Chapter 5 · Continuity and Differentiability

Exercise 5.2 · Q3: Differentiate $\sin(ax+b)$

Exercise 5.2Chain rule · Section 5.3.1

The question

Differentiate the function with respect to $x$ ⁠:

\sin\left(ax + b\right)

Exam answer

Let $y = \sin(ax + b)$ ⁠. Put $t = ax + b$ ⁠, so $y = \sin t$ ⁠.

\frac{dy}{dt} = \cos t, \qquad \frac{dt}{dx} = a

By the chain rule,

\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} \\ &= \cos t\cdot a \end{aligned}

$\therefore\ \dfrac{dy}{dx} = a\cos(ax + b)$ ⁠.

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

The one idea this problem is built on

Sine is applied to the linear expression $ax+b$ ⁠. By the chain rule, the derivative is $\cos(ax+b)$ times the rate at which $ax+b$ changes, and a straight line $ax+b$ changes at the constant rate $a$ (its slope). So the answer is $a\cos(ax+b)$ ⁠: the constant $a$ stretches the rate, and the shift $b$ has no effect on it.

Given, to find, and the plan

Given: $y = \sin(ax+b)$ ⁠, where $a$ and $b$ are constants (fixed real numbers). The function is defined for every real $x$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Inner function $t = ax+b$ ⁠, outer function $y = \sin t$ ⁠. Differentiate each, multiply, substitute back.

Basics you need

1Constants versus the variable

The letters $a$ and $b$ stand for fixed numbers; only $x$ varies. So $\frac{d}{dx}(ax) = a$ (a constant multiple of $x$ ⁠) and $\frac{d}{dx}(b) = 0$ ⁠. Basics: derivative rules

2The chain rule

$\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx}$ (Theorem 4, Section 5.3.1). Lesson: the chain rule

3Slope of a straight line

The line $t = ax + b$ has slope $a$ everywhere: when $x$ increases by 1, $t$ increases by $a$ ⁠. That slope is exactly $\frac{dt}{dx}$ ⁠. Basics: slope at a point

Solution

Name the inner and the outer function
$\begin{aligned} t &= ax + b &&(\text{inner}), \\ y &= \sin t &&(\text{outer}). \end{aligned}$
Sine is applied to $ax+b$ ⁠, not to $x$ ⁠, so the plain rule $(\sin x)' = \cos x$ does not apply directly. Splitting off the inner expression lets us use it on the outer link.

A composite $y = v(u(x))$ is differentiated link by link. Basics: composite functions
Differentiate each link
$\begin{aligned} \frac{dy}{dt} &= \frac{d}{dt}(\sin t) = \cos t, \\ \frac{dt}{dx} &= \frac{d}{dx}(ax) + \frac{d}{dx}(b) \\ &= a + 0 = a. \end{aligned}$
For the inner link, $ax$ is the constant $a$ times $x$ ⁠, whose derivative is $a\cdot 1 = a$ ⁠; and $b$ is a constant, whose derivative is $0$ ⁠. Both derivatives exist for every real number.

Table 5.3: $(\sin t)' = \cos t$ and $(x)' = 1$ ⁠. A constant factor stays in front: $(c\,u)' = c\,u'$ (a special case of the product rule, since $c' = 0$ ⁠). Basics: derivative rules
Multiply (chain rule) and return to $x$
$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} \\ &= \cos t\cdot a \\ &= a\cos(ax+b). \end{aligned}$
The constant $a$ comes out in front because the inside runs $a$ times as fast as $x$ ⁠. The constant $b$ only shifts the graph sideways, which does not change any slope, so it survives only inside the cosine.

Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule
Sense-check special cases

If $a = 1$ and $b = 0$ ⁠, the function is $\sin x$ and the formula gives $\cos x$ ⁠, the known derivative. If $a = 0$ ⁠, the function is the constant $\sin b$ and the formula gives $0\cdot\cos b = 0$ ⁠, as it must for a constant.

A general formula with letters must reduce to known results when the letters take simple values. Both special cases agree, which guards against a misplaced factor such as $b\cos(ax+b)$ or $\cos(a)$ ⁠.

Result

For every real $x$ (and any constants $a$ ⁠, $b$ ⁠),

\frac{d}{dx}\sin(ax+b) = a\cos(ax+b).

(Textbook answer: $a\cos(ax+b)$ ⁠.)

Check it with numbers

Take $a = 2$ ⁠, $b = 1$ ⁠, $x = 0.5$ ⁠, so $ax + b = 2$ ⁠. The formula gives $2\cos 2$ ⁠, and $\cos 2 \approx -0.4161468$ ⁠, so the slope is about $-0.8322937$ ⁠. A difference quotient of $\sin(2x+1)$ with $h = 0.0001$ on each side of $0.5$ gives $-0.83229$ ⁠. They agree.

Common mistakes

Forgetting the factor $a$ ⁠. $\cos(ax+b)$ alone is the derivative with respect to $t = ax+b$ ⁠, not with respect to $x$ ⁠.
Treating $b$ as a factor. $b$ is added inside, so it differentiates to $0$ and never appears as a multiplier.
Differentiating with respect to $a$ or $b$ ⁠. They are constants here; only $x$ is the variable.

What to take away

For any differentiable $f$ ⁠: $\frac{d}{dx}f(ax+b) = a\,f'(ax+b)$ ⁠. A linear inside contributes just its slope $a$ ⁠.
So $(\cos(ax+b))' = -a\sin(ax+b)$ and $(\tan(ax+b))' = a\sec^2(ax+b)$ follow the same way. Question 5 uses the first of these.

Every fact used, and where it comes from

Fact	Used in	Source
$(\sin t)' = \cos t$ ⁠; $(x)' = 1$ ⁠; constant has derivative 0	Step 2	Table 5.3 and rule (1), Section 5.3
$(c\,u)' = c\,u'$	Step 2	Product rule (2), Section 5.3, with $c' = 0$
Chain rule	Step 3	Theorem 4, Section 5.3.1 (Lesson: the chain rule)

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Exercise 5.2 · Q3: Differentiate sin⁡(ax+b)\sin(ax+b)sin(ax+b)

Basics you need

Solution

Name the inner and the outer function

Differentiate each link

Multiply (chain rule) and return to xxx

Sense-check special cases

Every fact used, and where it comes from

Exercise 5.2 · Q3: Differentiate $\sin(ax+b)$

Multiply (chain rule) and return to $x$