Intuition The big pictureThe trig functions all "rotate into each other" when you differentiate. There are really only two facts you must earn from scratch :
d d x sin x = cos x d d x cos x = − sin x \frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x d x d sin x = cos x d x d cos x = − sin x tan , cot , sec , csc \tan, \cot, \sec, \csc tan , cot , sec , csc quotient rule , because every other trig function is a ratio of sin \sin sin cos \cos cos
WHY: Trig functions model anything that oscillates — springs, waves, AC current, circular motion. The derivative tells you the rate of that oscillation (velocity of a swinging pendulum, slope of a wave). If you can't differentiate sin \sin sin cos \cos cos
WHAT: A rule that takes any of the six trig functions and returns its instantaneous slope.
HOW: Build sin ′ \sin' sin ′ cos ′ \cos' cos ′ derive the other four with the quotient rule. Never memorize a table you can't rebuild.
Why the first is true (geometric squeeze): For a small angle h h h h h h sin h \sin h sin h tan h \tan h tan h sin h ≤ h ≤ tan h . \sin h \le h \le \tan h. sin h ≤ h ≤ tan h . sin h \sin h sin h cos h ≤ sin h h ≤ 1. \cos h \le \frac{\sin h}{h} \le 1. cos h ≤ h s i n h ≤ 1. h → 0 h\to 0 h → 0 cos h → 1 \cos h \to 1 cos h → 1 sin h h → 1 \frac{\sin h}{h}\to 1 h s i n h → 1
Why the second is true (algebra trick): Multiply by the conjugate:
cos h − 1 h ⋅ cos h + 1 cos h + 1 = cos 2 h − 1 h ( cos h + 1 ) = − sin 2 h h ( cos h + 1 ) = − sin h h ⋅ sin h cos h + 1 . \frac{\cos h - 1}{h}\cdot\frac{\cos h + 1}{\cos h + 1} = \frac{\cos^2 h - 1}{h(\cos h+1)} = \frac{-\sin^2 h}{h(\cos h + 1)} = -\frac{\sin h}{h}\cdot\frac{\sin h}{\cos h + 1}. h c o s h − 1 ⋅ c o s h + 1 c o s h + 1 = h ( c o s h + 1 ) c o s 2 h − 1 = h ( c o s h + 1 ) − s i n 2 h = − h s i n h ⋅ c o s h + 1 s i n h . h → 0 h\to 0 h → 0 sin h h → 1 \frac{\sin h}{h}\to 1 h s i n h → 1 sin h cos h + 1 → 0 2 = 0 \frac{\sin h}{\cos h+1}\to \frac{0}{2}=0 c o s h + 1 s i n h → 2 0 = 0 → 1 ⋅ 0 = 0 \to 1\cdot 0 = 0 → 1 ⋅ 0 = 0
Common mistake "Radians vs degrees" — the silent killerThe limit sin h h = 1 \frac{\sin h}{h}=1 h s i n h = 1 only works in radians . In degrees, sin h ≈ π 180 h \sin h \approx \frac{\pi}{180}h sin h ≈ 180 π h d d x sin x ∘ = π 180 cos x ∘ \frac{d}{dx}\sin x^\circ = \frac{\pi}{180}\cos x^\circ d x d sin x ∘ = 180 π cos x ∘ Why the mistake feels right: calculators happily take degrees, so students assume calculus does too.
Fix: In calculus, angles are always radians unless told otherwise.
d d x sin x = lim h → 0 sin ( x + h ) − sin x h \frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h} d x d sin x = lim h → 0 h s i n ( x + h ) − s i n x
Why this step? This is the definition of derivative; nothing trig yet.
Use the angle-addition identity sin ( x + h ) = sin x cos h + cos x sin h \sin(x+h)=\sin x\cos h + \cos x\sin h sin ( x + h ) = sin x cos h + cos x sin h = lim h → 0 sin x cos h + cos x sin h − sin x h =\lim_{h\to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = lim h → 0 h s i n x c o s h + c o s x s i n h − s i n x = lim h → 0 [ sin x ⋅ cos h − 1 h ⏟ → 0 + cos x ⋅ sin h h ⏟ → 1 ] =\lim_{h\to 0}\left[\sin x\cdot\underbrace{\frac{\cos h - 1}{h}}_{\to 0} + \cos x\cdot\underbrace{\frac{\sin h}{h}}_{\to 1}\right] = lim h → 0 sin x ⋅ → 0 h cos h − 1 + cos x ⋅ → 1 h sin h
Why this step? Group the sin x \sin x sin x cos x \cos x cos x
= sin x ⋅ 0 + cos x ⋅ 1 = cos x =\sin x\cdot 0 + \cos x\cdot 1 = \boxed{\cos x} = sin x ⋅ 0 + cos x ⋅ 1 = cos x
Using cos ( x + h ) = cos x cos h − sin x sin h \cos(x+h)=\cos x\cos h - \sin x\sin h cos ( x + h ) = cos x cos h − sin x sin h d d x cos x = lim h → 0 [ cos x ⋅ cos h − 1 h − sin x ⋅ sin h h ] = cos x ⋅ 0 − sin x ⋅ 1 = − sin x \frac{d}{dx}\cos x = \lim_{h\to 0}\left[\cos x\cdot\frac{\cos h-1}{h} - \sin x\cdot\frac{\sin h}{h}\right] = \cos x\cdot 0 - \sin x\cdot 1 = \boxed{-\sin x} d x d cos x = lim h → 0 [ cos x ⋅ h c o s h − 1 − sin x ⋅ h s i n h ] = cos x ⋅ 0 − sin x ⋅ 1 = − sin x
The minus sign is the only thing to remember: differentiating cos \cos cos
Recall the quotient rule: ( u v ) ′ = u ′ v − u v ′ v 2 \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′
d d x tan x = cos x ⋅ cos x − sin x ⋅ ( − sin x ) cos 2 x = cos 2 x + sin 2 x cos 2 x = 1 cos 2 x = sec 2 x \frac{d}{dx}\tan x = \frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \boxed{\sec^2 x} d x d tan x = c o s 2 x c o s x ⋅ c o s x − s i n x ⋅ ( − s i n x ) = c o s 2 x c o s 2 x + s i n 2 x = c o s 2 x 1 = sec 2 x Why this step? Numerator collapses by the Pythagorean identity sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1
d d x cot x = − sin x ⋅ sin x − cos x ⋅ cos x sin 2 x = − ( sin 2 x + cos 2 x ) sin 2 x = − csc 2 x \frac{d}{dx}\cot x = \frac{-\sin x\cdot\sin x - \cos x\cdot\cos x}{\sin^2 x} = \frac{-(\sin^2 x+\cos^2 x)}{\sin^2 x} = \boxed{-\csc^2 x} d x d cot x = s i n 2 x − s i n x ⋅ s i n x − c o s x ⋅ c o s x = s i n 2 x − ( s i n 2 x + c o s 2 x ) = − csc 2 x
d d x sec x = 0 ⋅ cos x − 1 ⋅ ( − sin x ) cos 2 x = sin x cos 2 x = 1 cos x ⋅ sin x cos x = sec x tan x \frac{d}{dx}\sec x = \frac{0\cdot\cos x - 1\cdot(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \boxed{\sec x\tan x} d x d sec x = c o s 2 x 0 ⋅ c o s x − 1 ⋅ ( − s i n x ) = c o s 2 x s i n x = c o s x 1 ⋅ c o s x s i n x = sec x tan x
d d x csc x = 0 ⋅ sin x − 1 ⋅ cos x sin 2 x = − cos x sin 2 x = − csc x cot x \frac{d}{dx}\csc x = \frac{0\cdot\sin x - 1\cdot\cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = \boxed{-\csc x\cot x} d x d csc x = s i n 2 x 0 ⋅ s i n x − 1 ⋅ c o s x = s i n 2 x − c o s x = − csc x cot x
Intuition The "co-" pattern (dual coding)Every function starting with "co" (cos , cot , csc \cos, \cot, \csc cos , cot , csc minus sign . And each "co"-derivative is the mirror of its partner. Pair them: ( sin , cos ) (\sin,\cos) ( sin , cos ) ( tan , cot ) (\tan,\cot) ( tan , cot ) ( sec , csc ) (\sec,\csc) ( sec , csc ) sin ↔ cos \sin\leftrightarrow\cos sin ↔ cos
Worked example 1) Differentiatef ( x ) = x 2 sin x f(x) = x^2\sin x f ( x ) = x 2 sin x Product rule ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ u = x 2 , v = sin x u=x^2,\ v=\sin x u = x 2 , v = sin x f ′ ( x ) = 2 x sin x + x 2 cos x f'(x) = 2x\sin x + x^2\cos x f ′ ( x ) = 2 x sin x + x 2 cos x Why this step? Two functions multiplied ⇒ \Rightarrow ⇒ v ′ = cos x v'=\cos x v ′ = cos x
Worked example 2) Differentiateg ( x ) = sec ( 3 x ) g(x) = \sec(3x) g ( x ) = sec ( 3 x ) Chain rule: outer sec u \sec u sec u sec u tan u \sec u\tan u sec u tan u u = 3 x u=3x u = 3 x 3 3 3 g ′ ( x ) = sec ( 3 x ) tan ( 3 x ) ⋅ 3 = 3 sec ( 3 x ) tan ( 3 x ) g'(x) = \sec(3x)\tan(3x)\cdot 3 = 3\sec(3x)\tan(3x) g ′ ( x ) = sec ( 3 x ) tan ( 3 x ) ⋅ 3 = 3 sec ( 3 x ) tan ( 3 x ) Why this step? The inner-function derivative 3 3 3
Worked example 3) Differentiateh ( x ) = tan x x h(x) = \dfrac{\tan x}{x} h ( x ) = x tan x Quotient rule , u = tan x u=\tan x u = tan x u ′ = sec 2 x u'=\sec^2 x u ′ = sec 2 x v = x v=x v = x v ′ = 1 v'=1 v ′ = 1 h ′ ( x ) = sec 2 x ⋅ x − tan x ⋅ 1 x 2 = x sec 2 x − tan x x 2 h'(x) = \frac{\sec^2 x\cdot x - \tan x\cdot 1}{x^2} = \frac{x\sec^2 x - \tan x}{x^2} h ′ ( x ) = x 2 s e c 2 x ⋅ x − t a n x ⋅ 1 = x 2 x s e c 2 x − t a n x
Worked example 4) Slope ofy = sin x y=\sin x y = sin x x = π / 3 x=\pi/3 x = π /3 y ′ = cos x y' = \cos x y ′ = cos x = cos ( π / 3 ) = 1 2 =\cos(\pi/3) = \tfrac12 = cos ( π /3 ) = 2 1
Common mistake Forgetting the chain rule "inside derivative"Writing d d x sin ( x 2 ) = cos ( x 2 ) \frac{d}{dx}\sin(x^2) = \cos(x^2) d x d sin ( x 2 ) = cos ( x 2 ) wrong .
Why it feels right: the table says sin ′ = cos \sin'=\cos sin ′ = cos Fix: the inside x 2 x^2 x 2 cos ( x 2 ) ⋅ 2 x \cos(x^2)\cdot 2x cos ( x 2 ) ⋅ 2 x
sec ′ \sec' sec ′ csc ′ \csc' csc ′ sec ′ = + sec tan \sec' = +\sec\tan sec ′ = + sec tan csc ′ = − csc cot \csc' = -\csc\cot csc ′ = − csc cot Fix: "co" ⇒ \Rightarrow ⇒ sec \sec sec
Recall Feynman: explain to a 12-year-oldImagine spinning a wheel. The shadow of a dot on it bobs up and down — that's the sin \sin sin different but related wave, cos \cos cos sin \sin sin cos \cos cos cos \cos cos − sin -\sin − sin sin \sin sin cos \cos cos their speeds we use the division rule for slopes. That's the whole trick — no magic, just two facts and a division rule.
Mnemonic Remember the whole table"All Co's are Negative." Then pair them:
sin → cos \sin\to\cos sin → cos tan → sec 2 \tan\to\sec^2 tan → sec 2 sec → sec tan \sec\to\sec\tan sec → sec tan cos → − sin \cos\to-\sin cos → − sin cot → − csc 2 \cot\to-\csc^2 cot → − csc 2 csc → − csc cot \csc\to-\csc\cot csc → − csc cot
What is d d x sin x \frac{d}{dx}\sin x d x d sin x What is d d x cos x \frac{d}{dx}\cos x d x d cos x What is d d x tan x \frac{d}{dx}\tan x d x d tan x What is d d x cot x \frac{d}{dx}\cot x d x d cot x What is d d x sec x \frac{d}{dx}\sec x d x d sec x sec x tan x \sec x\tan x sec x tan x What is d d x csc x \frac{d}{dx}\csc x d x d csc x − csc x cot x -\csc x\cot x − csc x cot x Which two limits underpin all trig derivatives? lim h → 0 sin h h = 1 \lim_{h\to0}\frac{\sin h}{h}=1 lim h → 0 h s i n h = 1 and
lim h → 0 cos h − 1 h = 0 \lim_{h\to0}\frac{\cos h-1}{h}=0 lim h → 0 h c o s h − 1 = 0 Why must angles be in radians? Because
sin h h → 1 \frac{\sin h}{h}\to1 h s i n h → 1 only holds in radians; degrees insert a factor
π 180 \frac{\pi}{180} 180 π How do you get tan ′ = sec 2 x \tan' = \sec^2 x tan ′ = sec 2 x Quotient rule on
sin / cos \sin/\cos sin / cos ; numerator becomes
cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 , over
cos 2 \cos^2 cos 2 Which trig derivatives carry a minus sign? The "co" ones:
cos , cot , csc \cos, \cot, \csc cos , cot , csc d d x sin ( x 2 ) = ? \frac{d}{dx}\sin(x^2) = ? d x d sin ( x 2 ) = ? 2 x cos ( x 2 ) 2x\cos(x^2) 2 x cos ( x 2 ) (chain rule)
d d x sec ( 3 x ) = ? \frac{d}{dx}\sec(3x) = ? d x d sec ( 3 x ) = ? 3 sec ( 3 x ) tan ( 3 x ) 3\sec(3x)\tan(3x) 3 sec ( 3 x ) tan ( 3 x )
Limit definition of derivative
d/dx csc x = -csc x cot x
Intuition Hinglish mein samjhoDekho, asli baat sirf do facts hai: d d x sin x = cos x \frac{d}{dx}\sin x = \cos x d x d sin x = cos x d d x cos x = − sin x \frac{d}{dx}\cos x = -\sin x d x d cos x = − sin x sin h h → 1 \frac{\sin h}{h}\to 1 h s i n h → 1 cos h − 1 h → 0 \frac{\cos h-1}{h}\to 0 h c o s h − 1 → 0 radians mein hi sach hai, degrees mein nahi.
Baaki chaar functions — tan , cot , sec , csc \tan, \cot, \sec, \csc tan , cot , sec , csc sin \sin sin cos \cos cos quotient rule lagao aur ho gaya. Jaise tan x = sin x / cos x \tan x = \sin x/\cos x tan x = sin x / cos x cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 sec 2 x \sec^2 x sec 2 x
Ek mast trick: jis function mein "co" hai (cos , cot , csc \cos, \cot, \csc cos , cot , csc minus sign aata hai. Aur partner pairs banao — ( sin , cos ) (\sin,\cos) ( sin , cos ) ( tan , cot ) (\tan,\cot) ( tan , cot ) ( sec , csc ) (\sec,\csc) ( sec , csc )
Sabse common galti: chain rule bhoolna. d d x sin ( x 2 ) \frac{d}{dx}\sin(x^2) d x d sin ( x 2 ) cos ( x 2 ) \cos(x^2) cos ( x 2 ) x 2 x^2 x 2 2 x 2x 2 x 2 x cos ( x 2 ) 2x\cos(x^2) 2 x cos ( x 2 )
