Derivatives of all six trig functions: Worked examples

The scenario matrix

Every trig-derivative problem lives in one of these boxes. The examples below are labelled with the box they fill.

Cell	Case class	What makes it tricky	Example
A	Single function × single rule	pick the right rule	Ex 1 (product)
B	Chain rule, inner ≠ $x$	don't forget the inside derivative	Ex 2
C	Nested chain (chain inside chain)	multiply all inner derivatives	Ex 3
D	Quotient of two trig functions	signs from "co"-functions	Ex 4
E	Sign / quadrant evaluation	slope can be $+$ , $-$ , or $0$	Ex 5
F	Degenerate / undefined input	function blows up; slope undefined	Ex 6
G	Limiting behaviour ( $x \to 0$ )	uses the foundation limit again	Ex 7
H	Real-world oscillation (word problem)	translate physics → derivative	Ex 8
I	Exam twist (simplify to a clean form)	identity collapses the mess	Ex 9

We now walk A → I so no box is left blank.

Worked examples

Worked example Ex 1 — Cell A: product of a power and a trig function

Differentiate $f (x) = x^{3} cos x$ .

Forecast: Two things multiplied → we'll get two terms. One keeps $cos$ , the other keeps $x^{3}$ and turns $cos$ into... something with a minus.

Name the pieces: $u = x^{3}$ , $v = cos x$ . Why this step? Product Rule needs the two factors named before we touch them.
Differentiate each: $u^{'} = 3 x^{2}$ , and $v^{'} = - sin x$ . Why this step? $\frac{d}{d x} cos x = - sin x$ from the parent table — the "co" minus sign appears.
Apply $(uv)^{'} = u^{'} v + u v^{'}$ : $f^{'} (x) = 3 x^{2} cos x + x^{3} (- sin x) = 3 x^{2} cos x - x^{3} sin x .$ Why this step? The product rule says "differentiate the first, keep the second; then keep the first, differentiate the second" — you cannot just multiply the two derivatives, so both terms are mandatory.

Verify: At $x = 0$ : $f^{'} (0) = 0 - 0 = 0$ . Sanity: near $x = 0$ , $f (x) = x^{3} cos x \approx x^{3}$ , whose slope at $0$ is $0$ . ✅

Worked example Ex 2 — Cell B: chain rule with a linear inside

Differentiate $g (x) = sin (5 x)$ .

Forecast: The table gives $sin^{'} = cos$ , but the inside is $5 x$ , not $x$ . A factor of $5$ must pop out.

Identify outer and inner: outer $= sin (□)$ , inner $= u = 5 x$ . Why this step? Chain Rule says differentiate outer-at-inner, then times inner's derivative.
Outer derivative at the inner: $cos (5 x)$ . Why this step? The chain rule differentiates the outer function first while leaving the inner block untouched — so $sin (□)$ becomes $cos (□)$ with $□ = 5 x$ still sitting inside.
Inner derivative: $\frac{d}{d x} (5 x) = 5$ . Why this step? The chain rule then asks "how fast does the inside change?" — the inside $5 x$ changes $5$ times as fast as $x$ , and that rate must scale the whole result.
Multiply: $g^{'} (x) = 5 cos (5 x)$ . Why this step? Forgetting the $5$ is the #1 chain-rule error flagged in the parent note.

Verify: At $x = 0$ : $g^{'} (0) = 5 cos 0 = 5$ . Sanity: $sin (5 x) \approx 5 x$ for small $x$ , slope $5$ . ✅

Worked example Ex 3 — Cell C: nested chain (chain inside a chain)

Differentiate $p (x) = tan (cos x)$ .

Forecast: Two layers. Outer is $tan$ , inner is $cos x$ . We'll multiply two inner derivatives.

Layers: outermost $tan (□)$ , its argument is $cos x$ . Why this step? Peel from outside in — Chain Rule applied twice.
Differentiate the outer at the inner: $\frac{d}{d x} tan (□) = sec^{2} (□)$ , so $sec^{2} (cos x)$ . Why this step? The chain rule differentiates the outermost layer first while freezing its argument — $tan^{'} = sec^{2}$ from the parent table, evaluated at the still-intact inner block $cos x$ .
Times the derivative of the inside $cos x$ , which is $- sin x$ . Why this step? The chain rule then multiplies by the rate of change of that inner block; $\frac{d}{d x} cos x = - sin x$ , and the "co" minus sign carries through.
Combine: $p^{'} (x) = sec^{2} (cos x) \cdot (- sin x) = - sin x sec^{2} (cos x) .$ Why this step? The chain rule's answer is the product of the outer-at-inner derivative and the inner derivative — dropping either factor gives the wrong slope.

Verify: At $x = 0$ : $p^{'} (0) = - sin 0 \cdot sec^{2} (1) = 0$ . Sanity: $sin 0 = 0$ kills it — a horizontal tangent at $x = 0$ . ✅

Worked example Ex 4 — Cell D: quotient of two trig functions

Differentiate $q (x) = \frac{sin x}{1 + cos x}$ .

Forecast: Quotient rule → messy numerator, but the Pythagorean identity will clean it up.

Name: $u = sin x$ ( $u^{'} = cos x$ ), $v = 1 + cos x$ ( $v^{'} = - sin x$ ). Why this step? Quotient Rule needs both parts and their derivatives.
Apply $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ : $q^{'} (x) = \frac{c o s x ( 1 + c o s x ) - s i n x ( - s i n x )}{( 1 + c o s x ) ^{2}} .$ Why this step? A ratio of two functions requires the quotient rule — "derivative of top times bottom, minus top times derivative of bottom, all over bottom squared"; you cannot differentiate top and bottom separately.
Expand the numerator: $cos x + cos^{2} x + sin^{2} x = cos x + 1$ . Why this step? $cos^{2} x + sin^{2} x = 1$ collapses two terms into one.
Cancel the shared factor: $q^{'} (x) = \frac{1 + c o s x}{( 1 + c o s x ) ^{2}} = \frac{1}{1 + c o s x} .$ Why this step? The numerator $1 + cos x$ is one copy of the factor sitting in the squared denominator, so one copy cancels — this is what turns the ugly fraction into a clean closed form.

Verify: At $x = 0$ : $q^{'} (0) = \frac{1}{2}$ . Sanity: numerator started as $cos 0 (2) - 0 = 2$ , denominator $2^{2} = 4$ , ratio $\frac{1}{2}$ . ✅

Worked example Ex 5 — Cell E: sign of the slope in every quadrant

For $y = sin x$ , find the slope at $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ (one angle from each quadrant of the circle). See the figure.

Forecast: Slope $= cos x$ . Cosine is $+$ in QI and QIV, $-$ in QII and QIII. So we expect $+, -, -, +$ .

Slope function: $y^{'} = cos x$ . Why this step? The slope of $sin$ is $cos$ — parent Step 1.
Plug each angle (all values are $\pm \frac{2}{2}$ ):
- $cos \frac{π}{4} = + \frac{2}{2}$ (QI: rising)
- $cos \frac{3 π}{4} = - \frac{2}{2}$ (QII: falling)
- $cos \frac{5 π}{4} = - \frac{2}{2}$ (QIII: falling)
- $cos \frac{7 π}{4} = + \frac{2}{2}$ (QIV: rising) Why this step? The sign of the slope is exactly the sign of $cos$ at that point, and $cos$ flips sign as we cross into QII and QIII — this is where a naive " $sin$ always rises" guess fails.
Read the geometry (look at the amber tangent lines in the figure): steep up, steep down, steep down, steep up. Why this step? A tangent line's tilt is the derivative made visible — pairing each computed number with its drawn slope confirms $+$ means "up-tilt" and $-$ means "down-tilt", so the algebra and the picture agree.

Verify: $\frac{2}{2} \approx 0.707$ ; signs are $+, -, -, +$ as forecast. The slope is zero exactly at the crest $x = \frac{π}{2}$ and trough $x = \frac{3 π}{2}$ , where $cos = 0$ . ✅

Worked example Ex 6 — Cell F: degenerate input where the function explodes

Where is $\frac{d}{d x} tan x = sec^{2} x$ undefined, and what does the graph do there?

Forecast: $tan x$ has vertical walls at $x = \frac{π}{2} + nπ$ . Its slope should blow up to $+ \infty$ there, never negative.

Recall $sec^{2} x = \frac{1}{cos ^{2} x}$ . Why this step? The derivative is written cleanly as $sec^{2}$ , but its danger points are hidden inside $cos$ .
$sec^{2} x$ is undefined exactly when $cos x = 0$ , i.e. $x = \frac{π}{2} + nπ$ for integer $n$ . Why this step? A fraction is undefined only when its denominator is zero, and $cos^{2} x = 0$ precisely at those angles — so those are the exact spots where the slope ceases to exist.
Because $cos^{2} x \geq 0$ always, $sec^{2} x = \frac{1}{c o s ^{2} x} \geq 1$ everywhere it exists — the slope of $tan$ is never below $1$ . Why this step? This tells us $tan$ always rises (never flattens, never dips).
As $x \to \frac{π}{2}^{-}$ , $cos x \to 0^{+}$ , so $sec^{2} x \to + \infty$ : a vertical tangent. Why this step? As the denominator $cos^{2} x$ shrinks toward $0$ , the reciprocal $\frac{1}{c o s ^{2} x}$ grows without bound — that unbounded positive slope is what draws the vertical wall (asymptote) of $tan$ .

Verify: At the "flattest" point $x = 0$ : $sec^{2} 0 = \frac{1}{1} = 1$ , the minimum slope. At $x = \frac{π}{3}$ : $cos = \frac{1}{2}$ , so $sec^{2} = \frac{1}{1/4} = 4 \geq 1$ . ✅

Worked example Ex 7 — Cell G: limiting behaviour as

$x \to 0$ Evaluate $x \to 0 lim \frac{tan x}{x}$ using derivatives.

Forecast: Both $tan x$ and $x$ head to $0$ , so it's a $\frac{0}{0}$ form. Because $tan x \approx x$ near zero, the answer should be $1$ .

Recognise $lim_{x \to 0} \frac{tan x - tan 0}{x - 0}$ is exactly the derivative of $tan$ at $0$ . Why this step? This is the Limit definition of the derivative read backwards — the difference quotient of $tan$ at $0$ (since $tan 0 = 0$ ).
That derivative is $tan^{'} (0) = sec^{2} 0$ . Why this step? Once we recognise the limit is a derivative, we can skip the messy limit and just quote the known slope $tan^{'} = sec^{2}$ from the parent table, evaluated at $0$ .
$sec^{2} 0 = \frac{1}{cos ^{2} 0} = \frac{1}{1} = 1$ . Why this step? $cos 0 = 1$ , so the reciprocal-square is $1$ — this pins the numerical value of the limit and confirms the forecast.

Verify: Numerically $tan (0.001) /0.001 \approx 1.0000003 \to 1$ . Matches the parent's foundation-limit spirit ( $sin h / h \to 1$ ). ✅

Worked example Ex 8 — Cell H: real-world oscillation (SHM)

A mass on a spring sits at height $s (t) = 4 sin (2 t)$ cm, where $t$ is in seconds. Find its velocity, and its maximum speed.

Forecast: Velocity is the derivative. Chain rule → a factor of $2$ appears, so amplitude of velocity is $4 \times 2 = 8$ . Max speed = $8$ cm/s. See the figure. This is Simple Harmonic Motion.

Velocity $v (t) = s^{'} (t)$ . Differentiate $4 sin (2 t)$ : outer $sin$ gives $cos$ , inner $2 t$ gives $2$ . Why this step? Velocity = rate of change of position; Chain Rule for the inner $2 t$ .
$v (t) = 4 cos (2 t) \cdot 2 = 8 cos (2 t)$ cm/s. Why this step? The constant $4$ rides along untouched, while the chain rule multiplies by the inner derivative $2$ — that is why the velocity amplitude ( $8$ ) is bigger than the position amplitude ( $4$ ): faster wiggling means bigger speeds.
$cos$ ranges in $[- 1, 1]$ , so $v$ ranges in $[- 8, 8]$ . Maximum speed (size of velocity) is $8$ cm/s. Why this step? Speed is $∣ v ∣$ ; the peak of $∣ cos ∣$ is $1$ .
Where does max speed happen? When $cos (2 t) = \pm 1$ , i.e. $2 t = nπ$ , i.e. at the middle of the swing $s = 0$ — exactly where the position graph crosses zero (green dots in figure). Why this step? $∣ cos ∣$ hits its maximum $1$ exactly when its argument is a multiple of $π$ ; at those same instants $sin (2 t) = 0$ so $s = 0$ — revealing the physics fact that the mass is fastest as it whips through the centre.

Verify: At $t = 0$ : $s (0) = 0$ (centre), $v (0) = 8 cos 0 = 8$ cm/s — fastest at the centre, as physics demands. At $t = \frac{π}{4}$ : $s = 4 sin (\frac{π}{2}) = 4$ (top), $v = 8 cos (\frac{π}{2}) = 0$ — momentarily still at the top. ✅

Worked example Ex 9 — Cell I: exam twist that simplifies to a clean form

Differentiate $r (x) = sec x (sec x + tan x)$ and simplify.

Forecast: Looks like it needs product rule and gets ugly — but $sec$ and $tan$ derivatives share pieces, so watch for a collapse.

Expand first: $r (x) = sec^{2} x + sec x tan x$ . Why this step? Expanding avoids the product rule entirely — differentiate term by term.
Differentiate $sec^{2} x$ : treat as $(sec x)^{2}$ , Chain Rule gives $2 sec x \cdot (sec x tan x) = 2 sec^{2} x tan x$ . Why this step? $(sec x)^{2}$ is an outer square wrapped around $sec x$ ; the chain rule gives $2 \cdot (inside) \cdot (inside’s derivative)$ , and $sec^{'} = sec x tan x$ from the parent table.
Differentiate $sec x tan x$ with Product Rule: $(sec x tan x)^{'} = sec x tan x \cdot tan x + sec x \cdot sec^{2} x = sec x tan^{2} x + sec^{3} x$ . Why this step? This term is two functions multiplied, so the product rule applies: differentiate $sec$ (gives $sec tan$ ) times $tan$ , plus $sec$ times the derivative of $tan$ (gives $sec^{2}$ ).
Add: $r^{'} (x) = 2 sec^{2} x tan x + sec x tan^{2} x + sec^{3} x .$ Factor out $sec x$ : $r^{'} (x) = sec x (2 sec x tan x + tan^{2} x + sec^{2} x)$ . Why this step? Every term contains at least one $sec x$ , so pulling it out is legal and gives the tidy "clean form" an exam wants — the shared factor is the collapse we forecast.

Verify: At $x = 0$ : $sec 0 = 1$ , $tan 0 = 0$ , so $r^{'} (0) = 1 (0 + 0 + 1) = 1$ . Cross-check by the original expanded form's derivative at $0$ : $2 sec^{2} 0 tan 0 + (sec 0 tan^{2} 0 + sec^{3} 0) = 0 + (0 + 1) = 1$ . ✅

Recall Which cell am I in?

Product of two functions? ::: Cell A — product rule Inner function isn't plain $x$ ? ::: Cell B/C — chain rule, times every inside derivative Function is a fraction of trig? ::: Cell D — quotient rule, then Pythagorean cleanup Asked for the sign of a slope? ::: Cell E — plug into the derivative, read $\pm$ from the quadrant Point where $cos x = 0$ or $sin x = 0$ ? ::: Cell F — derivative may be undefined (blows up) $\frac{0}{0}$ limit with trig? ::: Cell G — recognise the difference quotient = a known derivative "Velocity of an oscillation"? ::: Cell H — differentiate the position, chain rule for angular speed

Connections

Derivatives of all six trig functions (parent)
Product Rule
Quotient Rule
Chain Rule
Limit definition of the derivative
Pythagorean identity sin^2 + cos^2 = 1
Simple Harmonic Motion
Derivatives of inverse trig functions

Decision Map

The picture below is the same "which rule do I use?" flow, drawn as a diagram so you can see the branch you land on: read a problem, ask the one question in the amber diamond, and follow the arrow to the rule you need.

Figure — Derivatives of all six trig functions