Differentiating Trigonometric Expressions

5 min read

Reference: Stewart §2.4

Combining Trig Derivatives with Algebraic Rules

Knowing that $\frac{d}{dx}\sin x = \cos x$ is just the beginning. Real problems involve expressions like $x^2 \sin x$ (product) or $\frac{\sec x}{1 + \tan x}$ (quotient). This skill bridges the gap between memorizing formulas and solving actual calculus problems.

The key insight: trig functions are just functions. They follow the same product rule, quotient rule, and sum rule as polynomials. The only new ingredient is the six derivative formulas from the previous skill.

Prerequisite Map

Quick Reference

Property	Value
Concept	Derivatives
Chapter	2.4
Difficulty	Intermediate
Time	~25 minutes

Common Expression Types

Type 1: Trig + Polynomial

Simple sums/differences: differentiate term by term.

Example: $f(x) = x^3 + \sin x - 2\cos x$

$$f'(x) = 3x^2 + \cos x - 2(-\sin x) = 3x^2 + \cos x + 2\sin x$$

Type 2: Products with Trig

Use the Product Rule: $(fg)' = f'g + fg'$

Example: $y = x^2 \sin x$

Let $f = x^2$ and $g = \sin x$:

$$y' = (x^2)' \cdot \sin x + x^2 \cdot (\sin x)' = 2x \sin x + x^2 \cos x$$

Type 3: Quotients with Trig

Use the Quotient Rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

Example: $y = \frac{\sin x}{1 + \cos x}$

$$y' = \frac{(\cos x)(1 + \cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2}$$

$$= \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2} = \frac{\cos x + 1}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$$

Type 4: Multiple Trig Functions

Products of trig functions also need the product rule.

Example: $y = \sin x \cos x$

$$y' = (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x = \cos 2x$$

Worked Example: Finding Horizontal Tangents

Problem: For $f(x) = \frac{\sec x}{1 + \tan x}$, find where the graph has horizontal tangents.

Solution:

Step 1: Find $f'(x)$ using the quotient rule.

Let $u = \sec x$ and $v = 1 + \tan x$.

$u' = \sec x \tan x$
$v' = \sec^2 x$

$$f'(x) = \frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2}$$

Step 2: Simplify the numerator.

$$= \frac{\sec x \tan x + \sec x \tan^2 x - \sec^3 x}{(1 + \tan x)^2}$$

Factor out $\sec x$:

$$= \frac{\sec x(\tan x + \tan^2 x - \sec^2 x)}{(1 + \tan x)^2}$$

Using $\sec^2 x = 1 + \tan^2 x$:

$$\tan x + \tan^2 x - \sec^2 x = \tan x + \tan^2 x - 1 - \tan^2 x = \tan x - 1$$

So:

$$f'(x) = \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2}$$

Step 3: Solve $f'(x) = 0$.

Since $\sec x \neq 0$ for any $x$ in the domain, we need $\tan x - 1 = 0$, i.e., $\tan x = 1$.

Answer: Horizontal tangents occur at $x = \frac{\pi}{4} + n\pi$ for any integer $n$.

Application: Simple Harmonic Motion

When an object oscillates on a spring, its position might be:

$$s(t) = 4\cos t \quad \text{(cm)}$$

The velocity is the derivative of position:

$$v(t) = s'(t) = -4\sin t \quad \text{(cm/s)}$$

The acceleration is the derivative of velocity:

$$a(t) = v'(t) = -4\cos t \quad \text{(cm/s}^2\text{)}$$

Observations:

The object is at rest ($v = 0$) when $\sin t = 0$, i.e., at the extremes of motion
The acceleration is maximum when $\cos t = \pm 1$, i.e., at the extremes
Notice that $a(t) = -s(t)$: acceleration is proportional and opposite to displacement

Practice Problems

Level 1 Sum of Trig and Polynomial

Differentiate $f(x) = x^2 + \cot x$.

Thought Process

This is a sum, so differentiate term by term.

Recall: $\frac{d}{dx}(\cot x) = -\csc^2 x$

Show Answer

$$f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(\cot x)$$

$$= 2x + (-\csc^2 x)$$

$$= 2x - \csc^2 x$$

Level 2 Product with Power Function

Differentiate $h(\theta) = \theta^3 \sin \theta$.

Thought Process

This is a product of $\theta^3$ and $\sin\theta$.

Use the product rule: $(fg)' = f'g + fg'$

Show Answer

Using the product rule with $f = \theta^3$ and $g = \sin\theta$:

$$h'(\theta) = (3\theta^2)(\sin\theta) + (\theta^3)(\cos\theta)$$

$$= 3\theta^2 \sin\theta + \theta^3 \cos\theta$$

Factor out $\theta^2$:

$$= \theta^2(3\sin\theta + \theta\cos\theta)$$

Level 3 Quotient with Trig

Differentiate $y = \frac{\cos x}{1 - \sin x}$.

Thought Process

Use the quotient rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

Here $f = \cos x$ and $g = 1 - \sin x$.

Show Answer

With $f = \cos x$ and $g = 1 - \sin x$:

$f' = -\sin x$
$g' = -\cos x$

$$y' = \frac{(-\sin x)(1 - \sin x) - (\cos x)(-\cos x)}{(1 - \sin x)^2}$$

$$= \frac{-\sin x + \sin^2 x + \cos^2 x}{(1 - \sin x)^2}$$

Using $\sin^2 x + \cos^2 x = 1$:

$$= \frac{-\sin x + 1}{(1 - \sin x)^2} = \frac{1 - \sin x}{(1 - \sin x)^2} = \frac{1}{1 - \sin x}$$

Level 4 Tangent Line to a Trig Curve

Find an equation of the tangent line to $y = x + \sin x$ at the point $(\pi, \pi)$.

Thought Process

For a tangent line, we need:

The slope: $y'$ evaluated at $x = \pi$
A point: given as $(\pi, \pi)$

Then use point-slope form: $y - y_1 = m(x - x_1)$

Show Answer

Step 1: Find the derivative.

$$y' = 1 + \cos x$$

Step 2: Evaluate at $x = \pi$.

$$y'(\pi) = 1 + \cos \pi = 1 + (-1) = 0$$

Step 3: Write the tangent line equation.

The slope is 0, so the tangent line is horizontal.

Using point-slope form with point $(\pi, \pi)$ and slope $m = 0$:

$$y - \pi = 0(x - \pi)$$

$$y = \pi$$

The tangent line is $y = \pi$.

Level 5 Motion Analysis

A mass on a spring vibrates according to $x(t) = 8\sin t$ cm, where $t$ is in seconds.

Find the velocity and acceleration at time $t$.
Find the position, velocity, and acceleration at $t = 2\pi/3$.
At $t = 2\pi/3$, is the mass moving toward or away from equilibrium? Is it speeding up or slowing down?

Thought Process

Velocity = $x'(t)$, acceleration = $x''(t)$.

For part (c):

Moving toward equilibrium if position and velocity have opposite signs
Speeding up if velocity and acceleration have the same sign

Show Answer

Part (a):

$$v(t) = x'(t) = 8\cos t \text{ cm/s}$$

$$a(t) = v'(t) = -8\sin t \text{ cm/s}^2$$

Part (b): At $t = 2\pi/3$:

Position: $x(2\pi/3) = 8\sin(2\pi/3) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \approx 6.93$ cm

Velocity: $v(2\pi/3) = 8\cos(2\pi/3) = 8 \cdot (-\frac{1}{2}) = -4$ cm/s

Acceleration: $a(2\pi/3) = -8\sin(2\pi/3) = -8 \cdot \frac{\sqrt{3}}{2} = -4\sqrt{3} \approx -6.93$ cm/s²

Part (c):

Position is positive ($x > 0$): mass is displaced in the positive direction
Velocity is negative ($v < 0$): mass is moving in the negative direction
Since position and velocity have opposite signs: moving toward equilibrium ✓
Velocity is negative ($v < 0$)
Acceleration is negative ($a < 0$)
Same sign means: speeding up ✓

The mass is returning toward equilibrium and gaining speed (it will be fastest when it crosses equilibrium).

Conceptual Check (CCI-Style)

Conceptual Horizontal Tangents on Sine

The function $f(x) = x + 2\sin x$ has horizontal tangents at certain points.

Without computing, predict: are there finitely many or infinitely many horizontal tangents? Explain your reasoning.

Thought Process

Horizontal tangents occur where $f'(x) = 0$.

$f'(x) = 1 + 2\cos x$

This equals zero when $\cos x = -\frac{1}{2}$.

How many solutions does $\cos x = -\frac{1}{2}$ have?

Show Answer

Infinitely many horizontal tangents.

Since $f'(x) = 1 + 2\cos x$, horizontal tangents occur when:

$$\cos x = -\frac{1}{2}$$

The cosine function equals $-\frac{1}{2}$ at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ in the interval $[0, 2\pi]$.

But cosine is periodic with period $2\pi$, so solutions occur at:

$$x = \frac{2\pi}{3} + 2\pi n \quad \text{and} \quad x = \frac{4\pi}{3} + 2\pi n$$

for all integers $n$.

There are infinitely many horizontal tangents, two per period of the underlying cosine function.

Common Mistakes

Mistake	Correction
Forgetting the negative in $(\cos x)' = -\sin x$	The “co” functions all have negative derivatives
Writing $(x \sin x)' = 1 \cdot \cos x$	This is a product; use product rule: $\sin x + x\cos x$
Simplifying $\sin^2 x + \cos^2 x$ to something other than 1	This identity always equals 1
Forgetting $\tan x$ has domain restrictions	$\tan x$ and $\sec x$ are undefined where $\cos x = 0$

Common Misconceptions

Common misconception

$\frac{d}{dx}[\sin(x^2)] = \cos(x^2)$. The sine function is composed with $x^2$, so the chain rule applies. The outer derivative gives $\cos(x^2)$, and the inner derivative gives $2x$, producing $\frac{d}{dx}[\sin(x^2)] = 2x\cos(x^2)$. Omitting the inner derivative $2x$ is the chain rule error; it occurs because students apply the basic formula $(\sin u)' = \cos u$ while treating $u$ as if it were the final variable.

Common misconception

$(\sin x)^2$ and $\sin(x^2)$ have the same derivative. These are different functions composed in different ways. For $(\sin x)^2 = \sin^2 x$, the chain rule gives $2\sin x \cos x$. For $\sin(x^2)$, the chain rule gives $2x\cos(x^2)$. Neither is $2\sin x$ or $2x\cos x$ alone.

Mastery Checklist

I can differentiate sums involving trig and polynomial terms
I can apply the product rule with one trig factor
I can apply the quotient rule with trig functions
I can find horizontal tangents by solving $f'(x) = 0$
I can analyze velocity and acceleration for harmonic motion
I simplify using $\sin^2 x + \cos^2 x = 1$ when it appears

Mental Model

Trig functions are just functions.

The product rule, quotient rule, and sum rule work exactly the same way whether you’re dealing with polynomials or trig functions. The only difference is what you substitute for $f'$ and $g'$.

Think of the six trig derivative formulas as your “lookup table.” Once you know them, everything else follows from the algebraic rules you already know.

Connections

Looking back:

Trig Derivative Formulas provides the six building blocks
Product Rule and Quotient Rule are the combining mechanisms

Looking ahead:

Chain Rule extends this to compositions like $\sin(x^2)$
Related Rates applies these derivatives to real-world rate problems

Real-world connections:

Harmonic oscillators (springs, pendulums) have trig position functions
Circular motion involves sine and cosine of time
Signal processing uses trig functions extensively

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Trig Derivative Formulas	Skills Index	Chain Rule

Last updated: 2026-01-22