\).
- Differentiating the first derivative again with respect to \(x\) gives the second derivative: \[\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \quad \text{or} \quad f''(x) = \frac{d}{dx}[f'(x)]\]
- \(\frac{d^2y}{dx^2}\) is fundamentally distinct from \(\left(\frac{dy}{dx}\right)^2\).
- To evaluate \(f''(a)\), differentiate twice consecutively to establish the expression \(f''(x)\), then substitute \(x = a\).
Short Activity
Objective: Test your understanding of second derivatives, notation and evaluation.
A. Objective Quiz
1 Given \(f(x) = 2x^3 + 3x^2 - 7x - 9\), what is the value of \(f''(x)\) when \(x = -1\)?
2 Which of the following statements is true regarding \(\left(\frac{dy}{dx}\right)^2\) and \(\frac{d^2y}{dx^2}\)?
B. Fill in the Blanks
3 The operation of differentiating \(\frac{dy}{dx}\) with respect to \(x\) produces the second derivative, written in Leibniz notation as ______.
4 If \(g(x) = hx^3 - 4x^2 + 5x\) and \(g''(1) = 4\), the value of the constant \(h\) is ______.
C. Matching / Drag and Drop
Match each given function or derivative expression with its corresponding second derivative, \(f''(x)\) or \(\frac{d^2y}{dx^2}\). On a phone, tap a card first, then tap the matching box.
