Derivatives of all six trig functions: Worked examples

4.1.18 · D3 · Maths › Calculus I — Limits & Derivatives › Derivatives of all six trig functions

Scenario matrix

Har trig-derivative problem inme se kisi ek box mein rehta hai. Neeche ke examples mein uss box ka label hai jo woh fill karta hai.

Cell	Case class	Kya tricky hai	Example
A	Single function × single rule	sahi rule chunna	Ex 1 (product)
B	Chain rule, inner ≠ $x$	andar ka derivative mat bhoolo	Ex 2
C	Nested chain (chain ke andar chain)	saari inner derivatives multiply karo	Ex 3
D	Do trig functions ka quotient	"co"-functions ke signs	Ex 4
E	Sign / quadrant evaluation	slope $+$ , $-$ , ya $0$ ho sakta hai	Ex 5
F	Degenerate / undefined input	function blow up karta hai; slope undefined	Ex 6
G	Limiting behaviour ( $x \to 0$ )	foundation limit phir se use hoti hai	Ex 7
H	Real-world oscillation (word problem)	physics → derivative mein translate karo	Ex 8
I	Exam twist (clean form mein simplify karo)	identity poori gandagi saaf kar deti hai	Ex 9

Ab hum A → I walk karte hain taaki koi box khali na rahe.

Worked examples

Worked example Ex 1 — Cell A: ek power aur ek trig function ka product

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

Forecast: Do cheezein multiply hain → hume do terms milenge. Ek $cos$ rakhega, doosra $x^{3}$ rakhega aur $cos$ ko... minus waali cheez mein badlega.

Pieces name karo: $u = x^{3}$ , $v = cos x$ . Yeh step kyun? Product Rule ke liye dono factors ko pehle name karna padta hai before touch karo.
Har ek differentiate karo: $u^{'} = 3 x^{2}$ , aur $v^{'} = - sin x$ . Yeh step kyun? $\frac{d}{d x} cos x = - sin x$ parent table se — "co" minus sign yahan aata hai.
$(uv)^{'} = u^{'} v + u v^{'}$ apply karo: $f^{'} (x) = 3 x^{2} cos x + x^{3} (- sin x) = 3 x^{2} cos x - x^{3} sin x .$ Yeh step kyun? Product rule kehta hai "pehle differentiate karo, doosra rakhho; phir pehla rakhho, doosra differentiate karo" — tum sirf dono derivatives multiply nahi kar sakte, isliye dono terms mandatory hain.

Verify: $x = 0$ par: $f^{'} (0) = 0 - 0 = 0$ . Sanity: $x = 0$ ke paas, $f (x) = x^{3} cos x \approx x^{3}$ , jiska slope $0$ par $0$ hai. ✅

Worked example Ex 2 — Cell B: linear inside ke saath chain rule

$g (x) = sin (5 x)$ differentiate karo.

Forecast: Table deta hai $sin^{'} = cos$ , lekin inside $5 x$ hai, $x$ nahi. $5$ ka ek factor zaroor niklega.

Outer aur inner identify karo: outer $= sin (□)$ , inner $= u = 5 x$ . Yeh step kyun? Chain Rule kehta hai outer-at-inner differentiate karo, phir inner ka derivative multiply karo.
Inner par outer derivative: $cos (5 x)$ . Yeh step kyun? Chain rule pehle outer function differentiate karta hai jabki inner block ko untouched chhod deta hai — toh $sin (□)$ ban jaata hai $cos (□)$ with $□ = 5 x$ abhi bhi andar baitha hai.
Inner derivative: $\frac{d}{d x} (5 x) = 5$ . Yeh step kyun? Chain rule phir poochta hai "inside kitni tez change hoti hai?" — inside $5 x$ , $x$ se $5$ guna tez change hoti hai, aur woh rate poore result ko scale karna chahiye.
Multiply karo: $g^{'} (x) = 5 cos (5 x)$ . Yeh step kyun? $5$ bhoolna chain-rule ki #1 error hai jo parent note mein flag ki gayi hai.

Verify: $x = 0$ par: $g^{'} (0) = 5 cos 0 = 5$ . Sanity: chote $x$ ke liye $sin (5 x) \approx 5 x$ , slope $5$ . ✅

Worked example Ex 3 — Cell C: nested chain (chain ke andar chain)

$p (x) = tan (cos x)$ differentiate karo.

Forecast: Do layers. Outer hai $tan$ , inner hai $cos x$ . Hum do inner derivatives multiply karenge.

Layers: outermost $tan (□)$ , iska argument hai $cos x$ . Yeh step kyun? Bahar se andar ki taraf peel karo — Chain Rule do baar apply hogi.
Outer ko inner par differentiate karo: $\frac{d}{d x} tan (□) = sec^{2} (□)$ , toh $sec^{2} (cos x)$ . Yeh step kyun? Chain rule sabse bahari layer ko pehle differentiate karta hai jabki iska argument freeze karta hai — $tan^{'} = sec^{2}$ parent table se, still-intact inner block $cos x$ par evaluate kiya.
Inner $cos x$ ke derivative se multiply karo, jo hai $- sin x$ . Yeh step kyun? Chain rule phir us inner block ki rate of change se multiply karta hai; $\frac{d}{d x} cos x = - sin x$ , aur "co" minus sign carry through hoti hai.
Combine karo: $p^{'} (x) = sec^{2} (cos x) \cdot (- sin x) = - sin x sec^{2} (cos x) .$ Yeh step kyun? Chain rule ka answer outer-at-inner derivative aur inner derivative ka product hai — koi bhi factor drop karo toh galat slope milega.

Verify: $x = 0$ par: $p^{'} (0) = - sin 0 \cdot sec^{2} (1) = 0$ . Sanity: $sin 0 = 0$ ise zero kar deta hai — $x = 0$ par horizontal tangent. ✅

Worked example Ex 4 — Cell D: do trig functions ka quotient

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

Forecast: Quotient rule → messy numerator, lekin Pythagorean identity ise saaf kar degi.

Name karo: $u = sin x$ ( $u^{'} = cos x$ ), $v = 1 + cos x$ ( $v^{'} = - sin x$ ). Yeh step kyun? Quotient Rule ke liye dono parts aur unke derivatives chahiye.
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ apply karo: $q^{'} (x) = \frac{c o s x ( 1 + c o s x ) - s i n x ( - s i n x )}{( 1 + c o s x ) ^{2}} .$ Yeh step kyun? Do functions ke ratio ke liye quotient rule chahiye — "top ka derivative times bottom, minus top times bottom ka derivative, sab over bottom squared"; top aur bottom alag-alag differentiate nahi kar sakte.
Numerator expand karo: $cos x + cos^{2} x + sin^{2} x = cos x + 1$ . Yeh step kyun? $cos^{2} x + sin^{2} x = 1$ do terms ko ek mein collapse karta hai.
Shared factor cancel karo: $q^{'} (x) = \frac{1 + c o s x}{( 1 + c o s x ) ^{2}} = \frac{1}{1 + c o s x} .$ Yeh step kyun? Numerator $1 + cos x$ uss factor ki ek copy hai jo squared denominator mein baitha hai, toh ek copy cancel ho jaati hai — yahi ugly fraction ko clean closed form mein badalta hai.

Verify: $x = 0$ par: $q^{'} (0) = \frac{1}{2}$ . Sanity: numerator shuru mein $cos 0 (2) - 0 = 2$ tha, denominator $2^{2} = 4$ , ratio $\frac{1}{2}$ . ✅

Worked example Ex 5 — Cell E: har quadrant mein slope ka sign

$y = sin x$ ke liye, $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ par slope nikalo (circle ke har quadrant se ek angle). Figure dekho.

Forecast: Slope $= cos x$ . Cosine QI aur QIV mein $+$ hai, QII aur QIII mein $-$ . Toh hume $+, -, -, +$ expect karna chahiye.

Slope function: $y^{'} = cos x$ . Yeh step kyun? $sin$ ka slope hi $cos$ hai — parent Step 1.
Har angle plug karo (saari values $\pm \frac{2}{2}$ hain):
- $cos \frac{π}{4} = + \frac{2}{2}$ (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) Yeh step kyun? Slope ka sign bilkul wohi hai jo us point par $cos$ ka sign hai, aur $cos$ sign flip karta hai jab hum QII aur QIII mein enter karte hain — yahan naive " $sin$ hamesha rise karta hai" waala guess fail hota hai.
Geometry padho (figure mein amber tangent lines dekho): steep up, steep down, steep down, steep up. Yeh step kyun? Tangent line ki tilt hi derivative hai jo visible hai — har computed number ko uski drawn slope ke saath pair karna confirm karta hai ki $+$ matlab "up-tilt" aur $-$ matlab "down-tilt", toh algebra aur picture agree karte hain.

Verify: $\frac{2}{2} \approx 0.707$ ; signs hain $+, -, -, +$ jaise forecast kiya. Slope zero hota hai exactly crest $x = \frac{π}{2}$ aur trough $x = \frac{3 π}{2}$ par, jahaan $cos = 0$ hai. ✅

Worked example Ex 6 — Cell F: degenerate input jahaan function explode karta hai

$\frac{d}{d x} tan x = sec^{2} x$ undefined kahan hai, aur graph wahan kya karta hai?

Forecast: $tan x$ ke vertical walls hain $x = \frac{π}{2} + nπ$ par. Iska slope wahan $+ \infty$ tak blow up hona chahiye, kabhi negative nahi.

Yaad karo $sec^{2} x = \frac{1}{cos ^{2} x}$ . Yeh step kyun? Derivative cleanly $sec^{2}$ likha hai, lekin iske danger points $cos$ ke andar chhupe hain.
$sec^{2} x$ undefined hai exactly jab $cos x = 0$ , yaani $x = \frac{π}{2} + nπ$ integer $n$ ke liye. Yeh step kyun? Ek fraction undefined tab hoti hai jab denominator zero ho, aur $cos^{2} x = 0$ precisely usi angle par hota hai — toh wahi exact spots hain jahaan slope exist karna band ho jaata hai.
Kyunki $cos^{2} x \geq 0$ hamesha hota hai, $sec^{2} x = \frac{1}{c o s ^{2} x} \geq 1$ har jagah jahan exist karta hai — $tan$ ka slope kabhi $1$ se neeche nahi jaata. Yeh step kyun? Yeh humein batata hai ki $tan$ hamesha rise karta hai (kabhi flat nahi hota, kabhi dip nahi karta).
Jab $x \to \frac{π}{2}^{-}$ , $cos x \to 0^{+}$ , toh $sec^{2} x \to + \infty$ : ek vertical tangent. Yeh step kyun? Jaise denominator $cos^{2} x$ zero ki taraf shrink hota hai, reciprocal $\frac{1}{c o s ^{2} x}$ without bound grow karta hai — woh unbounded positive slope hi $tan$ ki vertical wall (asymptote) draw karta hai.

Verify: "Flattest" point $x = 0$ par: $sec^{2} 0 = \frac{1}{1} = 1$ , minimum slope. $x = \frac{π}{3}$ par: $cos = \frac{1}{2}$ , toh $sec^{2} = \frac{1}{1/4} = 4 \geq 1$ . ✅

Worked example Ex 7 — Cell G:

$x \to 0$ par limiting behaviour $x \to 0 lim \frac{tan x}{x}$ derivatives use karke evaluate karo.

Forecast: $tan x$ aur $x$ dono $0$ ki taraf jaate hain, toh yeh $\frac{0}{0}$ form hai. Kyunki zero ke paas $tan x \approx x$ , answer $1$ hona chahiye.

Pehchano $lim_{x \to 0} \frac{tan x - tan 0}{x - 0}$ exactly $0$ par $tan$ ka derivative hai. Yeh step kyun? Yeh Limit definition of the derivative ulta padha gaya hai — $0$ par $tan$ ka difference quotient (kyunki $tan 0 = 0$ ).
Woh derivative hai $tan^{'} (0) = sec^{2} 0$ . Yeh step kyun? Jab humne pehchaan liya ki limit ek derivative hai, toh hum messy limit skip karke seedha known slope $tan^{'} = sec^{2}$ quote kar sakte hain parent table se, $0$ par evaluate karke.
$sec^{2} 0 = \frac{1}{cos ^{2} 0} = \frac{1}{1} = 1$ . Yeh step kyun? $cos 0 = 1$ , toh reciprocal-square $1$ hai — yeh limit ki numerical value pin karta hai aur forecast confirm karta hai.

Verify: Numerically $tan (0.001) /0.001 \approx 1.0000003 \to 1$ . Parent ki foundation-limit spirit se match karta hai ( $sin h / h \to 1$ ). ✅

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

Ek spring par mass ki height $s (t) = 4 sin (2 t)$ cm hai, jahaan $t$ seconds mein hai. Iska velocity nikalo, aur maximum speed.

Forecast: Velocity derivative hai. Chain rule → $2$ ka factor aata hai, toh velocity ka amplitude hai $4 \times 2 = 8$ . Max speed = $8$ cm/s. Figure dekho. Yeh Simple Harmonic Motion hai.

Velocity $v (t) = s^{'} (t)$ . $4 sin (2 t)$ differentiate karo: outer $sin$ gives $cos$ , inner $2 t$ gives $2$ . Yeh step kyun? Velocity = position ki rate of change; inner $2 t$ ke liye Chain Rule.
$v (t) = 4 cos (2 t) \cdot 2 = 8 cos (2 t)$ cm/s. Yeh step kyun? Constant $4$ untouched ride karta hai, jabki chain rule inner derivative $2$ se multiply karta hai — isliye velocity amplitude ( $8$ ) position amplitude ( $4$ ) se bada hai: tez wiggling matlab badi speeds.
$cos$ ki range $[- 1, 1]$ mein hai, toh $v$ ki range $[- 8, 8]$ mein hai. Maximum speed (velocity ka size) $8$ cm/s hai. Yeh step kyun? Speed hai $∣ v ∣$ ; $∣ cos ∣$ ka peak $1$ hai.
Max speed kahan hoti hai? Jab $cos (2 t) = \pm 1$ , yaani $2 t = nπ$ , yaani swing ke middle mein $s = 0$ par — exactly jahaan position graph zero cross karta hai (figure mein green dots). Yeh step kyun? $∣ cos ∣$ apna maximum $1$ exactly tab hit karta hai jab iska argument $π$ ka multiple ho; usi instant par $sin (2 t) = 0$ toh $s = 0$ — yeh physics fact reveal karta hai ki mass centre se whip through karte waqt sabse tez hota hai.

Verify: $t = 0$ par: $s (0) = 0$ (centre), $v (0) = 8 cos 0 = 8$ cm/s — centre par sabse tez, jaise physics demand karta hai. $t = \frac{π}{4}$ par: $s = 4 sin (\frac{π}{2}) = 4$ (top), $v = 8 cos (\frac{π}{2}) = 0$ — top par momentarily still. ✅

Worked example Ex 9 — Cell I: exam twist jo clean form mein simplify hota hai

$r (x) = sec x (sec x + tan x)$ differentiate karo aur simplify karo.

Forecast: Lagta hai product rule chahiye aur ugly ho jaayega — lekin $sec$ aur $tan$ ke derivatives mein shared pieces hain, toh collapse ke liye watch karo.

Pehle expand karo: $r (x) = sec^{2} x + sec x tan x$ . Yeh step kyun? Expand karne se product rule bilkul avoid ho jaata hai — term by term differentiate karo.
$sec^{2} x$ differentiate karo: $(sec x)^{2}$ treat karo, Chain Rule deta hai $2 sec x \cdot (sec x tan x) = 2 sec^{2} x tan x$ . Yeh step kyun? $(sec x)^{2}$ ek outer square hai $sec x$ ke upar wrapped; chain rule deta hai $2 \cdot (inside) \cdot (inside’s derivative)$ , aur $sec^{'} = sec x tan x$ parent table se.
$sec x tan x$ ko Product Rule se differentiate karo: $(sec x tan x)^{'} = sec x tan x \cdot tan x + sec x \cdot sec^{2} x = sec x tan^{2} x + sec^{3} x$ . Yeh step kyun? Yeh term do functions multiply kiya hua hai, toh product rule apply hoti hai: $sec$ differentiate karo (deta hai $sec tan$ ) times $tan$ , plus $sec$ times $tan$ ka derivative (deta hai $sec^{2}$ ).
Add karo: $r^{'} (x) = 2 sec^{2} x tan x + sec x tan^{2} x + sec^{3} x .$ $sec x$ factor out karo: $r^{'} (x) = sec x (2 sec x tan x + tan^{2} x + sec^{2} x)$ . Yeh step kyun? Har term mein kam se kam ek $sec x$ hai, toh bahar kheenchna valid hai aur woh tidy "clean form" deta hai jo ek exam chahta hai — shared factor wahi collapse hai jo humne forecast kiya tha.

Verify: $x = 0$ par: $sec 0 = 1$ , $tan 0 = 0$ , toh $r^{'} (0) = 1 (0 + 0 + 1) = 1$ . Original expanded form ke derivative se cross-check $0$ par: $2 sec^{2} 0 tan 0 + (sec 0 tan^{2} 0 + sec^{3} 0) = 0 + (0 + 1) = 1$ . ✅

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

Neeche wali picture wohi "which rule do I use?" flow hai, ek diagram ki tarah draw kiya gaya taaki tum dekh sako ki tum kaunse branch par land karte ho: ek problem padho, amber diamond mein ek sawaal poochho, aur needed rule ki taraf arrow follow karo.

Figure — Derivatives of all six trig functions