Chapter 5 · Continuity and Differentiability

Example 21: Derivative of $\sin(x^2)$ by the chain rule

ExamplesChain rule · Section 5.3.1

The question

Find the derivative of the function given by $f(x) = \sin(x^2)$ ⁠.

Exam answer

Let $y = \sin(x^2)$ ⁠. By the chain rule,

\begin{aligned} \frac{dy}{dx} &= \cos(x^2)\cdot\frac{d}{dx}(x^2) \\ &= \cos(x^2)\cdot 2x \end{aligned}

$\therefore\ \dfrac{dy}{dx} = 2x\cos(x^2)$ ⁠.

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

The one idea this problem is built on

$\sin(x^2)$ is a function inside another function: first square $x$ ⁠, then take the sine of the result. The chain rule says the rate of change of such a composite is the product of the two separate rates: the rate of the outer function (sine) at the inner value, times the rate of the inner function (squaring). So the answer is $\cos(x^2)$ times $2x$ ⁠.

Given, to find, and the plan

Given: $f(x) = \sin(x^2)$ for every real $x$ ⁠.
To find: $f'(x) = \dfrac{df}{dx}$ ⁠.
Plan: (1) Split $f$ into an inner function $t = u(x) = x^2$ and an outer function $v(t) = \sin t$ ⁠. (2) Differentiate each one separately. (3) Multiply the two derivatives (chain rule). (4) Write the answer in terms of $x$ only.

Basics you need

1Composite functions

$v \circ u$ means "do $u$ first, then $v$ ": $(v\circ u)(x) = v(u(x))$ ⁠. The order matters: squaring then taking sine is not the same as taking sine then squaring. Basics: composite functions

2Two standard derivatives

From Table 5.3: $\dfrac{d}{dx}(x^n) = nx^{n-1}$ ⁠, so $\dfrac{d}{dx}(x^2) = 2x$ ⁠; and $\dfrac{d}{dt}(\sin t) = \cos t$ ⁠. Basics: derivative rules

3The chain rule (Theorem 4)

If $f = v\circ u$ ⁠, $t = u(x)$ ⁠, and both $\dfrac{dt}{dx}$ and $\dfrac{dv}{dt}$ exist, then $\dfrac{df}{dx} = \dfrac{dv}{dt}\cdot\dfrac{dt}{dx}$ ⁠. Basics: slope at a point

Solution

Recognise the composite: name the inner and outer functions

To compute $\sin(x^2)$ for a given $x$ ⁠, we do two jobs in order: first square $x$ ⁠, then take the sine. Give each job a name:
$\begin{gathered} t = u(x) = x^2 \ \ (\text{inner}), \\ v(t) = \sin t \ \ (\text{outer}). \end{gathered}$
Then $v(u(x)) = v(x^2) = \sin(x^2) = f(x)$ ⁠, so $f = v \circ u$ ⁠.

The chain rule only applies once we can see a function as "outer applied to inner". The test for which is inner: it is the job done first when you evaluate. With $x = 3$ ⁠: first $3^2 = 9$ ⁠, then $\sin 9$ ⁠. So squaring is inner and sine is outer.

$\sin(x^2)$ (often printed $\sin x^2$ ⁠) is different from $\sin^2 x = (\sin x)^2$ ⁠. The second one is $u\circ v$ ⁠: sine first, then square. Example: at $x = \pi$ ⁠, $\sin(\pi^2) \approx -0.4303$ but $(\sin\pi)^2 = 0$ ⁠. Basics: composite functions
Differentiate each piece separately

Inner: by the power rule with $n = 2$ ⁠,
$\frac{dt}{dx} = \frac{d}{dx}(x^2) = 2x^{2-1} = 2x.$
Outer, as a function of its own variable $t$ ⁠:
$\frac{dv}{dt} = \frac{d}{dt}(\sin t) = \cos t.$
Both derivatives exist for every real $x$ and every real $t$ ⁠, so the conditions of the chain rule hold everywhere.

Theorem 4 asks for two things before it may be used: $\frac{dt}{dx}$ and $\frac{dv}{dt}$ must exist. Checking them is part of the solution, not a formality. Here both functions are differentiable on all of $\R$ ⁠, so there are no exceptional points.

Power rule $\frac{d}{dx}x^n = nx^{n-1}$ and $\frac{d}{dt}\sin t = \cos t$ ⁠, both from Table 5.3 (Section 5.3). Basics: derivative rules
Multiply the two rates (chain rule)
$\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx} = \cos t \cdot 2x.$
The diagram shows why the rates multiply.

The input passes through two machines. Near $x = 1$ ⁠, $t$ changes about 2 times as fast as $x$ ⁠, and $f$ changes about $\cos 1 \approx 0.540$ times as fast as $t$ ⁠. So $f$ changes about $2 \times 0.540$ times as fast as $x$ ⁠.

If a small change $\Delta x$ in $x$ makes $t$ change by about $2x\,\Delta x$ ⁠, and a small change $\Delta t$ makes $\sin t$ change by about $\cos t\,\Delta t$ ⁠, then the change in $f$ is about $\cos t \cdot 2x\,\Delta x$ ⁠. The two "magnification factors" multiply. That is the content of the chain rule.

Chain rule: $\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx}$ (Theorem 4, Section 5.3.1). Lesson: chain rule
Write the answer in terms of $x$

The question is about $x$ ⁠, and $t$ was only a helper. Put $t = x^2$ back:
$\begin{aligned} \frac{df}{dx} &= \cos(x^2)\cdot 2x \\ &= 2x\cos(x^2), \qquad x \in \R. \end{aligned}$
An answer that still contains $t$ mixes two variables and cannot be evaluated from $x$ alone. The book calls expressing the final result in $x$ "normal practice". Writing the factor $2x$ in front avoids the ambiguity of $\cos(x^2)\,2x$ being misread as $\cos(2x^3)$ ⁠.

Substitution: wherever $t$ appears, write $x^2$ ⁠. Keep the brackets: $\cos(x^2)$ ⁠, not $\cos^2 x$ ⁠.

Result

$\dfrac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ for every real $x$ ⁠. (The book's solution gets the same: $\frac{df}{dx} = \cos t\cdot 2x = 2x\cos x^2$ ⁠.)

Check it with numbers

At $x = 1$ the formula gives $2\cos 1 = 1.0806046\ldots$ ⁠. A difference quotient straddling $x = 1$ ⁠:

\begin{aligned} &\frac{\sin(1.001^2) - \sin(0.999^2)}{0.002} \\ &\approx 1.0806022, \end{aligned}

which agrees to five significant figures (the small gap shrinks as the step shrinks). At $x = 0.5$ ⁠: $2(0.5)\cos(0.25)$ $= \cos 0.25 = 0.9689124\ldots$ ⁠. At $x = 0$ the slope is $0$ ⁠: the graph of $\sin(x^2)$ is flat at the origin, as $2x$ is zero there.

Common mistakes

Forgetting the inner derivative: writing $\cos(x^2)$ alone. The factor $2x$ is the inner rate and must be multiplied in.
Differentiating the inside of the cosine: writing $\cos(2x)$ ⁠. The outer derivative is evaluated at the inner value $x^2$ ⁠; the inner derivative $2x$ multiplies from outside.
Confusing $\sin(x^2)$ with $\sin^2 x$ ⁠. The derivative of $\sin^2 x$ is $2\sin x\cos x$ ⁠, a different function.
Leaving $t$ in the final answer. Always substitute back.

What to take away

For a composite, ask "what is done first?" That is the inner function $t$ ⁠.
Derivative = (derivative of outer, evaluated at $t$ ⁠) $\times$ (derivative of inner).
The same pattern gives $\frac{d}{dx}\sin(g(x)) = \cos(g(x))\,g'(x)$ for any differentiable $g$ ⁠, for example Exercise 5.2 Q1.

Every fact used, and where it comes from

Fact	Used in	Source
Composite $(v\circ u)(x) = v(u(x))$	Step 1	Chapter 1 (composition of functions); Theorem 4, Section 5.3.1
$\frac{d}{dx}x^n = nx^{n-1}$ ⁠; $\frac{d}{dx}\sin x = \cos x$	Step 2	Table 5.3, Section 5.3 (Lesson: chain rule)
Chain rule $\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx}$	Step 3	Theorem 4 (Chain Rule), Section 5.3.1
All numerical values	Check	Recomputed in the site's verification script (exact derivative and difference quotients)

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Example 21: Derivative of sin⁡(x2)\sin(x^2)sin(x2) by the chain rule

Basics you need

Solution

Recognise the composite: name the inner and outer functions

Differentiate each piece separately

Multiply the two rates (chain rule)