Chain rule for multivariable functions — all cases: Worked examples

4.4.7 · D3 · Maths › Multivariable Calculus › Chain rule for multivariable functions — all cases

Shuru karne se pehle, ek promise: hum koi bhi symbol bina explain kiye nahi likhenge.

$z = f (x, y)$ ka matlab hai ek machine $f$ jo do numbers $x, y$ khaati hai aur ek number $z$ ugalti hai.
$\frac{\partial z}{\partial x}$ (padho "partial dee z by dee x") woh rate hai jisse $z$ badalta hai jab tum sirf $x$ ko nudge karo aur $y$ ko freeze karo. Dekho Partial Derivatives.
$\frac{d z}{d t}$ (round-free $d$ ) woh rate hai jab $z$ ultimately sirf ek variable $t$ par depend karta hai.

Hum aur symbols ( $J$ Jacobian matrix ke liye, $\sum$ summation ke liye) baad mein introduce karenge, bilkul us moment par jab hum pehli baar unhe use karte hain.

The scenario matrix

Is topic ke har problem ka ghar in cells mein se ek hai:

#	Cell (kya cheez tricky banati hai)	Kaun sa example
A	Case 1: ek independent variable, clean	Ex 1
B	Case 2: do independent variables (polar)	Ex 2
C	Sign / quadrant trap (ek component negative ho jaata hai)	Ex 3
D	Zero input: ek link derivative $0$ hai kisi point par	Ex 4
E	Degenerate: do inputs collapse hokar same variable ban jaate hain	Ex 5
F	Explicit reappearance: $t$ directly aur $x, y$ ke through bhi appear karta hai	Ex 6
G	Limiting behaviour: rate jab $t \to$ koi special value	Ex 7
H	Real-world word problem (units matter karte hain)	Ex 8
I	Exam twist: Jacobian / matrix form	Ex 9

Hum har cell neeche hit karte hain.

Cell A — ek independent variable, clean

Compute karne se pehle, is problem ka dependency tree dekho. Yeh deep variable $t$ se output $z$ tak do raaste dikhata hai: ek raasta $x$ se hoke jaata hai, doosra $y$ se. Yahi "two paths" ke peechhe ki picture hai.

Figure — Chain rule for multivariable functions — all cases

Worked example Ex 1 (Cell A)

$z = x^{2} y$ , $x = cos t$ , $y = sin t$ . $\frac{d z}{d t}$ nikalo aur $t = \frac{π}{2}$ par evaluate karo.

Forecast: $t = π /2$ par hum unit circle ke top par baithe hain, $x = 0, y = 1$ . Guess: kyunki $x = 0$ hai wahan, kya $\frac{d z}{d t}$ vanish ho jaata hai? Yeh soch ke rakho.

$\frac{\partial z}{\partial x} = 2 x y$ , $\frac{\partial z}{\partial y} = x^{2}$ . Yeh step kyun? Ek variable freeze karo, doosre ko differentiate karo — yahi $\partial$ ka matlab hai.
$\frac{d x}{d t} = - sin t$ , $\frac{d y}{d t} = cos t$ . Kyun? Ordinary single-variable derivatives — dekho Single-variable Chain Rule.
$\frac{d z}{d t} = 2 x y (- sin t) + x^{2} cos t$ . Kyun? Upar ka tree exactly do paths rakhta hai ( $t \to x \to z$ aur $t \to y \to z$ ): har path ke link-derivatives multiply karo, phir dono paths add karo.
Step 3 mein $x = cos t, y = sin t$ substitute karo: $2 (cos t) (sin t) (- sin t) + (cos t)^{2} cos t = - 2 cos t sin^{2} t + cos^{3} t$ .
$t = π /2$ par: $cos t = 0$ , toh dono terms $0$ hain. $\frac{d z}{d t} = 0$ .

Verify: direct substitution $z = cos^{2} t sin t$ . Directly differentiate karo: $\frac{d z}{d t} = 2 cos t (- sin t) sin t + cos^{2} t cos t = - 2 cos t sin^{2} t + cos^{3} t$ . ✓ Step 4 se match karta hai. Forecast sahi nikla — $x = 0$ factor dono paths ko kill kar deta hai.

Cell B — do independent variables (polar), figure ke saath

Agla figure do kaam ek saath karta hai. Left par polar dependency tree hai: output $z$ har ultimate variable $r$ aur $θ$ tak dono doors $x$ aur $y$ se pahunchta hai — isliye har polar chain-rule sum mein do terms hote hain. Right par geometric picture hai: angle $θ$ par ek ray par point $(x, y)$ , aur red radial arrow jo woh direction dikhata hai jismein tum jaate ho jab $r$ badhate ho. $r$ badhane se point us arrow ke saath seedha bahar slide hota hai, yahi precisely wajah hai ki $\partial z / \partial r$ $f$ ka Directional Derivative radial direction mein nikalta hai.

Worked example Ex 2 (Cell B)

$z = f (x, y)$ with $x = r cos θ$ , $y = r sin θ$ . Concrete $f = x^{2} + y^{2}$ lo aur $\frac{\partial z}{\partial r}$ aur $\frac{\partial z}{\partial θ}$ nikalo.

Forecast: $z = x^{2} + y^{2}$ exactly $r^{2}$ hai, toh guess karo $\partial z / \partial r = 2 r$ aur $\partial z / \partial θ = 0$ (angle distance-squared nahi badalta).

$\frac{\partial z}{\partial x} = 2 x = 2 r cos θ$ , $\frac{\partial z}{\partial y} = 2 y = 2 r sin θ$ . Kyun? $f$ ke partials, phir polar mein rewrite $x = r cos θ, y = r sin θ$ use karke.
$\frac{\partial x}{\partial r} = cos θ$ , $\frac{\partial y}{\partial r} = sin θ$ . Kyun? $θ$ fix rakho, $r$ mein differentiate karo.
$\frac{\partial z}{\partial r} = (2 r cos θ) (cos θ) + (2 r sin θ) (sin θ) = 2 r cos^{2} θ + 2 r sin^{2} θ = 2 r (cos^{2} θ + sin^{2} θ) = 2 r$ . Kyun? Left tree ke do paths sum karo; phir $cos^{2} θ + sin^{2} θ = 1$ use karo.
$\frac{\partial x}{\partial θ} = - r sin θ$ , $\frac{\partial y}{\partial θ} = r cos θ$ ; toh $\frac{\partial z}{\partial θ} = (2 r cos θ) (- r sin θ) + (2 r sin θ) (r cos θ) = - 2 r^{2} cos θ sin θ + 2 r^{2} sin θ cos θ = 0$ .

Verify: direct $z = r^{2} \Rightarrow \partial z / \partial r = 2 r$ , $\partial z / \partial θ = 0$ . ✓ Dono forecast se match karte hain. Step 3 figure mein red radial arrow ke direction mein $f$ ka directional derivative hai.

Cell C — sign / quadrant trap

Neeche ka figure point ko unit circle par $t = 3 π /4$ par plant karta hai: yeh quadrant II mein rehta hai, jahan horizontal coordinate $x = cos t$ negative hai (red, left pointing) lekin vertical coordinate $y = sin t$ positive hai (mint, upar pointing). In signs ko theek rakhna hi poora game hai.

Worked example Ex 3 (Cell C)

$z = x y$ , $x = cos t$ , $y = sin t$ . $\frac{d z}{d t}$ nikalo $t = \frac{3 π}{4}$ par (second quadrant, jahan $x < 0$ , $y > 0$ ).

Forecast: quadrant II mein, $x = cos (3 π /4) < 0$ lekin $y > 0$ . Ek student jo signs bhool jaata hai woh sab kuch positive rakh sakta hai. Signs dhyan se dekho.

$\frac{\partial z}{\partial x} = y$ , $\frac{\partial z}{\partial y} = x$ . Kyun? $\partial (x y) / \partial x = y$ , $y$ freeze karke; aur $\partial (x y) / \partial y = x$ .
$\frac{d x}{d t} = - sin t$ , $\frac{d y}{d t} = cos t$ . Kyun? Ye $cos t$ aur $sin t$ ke ordinary single-variable derivatives hain; $\frac{d x}{d t}$ par minus sign exactly wahi trap hai jis baare mein yeh cell hai — cosine differentiate karne se sign flip hota hai, isliye yeh negative sum mein le jaana zaroori hai.
$\frac{d z}{d t} = y (- sin t) + x (cos t)$ . Ab $x = cos t$ aur $y = sin t$ substitute karo: $= (sin t) (- sin t) + (cos t) (cos t) = - sin^{2} t + cos^{2} t = cos 2 t$ . Aakhri equality kyun? Identity $cos^{2} t - sin^{2} t = cos 2 t$ .
$t = 3 π /4$ par: $2 t = 3 π /2$ , toh $cos (3 π /2) = 0$ . Thus $\frac{d z}{d t} = 0$ .

Verify: $z = cos t sin t = \frac{1}{2} sin 2 t$ , toh $\frac{d z}{d t} = cos 2 t$ . ✓ Trap: $t = 3 π /4$ par term $x cos t = cos^{2} (3 π /4) = + \frac{1}{2}$ (positive kyunki yeh $cos^{2}$ hai) jabki $- sin^{2} t = - \frac{1}{2}$ ; dono cancel hokar $0$ dete hain. Agar tumne $sin^{2}$ par negative sign drop kar diya hota, toh galat answer $+ 1$ aata.

Cell D — zero input

Worked example Ex 4 (Cell D)

$z = x^{2} + y^{3}$ , $x = t^{2}$ , $y = t$ . $\frac{d z}{d t}$ nikalo $t = 0$ par.

Forecast: $t = 0$ par, $\frac{d x}{d t} = 2 t = 0$ . Toh $x$ -path derivative switched off hai. Guess: sirf $y$ -path bachega.

$\frac{\partial z}{\partial x} = 2 x$ , $\frac{\partial z}{\partial y} = 3 y^{2}$ . Kyun? $x$ mein partial $y$ freeze karta hai, toh $y^{3}$ constant ho jaata hai aur drop ho jaata hai, $2 x$ bachta hai; $y$ mein partial $x$ freeze karta hai, toh $x^{2}$ drop ho jaata hai aur $y^{3}$ power rule se $3 y^{2}$ ban jaata hai.
$\frac{d x}{d t} = 2 t$ , $\frac{d y}{d t} = 1$ . Kyun? Ye ordinary single-variable derivatives hain: $\frac{d}{d t} t^{2} = 2 t$ aur $\frac{d}{d t} t = 1$ power rule se.
$\frac{d z}{d t} = 2 x \cdot 2 t + 3 y^{2} \cdot 1$ . $x = t^{2}, y = t$ substitute karo: $= 2 t^{2} \cdot 2 t + 3 t^{2} = 4 t^{3} + 3 t^{2}$ . Kyun? Dono paths sum karo, phir parametrisation plug in karo.
$t = 0$ par: $4 (0) + 3 (0) = 0$ . Dono paths yahan $0$ dete hain kyunki $x = 0$ aur $y = 0$ bhi hai.

Verify: direct $z = (t^{2})^{2} + t^{3} = t^{4} + t^{3}$ , toh $\frac{d z}{d t} = 4 t^{3} + 3 t^{2}$ . ✓ $t = 0$ par yeh $0$ hai. Lesson: jab ek link derivative $0$ ho tum poora rule drop nahi karte — dono terms rakhte ho, aur woh term bas kuch contribute nahi karta.

Cell E — degenerate (inputs collapse hokar same variable ban jaate hain)

Worked example Ex 5 (Cell E)

$z = f (x, y) = x^{2} + 3 x y + y^{2}$ lekin $x = t$ aur $y = t$ (dono inputs same variable hain). $\frac{d z}{d t}$ nikalo.

Forecast: dono doors ab same street par khulte hain. Tum phir bhi do paths likhte ho — ek $x$ ke through, ek $y$ ke through — phir add karo. " $t$ ko ek baar count karo" ki temptation mat lo.

$\frac{\partial z}{\partial x} = 2 x + 3 y$ , $\frac{\partial z}{\partial y} = 3 x + 2 y$ . Kyun? $x$ mein partial $y$ ko constant treat karta hai, aur vice versa — chahe baad mein dono $t$ ke barabar hon, partials independently liye jaate hain.
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = 1$ . Kyun? $x = t$ aur $y = t$ mein se har ek $t$ mein identity map hai, aur $\frac{d}{d t} t = 1$ ; dono links unit rate se change karte hain, yahi precisely wajah hai ki dono paths equal weight ke saath bachte hain.
$\frac{d z}{d t} = (2 x + 3 y) (1) + (3 x + 2 y) (1) = (2 x + 3 y) + (3 x + 2 y) = 5 x + 5 y$ . Kyun? Dono paths sum karo, phir like terms collect karo.
$x = y = t$ substitute karo: $\frac{d z}{d t} = 5 t + 5 t = 10 t$ .

Verify: direct $z = t^{2} + 3 t \cdot t + t^{2} = 5 t^{2}$ , toh $\frac{d z}{d t} = 10 t$ . ✓ Agar tumne " $t$ ko ek baar count kiya" aur sirf $\partial z / \partial x = 2 t + 3 t = 5 t$ use kiya, toh $5 t$ milta — aadhi sach. Inputs ka collapse paths ki sankhya ko collapse nahi karta.

Cell F — explicit reappearance ( $t$ direct aur $x, y$ ke through)

Worked example Ex 6 (Cell F)

$w = f (x, y, t) = x y + t^{2}$ , with $x = t$ , $y = t^{2}$ . $\frac{d w}{d t}$ nikalo.

Forecast: $t$ teen tareekon se appear karta hai: $x$ ke through, $y$ ke through, aur wahan seedha $t^{2}$ ke roop mein baith ke. Teen paths, teen products. Explicit wala miss karo toh galat hoga.

Confusion se bachne ke liye, hum ek output symbol rakhte hain: machine $f$ hai, aur $w = f$ uska output value hai. Neeche ke saare partials $f$ ke partials hain; left par total derivative $\frac{d w}{d t}$ likha hai kyunki $w$ wahi output hai.

$\frac{\partial f}{\partial x} = y$ , $\frac{\partial f}{\partial y} = x$ , $\frac{\partial f}{\partial t} = 2 t$ . Kyun? Har partial baki do slots freeze karta hai; aakhri wala, $x, y$ freeze karke, $t^{2}$ mein explicit $t$ ko capture karta hai.
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = 2 t$ . Kyun? Ye parametrisation ke ordinary single-variable derivatives hain: $\frac{d}{d t} t = 1$ aur $\frac{d}{d t} t^{2} = 2 t$ power rule se — ye measure karte hain ki jab $t$ move karta hai toh har intermediate variable kitni tez move karta hai.
$\frac{d w}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} + \frac{\partial f}{\partial t}$ . Kyun? Tree mein $t$ tak ek direct branch hai plus do indirect branches — dekho Implicit Differentiation (multivariable).
$= y (1) + x (2 t) + 2 t$ . $x = t, y = t^{2}$ substitute karo: $= t^{2} + t \cdot 2 t + 2 t = t^{2} + 2 t^{2} + 2 t = 3 t^{2} + 2 t$ .

Verify: direct $w = t \cdot t^{2} + t^{2} = t^{3} + t^{2}$ , toh $\frac{d w}{d t} = 3 t^{2} + 2 t$ . ✓ $+ 2 t$ explicit term drop karo toh galat $3 t^{2}$ milega.

Cell G — limiting behaviour

Worked example Ex 7 (Cell G)

$z = e^{x y}$ , $x = t$ , $y = \frac{1}{t}$ ( $t > 0$ ke liye). $\frac{d z}{d t}$ nikalo aur $t \to \infty$ par uski limit nikalo.

Forecast: yahan $x y = t \cdot \frac{1}{t} = 1$ hamesha, toh $z = e^{1} = e$ constant hai. Constant ki rate zero hoti hai. Guess $\frac{d z}{d t} = 0$ har jagah, toh limit $0$ hai.

$z = e^{x y}$ differentiate karne ke liye hum exponent ko ek single lump treat karte hain: $u = x y$ lo, toh $z = e^{u}$ . Single-variable fact $\frac{d}{d u} e^{u} = e^{u}$ (exponential apna khud ka derivative hota hai) phir ordinary chain rule se deta hai, $\frac{\partial z}{\partial x} = e^{u} \frac{\partial u}{\partial x} = e^{x y} \cdot y$ aur $\frac{\partial z}{\partial y} = e^{u} \frac{\partial u}{\partial y} = e^{x y} \cdot x$ . Yahan $e$ kyun? Kyunki variable exponent mein baitha hai, aur $e^{u}$ woh ek function hai jiska derivative khud hi hai — yahi tool hai jo jawab deta hai "kaise $e^{kuch}$ change hota hai?"
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = - \frac{1}{t ^{2}}$ . Minus kyun? $\frac{1}{t} = t^{- 1}$ ka derivative $- t^{- 2}$ hai.
$\frac{d z}{d t} = y e^{x y} (1) + x e^{x y} (- \frac{1}{t ^{2}})$ . Yeh step kyun? Do paths $t$ tak pahunchte hain ( $x$ ke through aur $y$ ke through), toh hum har path ki links multiply karte hain aur add karte hain — yeh Cell A ka chain rule sum hai. Ab $x = t, y = 1/ t, x y = 1$ substitute karo (woh constraint jo exponent ko constant banata hai): $= \frac{1}{t} e - t \cdot e \cdot \frac{1}{t ^{2}} = \frac{e}{t} - \frac{e}{t} = 0$ .
$t \to \infty$ par limit: yeh identically $0$ hai, toh limit $0$ hai.

Verify: direct $z = e^{t \cdot (1/ t)} = e^{1} = e$ , constant, derivative $0$ . ✓ Dono paths exactly cancel ho jaate hain — ek khoobsurat check ki chain rule hidden constraints ko respect karta hai.

Cell H — real-world word problem (units)

Worked example Ex 8 (Cell H)

Ek metal plate ka temperature $T (x, y) = x^{2} + 2 y$ degrees Celsius hai, jahan $x, y$ metres mein hain. Ek bug $x = 1 + t$ , $y = 2 - t$ ( $t$ seconds mein) ke saath walk karta hai. Bug $t = 0$ par temperature kitni tez feel kar raha hai?

Forecast: $+ x$ mein move karna garam karta hai (kyunki $T$ $x^{2}$ ke saath badhta hai), $- y$ mein move karna thanda karta hai (kyunki $T$ $y$ ke saath badhta hai). Net kisi bhi sign ka ho sakta hai — compute karo.

$\frac{\partial T}{\partial x} = 2 x (°C/m)$ , $\frac{\partial T}{\partial y} = 2 (°C/m)$ . Kyun? $x$ mein partial $y$ freeze karta hai: $2 y$ constant hai aur drop ho jaata hai, $2 x$ bachta hai; $y$ mein partial $x$ freeze karta hai: $x^{2}$ drop ho jaata hai aur $2 y$ $2$ ban jaata hai. Units °C per metre hain kyunki $T$ °C mein hai aur $x, y$ metres mein.
$\frac{d x}{d t} = 1 (m/s)$ , $\frac{d y}{d t} = - 1 (m/s)$ . Kyun? Ye bug ke path ke ordinary derivatives hain: $\frac{d}{d t} (1 + t) = 1$ aur $\frac{d}{d t} (2 - t) = - 1$ ; minus keh raha hai bug $- y$ direction mein move kar raha hai, aur units metres per second hain.
$\frac{d T}{d t} = 2 x (1) + 2 (- 1) = 2 x - 2 (°C/s)$ . Kyun? Dono paths sum karo; note karo ki units $\frac{°C}{m} \cdot \frac{m}{s} = \frac{°C}{s}$ line up karte hain.
$t = 0$ par: $x = 1 + 0 = 1$ , toh $\frac{d T}{d t} = 2 (1) - 2 = 0$ °C/s.

Verify: direct $T = (1 + t)^{2} + 2 (2 - t) = 1 + 2 t + t^{2} + 4 - 2 t = t^{2} + 5$ , toh $\frac{d T}{d t} = 2 t$ ; $t = 0$ par woh $0$ °C/s hai. ✓ Warming aur cooling shuru mein exactly balance karte hain.

Cell I — exam twist: Jacobian / matrix form

Pehle, notation ke do pieces, har ek use se pehle define kiye hue.

Summation symbol $\sum$ . Sign $k = 1 \sum n a_{k}$ shorthand hai "terms $a_{k}$ add karo jab counter $k$ $1$ se $n$ tak run kare": iska matlab bas $a_{1} + a_{2} + \dots + a_{n}$ hai. Hum ise isliye use karte hain kyunki chain rule ka "sab paths par add karo" exactly aisa hi running sum hai, ek term per intermediate variable $k$ .

Jacobian matrix $J$ . Ek Jacobian matrix ek table hai jo vector-valued map ke saare first partials ko ek grid mein collect karta hai: row $i$ hai "i-th output", column $k$ hai "k-th input ke respect mein". Hum ise $J$ likhte hain. Map $(x, y) \mapsto (u, v)$ ke liye,

$J_{(u, v)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} .$

Deep fact (prove kiya gaya hai parent note mein): do maps compose karne se unke Jacobians matrix multiplication se chain hote hain, $J_{outer \circ inner} = J_{outer} J_{inner}$ . Aur matrix multiplication exactly "paths par sum karo" hai: product ki entry $(i, j)$ hai $k \sum (J_{outer})_{ik} (J_{inner})_{k j}$ — har intermediate variable $k$ par sum, "outer rate through $k$ " times "inner rate to $k$ ". Woh sum-over- $k$ hi tree ke paths add up hain. Figure yeh dikhata hai: inner map ek input change ko intermediate slots mein push karta hai, aur outer map ki rows un pushes ko gather karti hain.

Worked example Ex 9 (Cell I)

$u = x + y$ , $v = x - y$ , with $x = r cos θ$ , $y = r sin θ$ . Matrix product $J_{(u, v)} J_{(x, y)}$ use karke Jacobian $\frac{\partial ( u , v )}{\partial ( r , θ )}$ nikalo.

Forecast: general chain rule hai $J_{u \circ x} = J_{u} J_{x}$ — matrix multiply karo. Dekho Gradient and Jacobian. Guess karo ki outer Jacobian constant matrix hai.

Inner map partials: $\frac{\partial x}{\partial r} = cos θ, \frac{\partial x}{\partial θ} = - r sin θ, \frac{\partial y}{\partial r} = sin θ, \frac{\partial y}{\partial θ} = r cos θ$ . Kyun? Har ek $x = r cos θ$ ya $y = r sin θ$ ka ordinary partial hai, doosre variable ko freeze karke. Toh $J_{(x, y)} = (cos θ sin θ - r sin θ r cos θ) .$
Outer map: $u = x + y$ , $v = x - y$ deta hai $J_{(u, v)} = (11 1 - 1)$ . Constant kyun? $u, v$ $x, y$ mein linear hain, toh unke partials numbers hain ( $\partial u / \partial x = 1$ , $\partial u / \partial y = 1$ , $\partial v / \partial x = 1$ , $\partial v / \partial y = - 1$ ).
Multiply karo — har entry intermediate variable par sum hai, yaani path sum $\sum_{k}$ : $J = (11 1 - 1) (cos θ sin θ - r sin θ r cos θ) = (cos θ + sin θ cos θ - sin θ - r sin θ + r cos θ - r sin θ - r cos θ) .$ Yeh step kyun? Outer matrix ki row $i$ times inner matrix ka column $j$ "outer rate through $x$ " aur "outer rate through $y$ " add karta hai — exactly input $j$ se output $i$ tak do paths.

Verify: directly substitute karo. $u = r cos θ + r sin θ$ , toh $\partial u / \partial r = cos θ + sin θ$ ✓ aur $\partial u / \partial θ = - r sin θ + r cos θ$ ✓. $v = r cos θ - r sin θ$ , toh $\partial v / \partial r = cos θ - sin θ$ ✓ aur $\partial v / \partial θ = - r sin θ - r cos θ$ ✓. Matrix product hi chain rule hai.

Recap

Recall Which cell sabse hard tha, aur kyun?

Cell E (degenerate) aur Cell F (explicit reappearance) zyaadatar logon ko trip karti hain. Tree mein paths ki sankhya kabhi nahi ghatti sirf isliye ki do inputs same variable ke barabar hain ::: sach — regardless of sab paths sum karo $w = f (x, y, t)$ mein jab $x, y$ $t$ par depend karte hain, kitne terms? ::: teen — do indirect plus ek explicit $\partial f / \partial t$

Connections

Parent topic
Partial Derivatives — $\partial z / \partial x$ links.
Total Differential — terms kyun add hote hain.
Directional Derivative — Cell B radial form.
Gradient and Jacobian — Cell I matrix product.
Implicit Differentiation (multivariable) — Cell F explicit term.
Single-variable Chain Rule — one-path base case.