Higher-order derivatives — notation, physical meaning: Worked examples

4.1.24 · D3 · Maths › Calculus I — Limits & Derivatives › Higher-order derivatives — notation, physical meaning

Yeh page higher-order derivatives ka exhaustive drill room hai. Koi bhi example touch karne se pehle, hum ek scenario matrix banate hain: ek checklist jisme har tarah ki situation listed hai jo higher-order derivatives mein aa sakti hai. Phir neeche har example pe us cell ka stamp lagaya gaya hai jise woh fill karta hai, taaki end tak tumne sab dekh liya ho.

Scenario matrix

Cell	Case class	Kya cheez isse alag banati hai	Kisme fill hota hai
A	Polynomial — saare orders jab tak woh khatam na ho jaaye	Har derivative degree kam karta hai; eventually $0$	Ex 1
B	Product / zyada complex algebra	Power rule akela kaafi nahi; degree abhi bhi finite hai	Ex 2
C	Trig — kabhi khatam nahi hota, cycles karta hai	Derivatives period 4 ke saath repeat hote hain	Ex 3
D	Kinematics word problem (units!)	position → velocity → acceleration metres/seconds ke saath	Ex 4
E	Sign of $f^{''}$ saare regions mein (concave up / down / zero)	Har sign case + genuine inflection	Ex 5
E′	False candidate: $f^{''} = 0$ ek point pe bina sign change ke	$f^{''}$ ka zero jo inflection point nahi hai	Ex 6
F	Degenerate input: linear aur constant functions	$f^{''} = 0$ everywhere — koi curvature nahi	Ex 7
G	Limiting / asymptotic behaviour of high orders	Kya hota hai jab $n \to \infty$ (exponential vs. polynomial)	Ex 8
H	Exam twist: notation trap $\frac{d ^{2} y}{d x ^{2}}$ vs $(\frac{d y}{d x})^{2}$	Symbol ko sahi se padhna	Ex 9

Har cell A–H (aur E′ bhi) neeche cover ki gayi hai. Signs sab directions mein handle kiye gaye hain (positive, negative, zero), including woh crucial trap jahan $f^{''} = 0$ ho lekin concavity flip na kare; degenerate aur limiting inputs ko bhi apne examples milte hain taaki tum kabhi koi unseen scenario na dekho.

Cell A — Polynomial, saare orders

Worked example Example 1 —

$x^{5}$ ko differentiate karo jab tak kuch bache na $f (x) = x^{5}$ ke liye $f^{'} (x)$ se $f^{(6)} (x)$ tak nikalo.

Forecast: guess karo — kitne derivatives ke baad $x^{5}$ number $0$ ban jaata hai? Aage padhne se pehle apna guess likho.

$f^{'} (x) = 5 x^{4}$ . Yeh step kyun? Power rule: exponent $5$ ko multiplier ke roop mein neeche laao, exponent ek kam karo.
$f^{''} (x) = 20 x^{3}$ . Kyun? $5 x^{4}$ ko differentiate karo: $5 \cdot 4 = 20$ , exponent $4 \to 3$ .
$f^{'''} (x) = 60 x^{2}$ . Kyun? $20 \cdot 3 = 60$ , exponent $3 \to 2$ .
$f^{(4)} (x) = 120 x$ . Kyun? $60 \cdot 2 = 120$ , exponent $2 \to 1$ .
$f^{(5)} (x) = 120$ . Kyun? $120 \cdot 1 = 120$ , aur $x^{1} \to x^{0} = 1$ , ek pure number $120$ bachta hai.
$f^{(6)} (x) = 0$ . Kyun? Kisi bhi constant ka derivative $0$ hota hai — ek flat line ki slope.

Verify: $f^{(5)} (x) = 120 = 5!$ (five factorial $= 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ ). Ek degree- $n$ polynomial mein $f^{(n)} = n!$ aur $f^{(n + 1)} = 0$ hota hai — yahan $n = 5$ , toh yeh 6th derivative pe khatam hota hai. ✓

Cell B — Zyada complex algebra, phir bhi finite

Worked example Example 2 — pehle expand kiya hua ek product

Maano $g (x) = x^{2} (x - 3) = x^{3} - 3 x^{2}$ . $g^{''} (x)$ aur $g^{'''} (x)$ nikalo, aur check karo ki $x = 1$ ek genuine inflection point hai ya nahi.

Forecast: yeh degree 3 hai — kitne derivatives mein yeh khatam ho jaata hai?

Expand karo: $g (x) = x^{3} - 3 x^{2}$ . Yeh step kyun? Expand karne se product ek sum mein badal jaata hai, toh hum term-by-term power rule use kar sakte hain product rule ki jagah — yahan simpler aur kam error-prone.
$g^{'} (x) = 3 x^{2} - 6 x$ . Kyun? Har term pe power rule: $x^{3} \to 3 x^{2}$ , $- 3 x^{2} \to - 6 x$ .
$g^{''} (x) = 6 x - 6$ . Kyun? Phir differentiate karo: $3 x^{2} \to 6 x$ , $- 6 x \to - 6$ .
$g^{'''} (x) = 6$ . Kyun? $6 x \to 6$ , aur $- 6$ (ek constant) $\to 0$ .
$g^{''} (x) = 6 x - 6 = 0 \Rightarrow x = 1$ solve karo, phir $g^{''}$ ka sign dono taraf test karo. Kyun? $g^{''}$ ka zero sirf ek candidate hai — yeh ek genuine inflection point tabhi hai jab $g^{''}$ actually sign change kare.
- $g^{''} (0) = 6 (0) - 6 = - 6 < 0$ → $1$ ke left mein concave down.
- $g^{''} (2) = 6 (2) - 6 = + 6 > 0$ → $1$ ke right mein concave up. Sign $- \to +$ flip hota hai, toh $x = 1$ ek real inflection point hai.

Verify: $g^{''} (0) = - 6 < 0$ aur $g^{''} (2) = + 6 > 0$ $x = 1$ pe sign change confirm karte hain ✓, toh inflection genuine hai (false candidate nahi). Aur $g^{(4)} = 0$ , confirm karta hai ki ek degree-3 polynomial 4 derivatives ke baad khatam hota hai. ✓

Cell C — Trig, forever cycle karta hai

Worked example Example 3 —

$cos x$ aur period-4 cycle Maano $h (x) = cos x$ . $h^{(1)}$ se $h^{(4)}$ tak nikalo aur $h^{(n)}$ ke liye ek formula dho.

Forecast: $x^{5}$ ke unlike, yeh wala kabhi $0$ nahi banta. Guess karo $h^{(4)}$ $h$ ke comparison mein kya hoga.

$h^{'} (x) = - sin x$ . Yeh step kyun? $\cos$ ka derivative $- sin$ hota hai (jaisa cosine wave girta hai, uski slope negative sine hoti hai).
$h^{''} (x) = - cos x$ . Kyun? $- sin x$ differentiate karo: $sin$ ka derivative $cos$ hai, minus rakhho.
$h^{'''} (x) = sin x$ . Kyun? $- cos x$ differentiate karo: $- (- sin x) = sin x$ .
$h^{(4)} (x) = cos x = h (x)$ . Kyun? $sin x \to cos x$ differentiate karo. 4 steps ke baad hum wapas shuru par aa gaye.
General formula: $h^{(n)} (x) = cos (x + \frac{nπ}{2})$ . Kyun? Har derivative ek cosine ko left mein ek quarter-turn $\frac{π}{2}$ shift karta hai; $n$ baar karne se $\frac{nπ}{2}$ ka shift hota hai.

Verify: $n = 2$ daalo: $cos (x + π) = - cos x$ ✓ (step 2 se match karta hai). $n = 4$ daalo: $cos (x + 2 π) = cos x$ ✓. Cycle ki period 4 hai.

Neeche wala figure $h = cos x$ (white) ko uske pehle teen derivatives $h^{'} = - sin x$ (cyan), $h^{''} = - cos x$ (amber), aur $h^{'''} = sin x$ (dashed cyan) ke saath plot karta hai. $x = 0$ pe vertical dotted white line derivative values $1, 0, - 1, 0$ mark karti hai jo upar use ki gayi hain; har curve pichli wali se ek quarter-turn left shift hai, aur amber arrow dikhata hai ki 4th derivative white $cos x$ curve pe wapas land karta hai — period-4 cycle visible ho gayi.

Figure — Higher-order derivatives — notation, physical meaning

Figure 1 — Cell C: cos x (white solid) aur uske derivatives −sin x (cyan solid), −cos x (amber solid), sin x (cyan dashed) ek blueprint grid pe 0 se 2π tak ek full turn mein plot kiye gaye hain; x=0 pe ek dotted vertical line repeating derivative values 1, 0, −1, 0 read off karti hai, aur ek amber arrow amber curve ke far-right se white cos-x curve tak point karta hai yeh dikhane ke liye ki fourth differentiation wapas start pe return karta hai — woh period-4 cycle jo cosine Taylor series ko power deti hai.

Cell D — Kinematics word problem (units matter karte hain)

Worked example Example 4 — position → velocity → acceleration

Ek drone height $s (t) = t^{3} - 9 t^{2} + 15 t$ (metres, $t$ seconds mein) ke saath utha. Velocity $v (t)$ , acceleration $a (t)$ nikalo, aur woh instant dhundho jab acceleration zero ho.

Forecast: acceleration doosra derivative hai. Guess karo ki kya yeh kisi whole-number time pe zero hota hai.

$v (t) = s^{'} (t) = 3 t^{2} - 18 t + 15$ (m/s). Yeh step kyun? Velocity position ka pehla time-derivative hai — height kitni tezi se change hoti hai. Units: metres per second.
$a (t) = v^{'} (t) = 6 t - 18$ (m/s²). Kyun? Acceleration velocity ka derivative hai = $s$ ka doosra derivative, $\overset{s}{¨}$ likha jaata hai. Units: (m/s) per second $=$ m/s².
$a (t) = 0$ set karo: $6 t - 18 = 0 \Rightarrow t = \frac{18}{6} = 3$ s. Kyun? Zero acceleration woh jagah hai jahan velocity momentarily change karna band kar deti hai (ek velocity extremum).

Verify: $t = 3$ pe: $a = 6 (3) - 18 = 18 - 18 = 0$ ✓. Units flow check karo: metres → m/s (÷ time) → m/s² (÷ time again). Har derivative ek aur second se divide karta hai. ✓

Cell E — Har region mein $f^{''}$ ka sign (genuine inflection)

Worked example Example 5 — saare signs mein concavity

$f (x) = x^{3} - 3 x$ ke liye, $x < 0$ , $x = 0$ , aur $x > 0$ ke liye concavity describe karo, aur inflection point dhundho.

Forecast: curve "frowning" $⌢$ se "smiling" $⌣$ ki taraf kahan switch karta hai?

$f^{'} (x) = 3 x^{2} - 3$ . Yeh step kyun? Term-by-term power rule; $f^{''}$ se pehle $f^{'}$ chahiye.
$f^{''} (x) = 6 x$ . Kyun? $3 x^{2} - 3$ differentiate karo: constant $- 3$ vanish ho jaata hai.
Case $x < 0$ : $f^{''} = 6 x < 0$ → concave down $⌢$ . Kyun? Negative second derivative ka matlab hai slope decrease ho raha hai — curve neeche bend karti hai.
Case $x > 0$ : $f^{''} = 6 x > 0$ → concave up $⌣$ . Kyun? Positive second derivative ka matlab hai slope increase ho rahi hai.
Case $x = 0$ : $f^{''} = 0$ aur yeh around mein sign switch karta hai → inflection point at $(0, 0)$ . Kyun? Ek inflection point exactly wahi hai jahan bend flip hota hai.

Verify: test points: $f^{''} (- 1) = - 6 < 0$ ✓ (down), $f^{''} (1) = 6 > 0$ ✓ (up). Sign genuinely $0$ pe change hota hai, toh $(0, 0)$ ek real inflection point hai. ✓

Neeche wala figure $f = x^{3} - 3 x$ white mein draw karta hai. $x < 0$ wala region cyan shaded hai (wahan $f^{''} = 6 x < 0$ , toh curve concave down hai, $⌢$ ) aur $x > 0$ wala region amber shaded hai (wahan $f^{''} = 6 x > 0$ , concave up, $⌣$ ). Origin pe amber dot inflection point mark karta hai jahan shading — aur bend — switch karta hai.

Figure 2 — Cell E: cubic f = x³ − 3x white mein ek blueprint grid pe draw kiya gaya hai; left half-plane (x<0) cyan shaded hai aur concave-DOWN label kiya gaya hai kyunki f''=6x wahan negative hai, right half-plane (x>0) amber shaded hai aur concave-UP label kiya gaya hai kyunki f'' wahan positive hai, aur origin (0,0) pe ek amber dot with arrow genuine inflection point mark karta hai jahan f''=0 hai aur concavity flip hoti hai.

Cell E′ — False candidate: $f^{''} = 0$ lekin sign change nahi

Worked example Example 6 — woh inflection point jo inflection nahi hai

$f (x) = x^{4}$ ke liye, hum $f^{''} (0) = 0$ paate hain. Kya $x = 0$ ek inflection point hai?

Forecast: ek common exam trap — students $f^{''} = 0$ dekhte hain aur declare kar dete hain inflection. Guess karo ki kya $x^{4}$ really origin pe apni concavity change karta hai.

$f^{'} (x) = 4 x^{3}$ . Yeh step kyun? Power rule; doosre se pehle ek derivative chahiye.
$f^{''} (x) = 12 x^{2}$ . Kyun? $4 x^{3}$ differentiate karo: $4 \cdot 3 = 12$ , exponent $3 \to 2$ .
$f^{''} (x) = 12 x^{2} = 0 \Rightarrow x = 0$ solve karo. Kyun? Yeh sirf ek candidate hai — $f^{''}$ ka zero. Kuch bhi claim karne se pehle hume dono sides pe sign test karna hoga.
- $f^{''} (- 1) = 12 (- 1)^{2} = + 12 > 0$ → left taraf concave up.
- $f^{''} (1) = 12 (1)^{2} = + 12 > 0$ → right taraf concave up.
Sign dono sides pe positive hai — yeh origin pe $0$ touch karta hai lekin flip nahi karta. Yeh matter kyun karta hai? Ek inflection point ke liye concavity ka actually change karna zaroori hai. Yahan $x^{4}$ dono sides pe concave up hai (ek wide flat-bottomed bowl), toh $x = 0$ ek false candidate hai, inflection point nahi.

Verify: $f^{''} (- 1) = 12$ aur $f^{''} (1) = 12$ — same sign, koi flip nahi. ✓ Example 5 se contrast karo jahan $f^{''} = 6 x$ genuinely sign change karta tha. Rule: $f^{''} = 0$ inflection point ke liye necessary but not sufficient hai; tumhe hamesha sign change confirm karna hoga.

Neeche wala figure genuine case $f = x^{3} - 3 x$ (Example 5 se, white) ko false case $f = x^{4}$ (amber) ke saath overlay karta hai. Origin pe white cubic concave-down se concave-up mein cross karta hai (real inflection), jabki amber quartic dono sides pe concave-up bowl se axis ko bas kiss karta hai (koi inflection nahi).

Figure 3 — Cell E′: ek blueprint grid pe origin ke paas do curves — white cubic x³−3x, jo genuinely concavity change karta hai down se up x=0 pe (ek real inflection, amber dot se mark kiya gaya), aur amber quartic x⁴, jiska f''=12x² dono sides pe positive hai toh yeh ek concave-up bowl hi rehta hai aur sirf x=0 pe apna minimum touch karta hai (cyan cross), illustrating karta hai ki f''(0)=0 akele inflection point guarantee nahi karta.

Cell F — Degenerate inputs (bilkul bhi curvature nahi)

Cell G — High orders ka limiting behaviour

Worked example Example 8 — exponential vs polynomial as

$n \to \infty$ $p (x) = x^{3}$ aur $E (x) = e^{x}$ ke $n$ -th derivatives compare karo jab $n$ bada hota hai.

Forecast: kaun sa ek infinite baar differentiate hone ke baad "bachta" hai?

Polynomial $p (x) = x^{3}$ : $p^{'''} (x) = 6$ , aur $p^{(4)} (x) = 0$ , toh $p^{(n)} (x) = 0$ saare $n \geq 4$ ke liye. Yeh step kyun? Har derivative degree kam karta hai; ek degree-3 polynomial 4 derivatives ke baad $0$ hit karta hai aur wahan reh jaata hai.
Exponential $E (x) = e^{x}$ : $E^{'} (x) = e^{x}$ , toh $E^{(n)} (x) = e^{x}$ har $n$ ke liye. Kyun? $e^{x}$ ki defining property yeh hai ki yeh apna khud ka derivative hai — differentiate karne se yeh kabhi change nahi hota.
Limit view: jab $n \to \infty$ , $p^{(n)} \to 0$ lekin $E^{(n)} = e^{x}$ fixed rehta hai. Yeh matter kyun karta hai? Ek Taylor series mein, har term ko value $f^{(n)} (0)$ chahiye; polynomials finitely many contribute karte hain, lekin $e^{x}$ forever contribute karta hai — isliye $e^{x} = \sum \frac{x ^{n}}{n !}$ ek infinite sum hai.

Verify: $p^{(4)} (x) = 0$ ✓ (constant $6$ differentiate ho ke $0$ deta hai). $E^{(n)} (0) = e^{0} = 1$ har $n$ ke liye, toh uske Taylor coefficients sab $\frac{1}{n !}$ hain — ek infinite, never-dying tail. ✓

Cell H — Exam-style notation trap

Active Recall

x^{5}

kitne derivatives ke baad

0

ho jaata hai?

6th derivative pe;

f^{(5)} = 120 = 5!

, phir

f^{(6)} = 0

cos x

ka 4th derivative equals

cos x

— trig derivatives period 4 ke saath cycle karte hain.

Position ke 2nd derivative (acceleration) ki units hain

metres per second squared, m/s².

Kisi bhi line

f (x) = m x + c

ke liye, second derivative hai

0

everywhere (constant slope → zero curvature).

Agar

f^{''} (x_{0}) = 0

ho, toh kya

x_{0}

ek inflection point hai?

Zaroor nahi — tumhe check karna hoga ki

f^{''}

sign change karta hai; e.g.

x^{4}

mein

f^{''} (0) = 0

hai lekin koi sign change nahi, toh koi inflection nahi.

Infinite differentiation ke under,

e^{x}

hamesha

e^{x}

rehta hai, jabki koi bhi polynomial eventually

0

ho jaata hai.

Connections

Higher-order derivatives — notation, physical meaning (parent)
Power Rule and differentiation rules
Derivative — definition as a limit
Concavity and the Second-Derivative Test
Inflection points
Kinematics — position, velocity, acceleration
Taylor series
Trigonometric derivatives