Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q1: Second derivative of $x^2+3x+2$

Exercise 5.7Second order derivative · Section 5.7

The question

Find the second order derivatives of the functions given in Exercises 1 to 10.

1. $x^2 + 3x + 2$

Exam answer

Let $y = x^2 + 3x + 2$ ⁠. Then

\frac{dy}{dx} = 2x + 3

$\therefore\ \dfrac{d^2y}{dx^2} = 2$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The second order derivative is the derivative of the derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).

So we differentiate twice, one term at a time. Each differentiation lowers every power of $x$ by one, so the quadratic becomes a linear expression and then a constant.

Given, to find, and the plan

Given: $y = x^2 + 3x + 2$ ⁠, a polynomial, defined and differentiable for every real $x$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: (1) State the domain. (2) Find $\frac{dy}{dx}$ term by term with the sum rule, the power rule and the constant rule. (3) Differentiate $\frac{dy}{dx}$ again with the same rules.

Basics you need

1Sum rule and constant multiples

$(u \pm v)' = u' \pm v'$ (Section 5.3, rule (1)), and a constant factor stays in front: $(cu)' = c\,u'$ ⁠. So a polynomial is differentiated one term at a time. Basics: derivative rules

2Power rule and constants

$\frac{d}{dx}x^n = nx^{n-1}$ (Table 5.3). In particular $\frac{d}{dx}x = 1$ ⁠. The derivative of a constant is $0$ ⁠, because a constant does not change.

3Second order derivative (Section 5.7)

If $y = f(x)$ and $\frac{dy}{dx} = f'(x)$ is itself differentiable, then $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠, also written $f''(x)$ ⁠, $y''$ or $y_2$ ⁠. Lesson: second order derivative

Solution

Name the function and its domain

Let $y = x^2 + 3x + 2$ ⁠. This is a polynomial, so it is defined for every real $x$ ⁠, and so are all its derivatives. The answer will therefore hold for every $x \in \mathbf{R}$ ⁠.

A derivative only makes sense where the function is defined. Polynomials have no denominators, square roots or logarithms, so there is nothing to exclude.

Polynomials are differentiable at every real number. Basics: derivative rules
First derivative, term by term

By the sum rule, differentiate each of the three terms separately:
$\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(2) \\ &= 2x + 3\cdot 1 + 0 \\ &= 2x + 3. \end{aligned}$
The power rule with $n = 2$ gives $2x^{1} = 2x$ ⁠. For $3x$ ⁠, the constant 3 stays in front and $\frac{d}{dx}x = 1$ ⁠. The constant 2 has derivative 0: it does not change as $x$ changes, so its rate of change is zero.

Sum rule (Section 5.3, rule (1)); power rule $\frac{d}{dx}x^n = nx^{n-1}$ (Table 5.3).
Second derivative: differentiate $\frac{dy}{dx}$ again

Now apply the same rules to $\frac{dy}{dx} = 2x + 3$ ⁠:
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(2x + 3) \\ &= \frac{d}{dx}(2x) + \frac{d}{dx}(3) \\ &= 2\cdot 1 + 0 = 2. \end{aligned}$
So $\dfrac{d^2y}{dx^2} = 2$ for every real $x$ ⁠.

The second derivative is by definition the derivative of the first derivative, so we differentiate the expression found in Step 2, not the original $y$ ⁠. Sum rule again; $2x$ gives 2 (constant times $\frac{d}{dx}x = 1$ ⁠); the constant 3 gives 0.

Section 5.7: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative
What the answer means

$\frac{dy}{dx} = 2x + 3$ is the slope of the graph of $y$ ⁠. The second derivative 2 says this slope increases at the steady rate of 2 per unit of $x$ ⁠: for example the slope is $3$ at $x = 0$ ⁠, $5$ at $x = 1$ and $7$ at $x = 2$ ⁠. A positive second derivative everywhere matches the shape of this parabola, which opens upward.

For any quadratic $ax^2 + bx + c$ the same two steps give $2ax + b$ and then $2a$ ⁠: the second derivative of a quadratic is always the constant $2a$ ⁠. Here $a = 1$ ⁠.

Result

For $y = x^2 + 3x + 2$ (all real $x$ ⁠):

\frac{dy}{dx} = 2x + 3, \qquad \frac{d^2y}{dx^2} = 2.

(Textbook answer: 2.)

Check it with numbers

The second difference $\frac{y(a+h) - 2y(a) + y(a-h)}{h^2}$ approximates $y''(a)$ ⁠. At $a = 1$ ⁠, $h = 0.001$ ⁠: $y(1.001) = 6.005001$ ⁠, $y(1) = 6$ ⁠, $y(0.999) = 5.995001$ ⁠, so

\begin{aligned} &\frac{6.005001 - 12 + 5.995001}{0.000001} \\ &= \frac{0.000002}{0.000001} = 2. \end{aligned}

For a quadratic this is exact for every $a$ and $h$ ⁠: the $3x$ and $2$ parts cancel and $(a+h)^2 - 2a^2 + (a-h)^2 = 2h^2$ ⁠. Also the first difference $\frac{y(1.001) - y(0.999)}{0.002}$ equals $\frac{0.01}{0.002} = 5$ ⁠, which is $2(1) + 3$ ⁠. ✓

Common mistakes

Differentiating the original function twice "in one go" and losing a term. Write down $\frac{dy}{dx}$ in full first, then differentiate that expression.
Treating the constant 2 as having derivative 2 (or 1). The derivative of any constant is 0.
Writing $\frac{d^2y}{dx^2}$ as $\left(\frac{dy}{dx}\right)^2$ ⁠. $(2x+3)^2$ is the square of the slope, not the second derivative.

What to take away

Second derivative = derivative of the first derivative; always find $\frac{dy}{dx}$ completely first.
For a quadratic $ax^2 + bx + c$ ⁠, $y'' = 2a$ ⁠, a constant.
Second differences $\frac{y(a+h) - 2y(a) + y(a-h)}{h^2}$ give a quick numerical check of $y''(a)$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$(u \pm v)' = u' \pm v'$ ⁠; $(cu)' = cu'$	Steps 2, 3	Section 5.3, rule (1) (algebra of derivatives, Class XI)
$\frac{d}{dx}x^n = nx^{n-1}$ ⁠; derivative of a constant is 0	Steps 2, 3	Table 5.3, Section 5.3 (Basics: derivative rules)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 3	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q1: Second derivative of x2+3x+2x^2+3x+2x2+3x+2

Basics you need

Solution

Name the function and its domain

First derivative, term by term

Second derivative: differentiate dydx\frac{dy}{dx}dxdy​ again

What the answer means

Every fact used, and where it comes from

Exercise 5.7 · Q1: Second derivative of $x^2+3x+2$

Second derivative: differentiate $\frac{dy}{dx}$ again