Chain rule — proof, composite function derivatives: Worked examples

4.1.16 · D3 · Maths › Calculus I — Limits & Derivatives › Chain rule — proof, composite function derivatives

Neeche poori jagah, "the machine" language yeh hai: outer function $f$ , inner function $g$ , composite $h (x) = f (g (x))$ , aur rule hai $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ — bahar wale ka derivative (andar wala frozen rakho) times andar wale ka derivative.

Scenario matrix

Kuch bhi work karne se pehle, aao har class of case list karein jo chain-rule problem mein ho sakta hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Kya special hai / kya galat ho sakta hai	Example
A	Power-of-polynomial (positive inner)	Basic peel; answer ka sign $x$ follow karta hai	Ex 1
B	Trig inside polynomial	Inside intact rakhna zaroori hai ( $cos (x^{2}) \neq = cos 2 x$ )	Ex 2
C	Triple / multi-nesting	Har layer ka ek factor	Ex 3
D	Inside ek point par zero evaluate hota hai	Kya $h^{'}$ phir bhi defined hai? Outer ko $0$ par check karo	Ex 4
E	Degenerate: ek gear ki rate $= 0$ us point par	Poora product $0$ ho jaata hai	Ex 5
F	Limiting / blow-up (denominator $\to 0$ )	Derivative $\to \pm \infty$ ; kaun sa sign?	Ex 6
G	Negative / all-sign inner (quadrant sweep)	Answer regions mein sign change karta hai	Ex 7
H	Word problem — related rates	Chain rule in time; units carry karo	Ex 8
I	Exam twist — abstract $f$ with a table	$f$ ka koi formula nahi; diye gaye values use karo	Ex 9
J	Trig-inside-trig (non-power nesting)	Dono layers transcendental hain	Ex 10

Page ka baaki hissa har cell fill karta hai.

Cell A — polynomial ki power

Worked example Ex 1 (Cell A):

$h (x) = (2 x^{3} - 5)^{4}$ Forecast: answer ki shape guess karo padhne se pehle. Kya $x^{2}$ ka ek factor hoga? Kya power $3$ tak drop hogi? Likh lo.

Layers name karo. Outer $f (u) = u^{4}$ , inner $g (x) = 2 x^{3} - 5$ , toh $u = g (x)$ . Yeh step kyun? Tum peel nahi kar sakte jab tak pata na ho kaun sa part "shell" hai aur kaun sa "filling."
Outer differentiate karo, inside frozen. $f^{'} (u) = 4 u^{3} \Rightarrow f^{'} (g (x)) = 4 (2 x^{3} - 5)^{3}$ . Yeh step kyun? Shell par power rule; hum poore blob $2 x^{3} - 5$ ko ek single lump $u$ maante hain.
Inner differentiate karo. $g^{'} (x) = 6 x^{2}$ . Yeh step kyun? Yeh inside ka "gear ratio" hai — filling kitni fast move karti hai.
Multiply karo (DOTI). $h^{'} (x) = 4 (2 x^{3} - 5)^{3} \cdot 6 x^{2} = 24 x^{2} (2 x^{3} - 5)^{3} .$ Yeh step kyun? DOTI: outer rate (step 2) ko inner rate (step 3) se multiply karo — dono amplifications compound hoti hain.

Verify: $x = 1$ par, $g (1) = 2 - 5 = - 3$ , toh $h^{'} (1) = 24 \cdot 1 \cdot (- 3)^{3} = 24 \cdot (- 27) = - 648$ . Ek numerical slope $\frac{h ( 1.001 ) - h ( 0.999 )}{0.002} \approx - 648$ confirm karta hai. ✓ $x^{2}$ ka factor: haan; power $3$ tak drop: haan — forecast se match karta hai.

Cell B — polynomial ke andar trig (inside intact rakho)

Worked example Ex 2 (Cell B):

$h (x) = cos (x^{2} + 1)$ Forecast: kya answer mein $sin (2 x)$ hoga ya $sin (x^{2} + 1)$ ? Yeh classic trap hai — abhi decide karo.

Layers. Outer $f (u) = cos u$ , inner $g (x) = x^{2} + 1$ . Yeh step kyun? Outer $cos$ aur inner $x^{2} + 1$ ko alag tag karna zaroori hai taaki hum unhe kabhi accidentally saath differentiate na karein.
Outer derivative, inside frozen. $f^{'} (u) = - sin u \Rightarrow f^{'} (g (x)) = - sin (x^{2} + 1)$ . Yeh step kyun? $cos$ ka derivative $- sin$ hota hai; hum argument ko abhi nahi touch karte.
Inner derivative. $g^{'} (x) = 2 x$ . Yeh step kyun? Yeh inner gear ratio hai; yeh ek separate multiplied factor ke roop mein aayega, $sin$ ke andar nahi.
Multiply karo (DOTI). $h^{'} (x) = - sin (x^{2} + 1) \cdot 2 x = - 2 x sin (x^{2} + 1) .$ Yeh step kyun? DOTI: outer rate (step 2) times inner rate (step 3); inner $2 x$ $sin$ ke bahar aata hai.

Verify: $x = 0$ par, $h^{'} (0) = - 2 \cdot 0 \cdot sin (1) = 0$ . Sensible lagta hai: $x^{2} + 1$ ka $x = 0$ par zero slope hai (parabola ka bottom), toh composite momentarily flat hai. ✓

Cell C — triple nesting

Worked example Ex 3 (Cell C):

$h (x) = sin^{3} (4 x) = (sin (4 x))^{3}$ Forecast: answer mein kitne factors honge? (Hint: layers count karo.)

Teen layers hain, outside-in peel karo:

Layer 1 (outermost): cube, $(\cdot)^{3}$ .
Layer 2: $sin (\cdot)$ .
Layer 3 (innermost): $4 x$ .

Cube ka derivative: $3 (sin 4 x)^{2}$ . Kyun? Power rule, inside ( $sin 4 x$ ) frozen.
$sin$ ka derivative: $cos (4 x)$ . Kyun? Agla shell peel karo, uska inside ( $4 x$ ) frozen.
Inner ka derivative: $\frac{d}{d x} (4 x) = 4$ . Yeh step kyun? Innermost gear ratio: $4 x$ , $x$ se $4$ times fast change karta hai; yeh last multiplied factor ban jaata hai.
Teeno factors multiply karo (DOTI, teen baar): $h^{'} (x) = 3 (sin 4 x)^{2} \cdot cos (4 x) \cdot 4 = 12 sin^{2} (4 x) cos (4 x) .$ Yeh step kyun? DOTI har layer boundary par apply hota hai: har shell ka ek factor, sab multiply hote hain.

Verify: teen layers → teen factors → forecast se match. $x = \frac{π}{8}$ par: $4 x = \frac{π}{2}$ , $sin = 1, cos = 0$ , toh $h^{'} = 12 \cdot 1 \cdot 0 = 0$ . Correct — $sin^{3} (4 x)$ $4 x = π /2$ par peak karta hai, toh wahan slope $0$ hai. ✓

Cell D — inside zero hit karta hai

Yeh woh case hai jise log proof se darate hain (" $Δ u = 0$ " ki chinta). Isko concretely dekhte hain.

Worked example Ex 4 (Cell D):

$h (x) = e^{x^{2} - 4}$ , $h^{'}$ evaluate karo jahan inside $0$ ho. Forecast: inside $g (x) = x^{2} - 4$ , $x = \pm 2$ par zero hai. Kya $h^{'}$ wahan phir bhi defined hai? Haan/nahi guess karo.

Layers. Outer $f (u) = e^{u}$ , inner $g (x) = x^{2} - 4$ . Yeh step kyun? Shell ( $e^{u}$ ) aur filling ( $x^{2} - 4$ ) identify karo peel karne se pehle — exponential outer machine hai.
Outer derivative: $f^{'} (u) = e^{u}$ (apna khud ka derivative) $\Rightarrow f^{'} (g (x)) = e^{x^{2} - 4}$ . Yeh step kyun? $e^{u}$ woh unique function hai jo apne aap ke derivative ke barabar hai; hum ise inner value $u = x^{2} - 4$ par evaluate karte hain.
Inner derivative: $g^{'} (x) = 2 x$ . Yeh step kyun? $x^{2} - 4$ par power rule inner gear ratio $2 x$ deta hai.
Multiply karo (DOTI): $h^{'} (x) = e^{x^{2} - 4} \cdot 2 x = 2 x e^{x^{2} - 4} .$ Yeh step kyun? DOTI: outer rate (step 2) ko inner rate (step 3) se compound karo.
Zero-inside point $x = 2$ plug in karo: $g (2) = 0$ , toh $h^{'} (2) = 2 \cdot 2 \cdot e^{0} = 4 \cdot 1 = 4.$ Yeh step kyun? Yahi is cell ka poora point hai — test karna ki "inside $= 0$ " kuch todi hai ya nahi. Nahi tooda.

Verify: numerically $\frac{h ( 2.001 ) - h ( 1.999 )}{0.002} \approx 4.00$ . ✓ Aur $h^{'} (2) > 0$ : $x = 2$ par inside increase ho raha hai ( $2 x = 4 > 0$ ), toh composite rise karta hai. Sign sahi hai.

Cell E — degenerate: ek gear ki rate zero hai

Worked example Ex 5 (Cell E):

$h (x) = cos (x^{3})$ , $h^{'}$ evaluate karo $x = 0$ par. Forecast: $x = 0$ par inner rate $g^{'} (x) = 3 x^{2}$ vanish hoti hai — aur aisa hota hai ki outer rate bhi. $h^{'} (0)$ predict karo.

Layers. Outer $f (u) = cos u$ , inner $g (x) = x^{3}$ . Yeh step kyun? Shell ( $cos$ ) aur filling ( $x^{3}$ ) tag karo taaki hum track kar sakein kaun sa gear rukta hai.
Outer derivative: $- sin (x^{3})$ . Yeh step kyun? $cos$ ka derivative $- sin$ hai, inner value $x^{3}$ par evaluate kiya (inside frozen).
Inner derivative: $3 x^{2}$ . Yeh step kyun? $x^{3}$ par power rule inner gear ratio deta hai — yahi factor $x = 0$ par vanish hoga.
Multiply karo (DOTI): $h^{'} (x) = - 3 x^{2} sin (x^{3})$ . Yeh step kyun? DOTI: outer rate times inner rate, dono ek product mein.
$x = 0$ par: $h^{'} (0) = - 3 \cdot 0^{2} \cdot sin (0) = 0$ . Yahan inner gear $3 x^{2}$ zero hai aur outer $- sin (x^{3}) = - sin 0 = 0$ — dono factors vanish hote hain, toh product emphatic taur par $0$ hai.

Verify: $\frac{h ( 0.001 ) - h ( - 0.001 )}{0.002} \approx 0$ . ✓ (Symmetry se $cos (x^{3})$ even hai, toh uska slope $0$ par zaroori $0$ hai — sanity confirmed.)

Cell F — limiting / blow-up (derivative $\to \pm \infty$ )

Worked example Ex 6 (Cell F):

$h (x) = 1 - x^{2}$ (upper unit semicircle), $x \to 1^{-}$ ke paas behaviour. Forecast: jaise $x$ neeche se $1$ ke paas jaata hai, semicircle axis ki taraf gire jaata hai. Kya slope $0$ jaata hai, koi finite number, ya $\pm \infty$ ? Kaun sa sign?

Layers. Outer $f (u) = u^{1/2} = u$ , inner $g (x) = 1 - x^{2}$ .
Outer derivative: $f^{'} (u) = \frac{1}{2} u^{- 1/2} = \frac{1}{2 u}$ . Yeh step kyun? Exponent $\frac{1}{2}$ ke saath power rule; $u^{- 1/2}$ ki wajah se blow-up possible hai — jaise $u \to 0$ , yeh factor explode karta hai.
Inner derivative: $g^{'} (x) = - 2 x$ . Yeh step kyun? $1 - x^{2}$ par power rule; minus sign yahi wajah hai ki final slope negative hogi (curve girna).
Multiply karo (DOTI): $h^{'} (x) = \frac{1}{2 1 - x ^{2}} \cdot (- 2 x) = \frac{- x}{1 - x ^{2}} .$ Yeh step kyun? DOTI: outer rate (step 2) times inner rate (step 3), phir simplify.
Limit $x \to 1^{-}$ : numerator $\to - 1$ ; denominator $1 - x^{2} \to 0^{+}$ (ek shrinking positive number). Toh $h^{'} (x) \to \frac{- 1}{0 ^{+}} = - \infty.$

Neeche wali figure (s01): upper semicircle violet mein, teen dashed tangent lines $x = 0, 0.6, 0.9$ par jo steeper aur steeper hoti jaati hain, $x = 1$ ke paas vertical ki taraf tilt karti hain. Minus sign kehta hai har tangent downward tilt karta hai (curve girna); magnitude $\to \infty$ matlab vertical ho jaata hai.

Verify: $x = 0.99$ par, $h^{'} = \frac{- 0.99}{1 - 0.9801} = \frac{- 0.99}{0.0199} \approx \frac{- 0.99}{0.1411} \approx - 7.02$ — bada aur negative, $- \infty$ ki taraf. ✓

Cell G — all-sign inner (quadrant/region sweep)

Worked example Ex 7 (Cell G):

$h (x) = (x^{2} - 1)^{3}$ ; $x$ ke saare regions mein $h^{'}$ ka sign describe karo. Forecast: composite kahan increase ho raha hai kahan decrease? Teen break-points ke saath guess sketch karo.

Layers. Outer $f (u) = u^{3}$ , inner $g (x) = x^{2} - 1$ .
Outer derivative: $3 u^{2} = 3 (x^{2} - 1)^{2}$ . Yeh step kyun? Cube par power rule; crucially yeh ek square hai, toh yeh factor $\geq 0$ hamesha hai — zero touch kar sakta hai lekin sign kabhi flip nahi karta.
Inner derivative: $2 x$ . Yeh step kyun? $x^{2} - 1$ par power rule; yahi akela factor sign change karta hai, toh yeh control karta hai kahan $h$ rise ya fall karta hai.
Multiply karo (DOTI): $h^{'} (x) = 3 (x^{2} - 1)^{2} \cdot 2 x = 6 x (x^{2} - 1)^{2} .$ Yeh step kyun? DOTI: outer rate times inner rate; sign-carrying factor inner $2 x$ hai.
Sign analysis — har region cover karo. Factor $(x^{2} - 1)^{2} \geq 0$ kabhi sign flip nahi karta; yeh $x = \pm 1$ par $0$ hai. Toh $h^{'}$ ka sign $6 x$ ka sign hai:

Region	$6 x$	$(x^{2} - 1)^{2}$	$h^{'} (x)$	Behaviour
$x < - 1$	$-$	$+$	$-$	decreasing
$x = - 1$	$-$	$0$	$0$	flat (touch)
$- 1 < x < 0$	$-$	$+$	$-$	decreasing
$x = 0$	$0$	$+$	$0$	flat (min)
$0 < x < 1$	$+$	$+$	$+$	increasing
$x = 1$	$+$	$0$	$0$	flat (touch)
$x > 1$	$+$	$+$	$+$	increasing

Neeche wali figure (s02): curve $(x^{2} - 1)^{3}$ violet mein, teen horizontal tangents $x = - 1, 0, 1$ par mark kiye gaye hain (orange dots, magenta level segments); labels $x = 0$ ko true minimum aur $x = \pm 1$ ko saddle-flats flag karte hain.

Verify: $h^{'} (0.5) = 6 (0.5) (0.25 - 1)^{2} = 3 \cdot 0.5625 = 1.6875 > 0$ ✓ (increasing). $h^{'} (- 0.5) = 6 (- 0.5) (0.5625) = - 1.6875 < 0$ ✓ (decreasing). $h^{'} (\pm 1) = 0$ ✓.

Worked example Ex 8 (Cell H): Ek spherical balloon inflate ho raha hai taaki uski

radius $\frac{d r}{d t} = 2 cm/s$ ki rate se badhti hai. Jab $r = 5 cm$ ho toh volume kitni fast badh raha hai? Forecast: volume 3D hai, radius 1D — kya tum expect karte ho ki volume rate $2$ se badi hogi ya chhoti? Ballpark number guess karo.

Relationship likho. $V = \frac{4}{3} π r^{3}$ , aur $r$ time $t$ par depend karta hai. Yeh step kyun? Hum $\frac{d V}{d t}$ chahte hain lekin sirf $\frac{d r}{d t}$ jaante hain; pehle hum ek formula chahte hain jo $V$ ko shared variable $r$ se tie kare, warna differentiate karne ko kuch nahi hoga.
Chain rule in time. $V = f (r)$ aur $r = r (t)$ ke saath, $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t} .$ Yeh step kyun? Yeh exactly $\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ hai parent note se, $u = r$ , $x = t$ ke saath — topic Related rates.
Outer rate. $\frac{d V}{d r} = \frac{4}{3} π \cdot 3 r^{2} = 4 π r^{2}$ . Yeh step kyun? $r^{3}$ par power rule; $\frac{4}{3} π \cdot 3$ collapse hokar $4 π r^{2}$ ban jaata hai — jo exactly sphere ka surface area hai, ek handy sanity anchor.
Values ke saath assemble karo (DOTI). $r = 5$ par: $\frac{d V}{d t} = d V / d r 4 π (5)^{2} \cdot d r / d t 2 = 4 π \cdot 25 \cdot 2 = 200 π \approx 628.3.$ Yeh step kyun? DOTI in time: outer rate $\frac{d V}{d r}$ (step 3) ko inner rate $\frac{d r}{d t}$ (diya hua) se multiply karo.
Units attach karo. $\frac{d V}{d r}$ ki units $cm^{2}$ (area) hain, $\frac{d r}{d t}$ ki $cm/s$ , toh product ki units $cm^{2} \cdot cm/s = cm^{3} / s$ hain. Isliye $\frac{d V}{d t} \approx 628.3 cm^{3} / s .$ Yeh step kyun? Word problem mein units answer ka hissa hain — volume change ki rate volume-per-time mein aani chahiye.

Verify: units sahi multiply hoti hain ( $cm^{2} \cdot cm/s = cm^{3} / s$ ✓). Forecast revisit: haan, $628.3$ $2$ se bahut bada hai — exactly jaisa predict kiya tha, kyunki ek bada sphere radius ke har extra cm ke liye bahut saara volume gain karta hai. Numeric: $200 π = 628.318 \dots$ ✓ Takeaway: related rates sirf chain rule hai jisme time driving variable hai.

Cell I — exam twist (table se abstract $f$ , koi formula nahi)

Worked example Ex 9 (Cell I): Tumhe bataya gaya hai

$h (x) = f (g (x))$ aur sirf yeh table diya gaya hai. $h^{'} (1)$ find karo.

$x$	$g (x)$	$g^{'} (x)$	$f (x)$	$f^{'} (x)$
$1$	$3$	$4$	$6$	$- 2$
$3$	$2$	$5$	$7$	$10$

Forecast: is table se actually kaun se do numbers chahiye? Compute karne se pehle unhe circle karo.

Point par rule likho. $h^{'} (1) = f^{'} (g (1)) \cdot g^{'} (1)$ . Yeh step kyun? Yeh poora chain rule $x = 1$ par instantiate kiya gaya hai; twist sirf yeh hai ki $f, g$ ka koi formula nahi, toh har quantity table se padhni hogi.
Inner value find karo. $g (1) = 3$ (row $x = 1$ ). Yeh step kyun? Outer derivative $u = g (1)$ par evaluate hoti hai, toh pehle hum jaanna chahte hain kahan dekhna hai. Answer $3$ hai, $1$ nahi — yeh #1 exam trap hai.
Us inner value par $f^{'}$ padho. $f^{'} (3) = 10$ (row $x = 3$ ). Yeh step kyun? Hume $f^{'}$ chahiye $u = 3$ par (step 2 se), jo $x = 3$ row mein hai — $x = 1$ row mein nahi.
Inner rate padho. $g^{'} (1) = 4$ (row $x = 1$ ). Yeh step kyun? Doosra chain-rule factor $g^{'} (1)$ seedha original point $x = 1$ par padha jaata hai.
Multiply karo (DOTI): $h^{'} (1) = f^{'} (g (1)) \cdot g^{'} (1) = f^{'} (3) \cdot g^{'} (1) = 10 \cdot 4 = 40.$ Yeh step kyun? DOTI: outer rate $f^{'} (3)$ times inner rate $g^{'} (1)$ — sirf do table entries jo actually matter karti hain.

Verify: do zaroori numbers $f^{'} (3) = 10$ aur $g^{'} (1) = 4$ the; product $= 40$ . Forecast se consistent agar tune unhe circle kiya. ✓

Cell J — trig-inside-trig (dono layers transcendental)

Worked example Ex 10 (Cell J):

$h (x) = sin (cos x)$ Forecast: koi bhi layer power nahi hai — dono trig hain. Kya answer $cos (cos x)$ times kuch hoga, ya $sin (sin x)$ ? Aur extra factor kaun sa sign layega? Pehle guess karo.

Layers. Outer $f (u) = sin u$ , inner $g (x) = cos x$ . Yeh step kyun? Do trig layers ke saath inhe name karna doubly important hai — inner ko accidentally outer ki tarah differentiate karna easy hai.
Outer derivative, inside frozen. $f^{'} (u) = cos u \Rightarrow f^{'} (g (x)) = cos (cos x)$ . Yeh step kyun? $sin$ ka derivative $cos$ hai, inner value $u = cos x$ par evaluate kiya — argument $cos x$ outer $cos$ ke andar intact rehta hai.
Inner derivative. $g^{'} (x) = - sin x$ . Yeh step kyun? $cos x$ ka derivative $- sin x$ hai; minus sign hi wajah hai ki final answer mein leading negative aata hai.
Multiply karo (DOTI): $h^{'} (x) = cos (cos x) \cdot (- sin x) = - sin x cos (cos x) .$ Yeh step kyun? DOTI: outer rate (step 2) times inner rate (step 3).

Verify: $x = 0$ par: $h^{'} (0) = - sin 0 \cdot cos (cos 0) = 0 \cdot cos 1 = 0$ . Sanity: $cos x$ ka $x = 0$ par zero slope hai (cosine wave ka top), toh composite wahan flat hai. ✓ $x = \frac{π}{2}$ par: $h^{'} = - sin \frac{π}{2} cos (cos \frac{π}{2}) = - 1 \cdot cos 0 = - 1$ . ✓

Recall Self-test: cell name karo, phir solve karo

Worked solutions cover karo. Har ke liye, pehle bolo kaun sa matrix cell hai, phir differentiate karo.

$(2 x^{3} - 5)^{4}$ differentiate karo. ::: Cell A: $24 x^{2} (2 x^{3} - 5)^{3}$ . $cos (x^{2} + 1)$ differentiate karo (trap dekho). ::: Cell B: $- 2 x sin (x^{2} + 1)$ , nahi $- sin (2 x)$ . $sin^{3} (4 x)$ differentiate karo; kitne factors? ::: Cell C: $12 sin^{2} (4 x) cos (4 x)$ ; teen layers → teen factors. Kya $h^{'}$ defined hai jahan $e^{x^{2} - 4}$ ka inside zero ho? $x = 2$ par value? ::: Cell D: haan; $h^{'} (2) = 4$ . $cos (x^{3})$ ke liye $h^{'} (0) = 0$ kyun hai? ::: Cell E: inner rate $3 x^{2} = 0$ at $x = 0$ , product collapse ho jaata hai. $lim_{x \to 1^{-}} \frac{d}{d x} 1 - x^{2}$ kya hai? ::: Cell F: $- \infty$ (tangent vertical ho jaata hai, girna). $(x^{2} - 1)^{3}$ ke horizontal tangents kahan hain, aur kaun sa true min hai? ::: Cell G: $x = - 1, 0, 1$ ; sirf $x = 0$ min hai (baaki saddle-flats hain). Balloon: $\frac{d V}{d t}$ at $r = 5$ , $\frac{d r}{d t} = 2$ ? ::: Cell H: $200 π \approx 628.3 cm^{3} / s$ . Table se, $h^{'} (1)$ jahan $h = f (g (x))$ ? ::: Cell I: $f^{'} (3) \cdot g^{'} (1) = 10 \cdot 4 = 40$ . $sin (cos x)$ differentiate karo. ::: Cell J: $- sin x cos (cos x)$ .

Connections

Chain rule — proof, composite function derivatives — parent; yeh page uska applied stress-test hai.
Related rates — Cell H seedha $\frac{d V}{d t} = \frac{d V}{d r} \frac{d r}{d t}$ ka related-rates use hai.
Derivative — limit definition — Cells D–F derivative ke limit meaning par lean karte hain (blow-up, zero rate).
Product rule aur Quotient rule — tougher exam problems mein aksar chain rule ke saath combine hote hain.
Implicit differentiation — Cell I ka abstract- $f$ mindset yahan generalise hota hai.
Inverse function derivative aur Continuity implies differentiability fails converse — related follow-ups.

Worked examples — Chain rule — proof, composite function derivatives