Derivatives of all six trig functions: Question bank

Before we start, three words that recur below get pinned down so nothing is used unearned:

True or false — justify

Answer T/F, then give the reason in your own words before revealing.

\frac{d}{d x} sin x = cos x

holds whether

x

is in degrees or radians.

False. It only holds in radians. In degrees

sin h \approx \frac{π}{180} h

, so the derivative picks up a factor

\frac{π}{180}

\frac{d}{d x} sin x^{\circ} = \frac{π}{180} cos x^{\circ}

Since

\frac{d}{d x} sin x = cos x

and

\frac{d}{d x} cos x = - sin x

, the pair "rotates" — applying

\frac{d}{d x}

four times returns you to

sin x

True.

sin \to cos \to - sin \to - cos \to sin

. Four derivatives is a full cycle, which is exactly why

sin

and

cos

solve the oscillator equation in Simple Harmonic Motion.

\frac{d}{d x} tan x = sec^{2} x

is valid for all real

x

False. Both

tan x

and

sec^{2} x

blow up where

cos x = 0

(at

x = \frac{π}{2} + k π

). The rule holds only on the domain where

tan x

exists.

The Pythagorean identity is used in deriving

tan^{'}

but not in deriving

sin^{'}

True.

tan^{'}

needs

cos^{2} + sin^{2} = 1

to collapse the numerator. The derivation of

sin^{'}

uses the two foundation limits and Angle addition formulas instead — no Pythagoras.

sec^{2} x \geq 1

for every

x

in the domain, so

tan x

is always increasing.

True.

sec^{2} x = \frac{1}{c o s ^{2} x} \geq 1

since

cos^{2} x \leq 1

. A slope that never drops below

1

means

tan

climbs on every interval between its vertical asymptotes.

Because

cos

is an even function (

cos (- x) = cos x

), its derivative

- sin x

is also even.

False. The derivative of an even function is odd.

- sin (- x) = sin x = - (- sin x)

, so

- sin

is odd — matching the rule that differentiating flips even

\leftrightarrow

odd parity.

\frac{d}{d x} csc x = - csc x cot x

can be checked at

x = \frac{π}{2}

: it should give

0

True. At

x = \frac{π}{2}

cot (\frac{π}{2}) = \frac{c o s}{s i n} = \frac{0}{1} = 0

, so the derivative is

0

. Geometrically

csc

has a minimum there (value

1

), and a minimum has zero slope.

Spot the error

Each line contains a flawed step. Find it, then reveal the fix.

\frac{d}{d x} sin (x^{2}) = cos (x^{2})

Missing the inside derivative (Chain Rule). The inside

x^{2}

also gets differentiated: correct answer is

cos (x^{2}) \cdot 2 x = 2 x cos (x^{2})

\frac{d}{d x} cos (5 x) = - sin (5 x)

Forgot the factor

5

from the inner function. Correct:

- 5 sin (5 x)

. The table rule

cos^{'} = - sin

only applies when the inside is plain

x

\frac{d}{d x} sec x = - sec x tan x

Wrong sign.

sec

has no "co", so it is positive:

\frac{d}{d x} sec x = + sec x tan x

. The minus belongs to

csc

\frac{d}{d x} tan x = sec x^{2}

Notation abuse:

sec x^{2}

reads as

sec (x^{2})

. The answer is

sec^{2} x

, meaning

(sec x)^{2}

. Write it as

(sec x)^{2}

if unsure.

\frac{d}{d x} (sin x cos x) = cos x \cdot (- sin x)

This treated it as chain rule; it's a product (Product Rule):

cos x cos x + sin x (- sin x) = cos^{2} x - sin^{2} x

"For

\frac{d}{d x} cot x

: quotient rule on

\frac{c o s x}{s i n x}

gives

\frac{- s i n x s i n x + c o s x c o s x}{s i n ^{2} x} = csc^{2} x

Sign slip in the numerator. Quotient Rule gives

\frac{( - s i n x ) ( s i n x ) - ( c o s x ) ( c o s x )}{s i n ^{2} x} = \frac{- ( s i n ^{2} + c o s ^{2} )}{s i n ^{2} x} = - csc^{2} x

\frac{d}{d x} sin x = cos x

, so

\frac{d}{d x} sin (3 0^{\circ}) = cos (3 0^{\circ})

Two errors:

sin (3 0^{\circ})

is a constant, so its derivative is

0

; and even a variable-degree input would need the

\frac{π}{180}

factor. Derivative of any constant is

0

Why questions

Why must the angle be in radians for

\frac{d}{d x} sin x = cos x

The whole derivation rests on

lim_{h \to 0} \frac{s i n h}{h} = 1

, and that limit equals

1

only when arc length equals angle measure — which is the definition of radians. Degrees rescale the angle, inserting

\frac{π}{180}

Why do exactly three of the six derivatives carry a minus sign?

The three "co-" functions (

cos, cot, csc

) decrease as their partners increase near

0

. Structurally, each "co" derivative comes from differentiating a

cos

in a numerator (giving

- sin

) or a reciprocal of

sin

, and that

cos^{'} = - sin

minus propagates through.

Why can we derive

tan, cot, sec, csc

without ever using the limit definition again?

Each is a ratio of

sin

and

cos

, whose derivatives we already earned. The Quotient Rule converts "slope of a ratio" into known slopes — no fresh limits needed.

Why does

\frac{d}{d x} tan x = sec^{2} x

never take a negative value?

sec^{2} x = 1/ cos^{2} x

is a square divided into

1

, so it's always positive. This encodes that

tan

only ever rises — it never turns around between asymptotes.

Why is

cos x

the rate of

sin x

rather than just a "related wave"?

Because

cos x

literally is the limit of

\frac{s i n ( x + h ) - s i n x}{h}

— the instantaneous change in

sin

. See Limit definition of the derivative. The phase shift (

cos

leads

sin

\frac{π}{2}

) is the visual of this rate.

Why do the inverse trig derivatives (e.g.

\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}

) look algebraic, with no trig at all?

Inverting the relationship forces

sec^{2}

and friends to be rewritten via Pythagorean identity sin^2 + cos^2 = 1 back in terms of

x

, dissolving the trig. Details in Derivatives of inverse trig functions.

Edge cases

x = 0

, what is the slope of

sin x

, and what does it look like?

cos (0) = 1

— the steepest the

sin

wave ever climbs. Near the origin

sin x \approx x

, a line of slope

1

, which is exactly the foundation limit made visible.

What happens to

\frac{d}{d x} tan x

x \to \frac{π}{2}^{-}

sec^{2} x \to + \infty

. The tangent curve becomes vertical approaching its asymptote; an infinite slope matches the graph shooting up.

\frac{d}{d x} cot x

defined at

x = 0

No.

cot x = cos x / sin x

is undefined where

sin x = 0

(including

x = 0

), and

- csc^{2} x = - 1/ sin^{2} x

blows up there. The derivative shares the parent's domain gaps.

\frac{d}{d x} sec x

x = 0

— value and meaning?

sec (0) tan (0) = 1 \cdot 0 = 0

sec

has a local minimum of

1

x = 0

, and a minimum has zero slope — consistent.

Does

\frac{d}{d x} sin x = cos x

still hold at the peaks

x = \frac{π}{2}

where

sin

is flat?

Yes.

cos (\frac{π}{2}) = 0

, correctly reporting zero slope at the crest. The rule handles maxima/minima automatically — no special case needed.

What is

\frac{d}{d x} csc x

x \to 0^{+}

- csc x cot x = - \frac{c o s x}{s i n ^{2} x} \to - \infty

. As

csc x

rockets to

+ \infty

near

0^{+}

, its slope is steeply negative just before — a huge downward drop into the asymptote from the right's mirror side.

If a function is a constant like

y = tan (π /4) = 1

, what is its derivative?

0

. Once the input is a fixed number, the output is a constant, and constants have zero slope — no trig rule fires at all.

Recall One-line summary of every trap

Radians only · chain rule needs the inside · "co" ⇒ minus · derivative rules inherit the parent's domain gaps · constants differentiate to $0$ .

Connections

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