Chain rule for multivariable functions — all cases: Question bank

4.4.7 · D5 · Maths › Multivariable Calculus › Chain rule for multivariable functions — all cases

Shuru karne se pehle, teen pictures apne dimaag mein rakho:

Ek dependency tree — output upar, arrows neeche "kis par depend karta hai" ki taraf.
Paths add hote hain, links multiply hote hain — output se target variable tak har path ke liye ek product.
$d$ vs $\partial$ — $d$ jab end result ek variable par depend kare, $\partial$ jab kaafi variables bachein.

Sahi hai ya galat — justify karo

Neeche har statement ya to sahi hai ya galat. Reveal mein reason diya hai, jo verdict se zyada zaroori hai.

z = f (x, y)

ke liye jab

x = x (t), y = y (t)

ho, to

\frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t}

hota hai.

Galat.

t

tak pahunchne ke do raaste hain (

x

ke through aur

y

ke through), isliye doosra product

\frac{\partial z}{\partial y} \frac{d y}{d t}

bhi add karna padega. Ek path bhool jaana sabse common galti hai.

Chain rule path-contributions isliye add karta hai kyunki total differential

d z = z_{x} d x + z_{y} d y

linear hai.

Sahi. Linear approximation ka matlab hai ki independent inputs swatantra roop se contribute karte hain, isliye unke effects superpose hote hain — differential mein jo "+" hai wahi chain rule mein "+" ban jaata hai.

Case 2 mein (

z = f (x, y)

x = x (s, t)

y = y (s, t)

) har derivative

\partial

se likhna zaroori hai.

Sahi. Final output abhi bhi do surviving variables

s

aur

t

par depend karta hai, isliye yahan kuch bhi single-variable derivative nahi hai —

\frac{\partial x}{\partial s}

ko bhi

\partial

chahiye kyunki

x

bhi

t

par depend karta hai.

Jacobian form

J_{u \circ x} = J_{u} J_{x}

mein factors ka order swap kiya ja sakta hai kyunki chain rule "bas multiply karna" hai.

Galat. Matrix multiplication commutative nahi hai, aur shapes sirf

J_{u} J_{x}

order mein fit hoti hain (outer map ka Jacobian left par). Swap karne par aksar dimension mismatch ya galat matrix milti hai.

Agar

w = f (x, y, t)

aur saath mein

x = x (t), y = y (t)

bhi ho, to

\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}

hoga.

Galat.

w

t

tak ek direct path bhi hai, isliye

+ \frac{\partial f}{\partial t}

term bhi add karni padegi.

t

explicitly aur

x, y

ke through dono taraf se

w

tak pahunchta hai.

Polar radial derivative

\frac{\partial z}{\partial r} = z_{x} cos θ + z_{y} sin θ

ek directional derivative hai.

Sahi.

(cos θ, sin θ)

unit radial direction hai, isliye yeh

\nabla f \cdot \overset{r}{^}

hai — chain rule quietly outward direction mein directional derivative compute kar deti hai.

Kyunki

\frac{d z}{d t}

mein

d

use hota hai, isliye iske andar ke saare derivatives mein bhi

d

hona chahiye.

Galat.

z

dono

x

aur

y

par depend karta hai, isliye

\frac{\partial z}{\partial x}

mein

\partial

hi rehta hai; sirf akele variable

t

ki final dependence (jaise

\frac{d x}{d t}

\frac{d z}{d t}

) ko

d

milta hai.

Agar

x

aur

y

kisi variable

s

par actually depend nahi karte, to woh path automatically zero contribute karta hai.

Sahi. Missing dependence se

\frac{\partial x}{\partial s} = 0

ho jaata hai, isliye woh product apne aap vanish ho jaata hai — general formula kaam karta rehta hai, bas dead paths ko khatam kar deta hai.

Chain rule ke liye

f

differentiable hona zaroori hai, sirf partial derivatives ka exist karna kaafi nahi.

Sahi. Derivation mein

Δ z = f_{x} Δ x + f_{y} Δ y + ε_{1} Δ x + ε_{2} Δ y

use hota hai jisme

ε \to 0

; yeh error-shrinking hi differentiability hai. Sirf

f_{x}, f_{y}

ka exist karna kaafi nahi.

Galti dhundho

Har line mein kaam ka ek (flawed) hissa hai. Reveal mein galti ka naam aur repair diya hai.

z = x^{2} y

x = cos t

y = sin t

, to

\frac{d z}{d t} = 2 x y \cdot (- sin t)

y

-path drop kar diya gaya. Poora answer

x^{2} cos t

bhi add karta hai, giving

\frac{d z}{d t} = 2 x y (- sin t) + x^{2} cos t

z = f (x, y)

ke liye jab

x = x (s, t)

ho, maine likha

\frac{d z}{d s} = \frac{d z}{d x} \frac{d x}{d s} + \frac{d z}{d y} \frac{d y}{d s}

Differential symbols galat hain. Kyunki

z

do survivors

s, t

par depend karta hai, yahan har derivative partial hai:

\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}

w = f (x, y, t)

x = x (t), y = y (t)

; answer hai

z_{x} x^{'} + z_{y} y^{'}

— bas itne hi paths hain."

t

ka direct explicit path miss ho gaya.

\frac{\partial f}{\partial t}

add karo; tree mein

t

tak pahunchne ke teen branches hain, do nahi.

z = r^{2}

ko chain rule se differentiate karte hue mujhe

2 r cos θ + 2 r sin θ

mila."

Pieces sahi se combine nahi ki gayin. Hona chahiye tha

2 r cos θ \cdot cos θ + 2 r sin θ \cdot sin θ = 2 r (cos^{2} θ + sin^{2} θ) = 2 r

— add karne se pehle har link ko uski apni inner derivative se multiply karna zaroori hai.

"Kyunki chain rule 'multiply karta hai', maine general case ke liye likha

\frac{\partial u}{\partial t _{j}} = \frac{\partial u}{\partial x _{i}} \frac{\partial x _{i}}{\partial t _{j}}

Sum missing hai. Har

x_{i}

ek alag path hai, isliye

\frac{\partial u}{\partial t _{j}} = \sum_{i = 1}^{n} \frac{\partial u}{\partial x _{i}} \frac{\partial x _{i}}{\partial t _{j}}

— ek path ke andar multiply karo aur paths ke across add karo.

"Maine differentiate karne se pehle

x = cos t

substitute kar diya, phir

\frac{\partial z}{\partial x}

use kiya."

Ek baar substitute karne ke baad,

z

pehle se hi

t

ka pure function ban jaata hai; ab reduced form se

\frac{\partial z}{\partial x}

ki baat nahi kar sakte. Ya to pehle partials lo (chain rule) ya substitute karke ek variable ki tarah differentiate karo — dono ko mix mat karo.

Kyun waale sawaal

Reason explain karo, sirf rule nahi.

Hum path contributions ko MULTIPLY karne ki jagah ADD kyun karte hain?

Kyunki total differential independent linear effects ka sum hai; multiply karna un contributions ko galat tarike se couple kar deta jo independently act karte hain. Hum multiply sirf ek single path ke andar karte hain (link by link).

General formula mein

\sum

j

ke upar nahi balki

i

(inner variables) ke upar kyun hai?

j

us ek target variable ko label karta hai jiske against differentiate karte hain (har equation ke liye fixed); sum

i

ke upar chhalta hai kyunki har inner variable

x_{i}

output se us target tak ek alag raasta hai.

Total differential ko

Δ t

se divide karke limit lene par "chain rule khud-ba-khud kyun mil jaata hai"?

Kyunki error terms

ε_{1} \frac{Δ x}{Δ t} + ε_{2} \frac{Δ y}{Δ t}

vanish ho jaate hain —

ε \to 0

jabki

\frac{Δ x}{Δ t}

finite rehta hai; jo bachta hai woh exactly

z_{x} x^{'} + z_{y} y^{'}

hai.

Jacobian view chain rule ko "bas matrix multiplication" kyun kehta hai?

Composition ki har layer locally ek linear map hai (uska Jacobian); maps ko compose karna unke linear parts ko compose karna hai, aur linear maps ko compose karna matrix multiplication hai — isliye

J_{u \circ x} = J_{u} J_{x}

Derivation mein

x

aur

y

continuous (sirf differentiable nahi) kyun hone chahiye?

Δ t \to 0

Δ x, Δ y \to 0

" wala step continuity hai. Iske bina, inner changes shrink nahi kar sakti, aur error terms vanish nahi honge.

Case 2,

s

ke against differentiate karne par Case 1 mein kyun reduce ho jaata hai?

t

ko fixed rakhne se map single variable

s

ka function ban jaata hai; us frozen slice ke along yeh literally ek-independent-variable chain rule hai jo do baar apply hoti hai (ek baar har inner variable ke liye).

Edge cases

Jahan rule zeros, degeneracies, aur boundaries se milta hai.

\frac{d z}{d t}

kya hoga agar

z = f (x, y)

ho lekin

x

constant ho (

t

par depend na kare)?

Tab

\frac{d x}{d t} = 0

ho jaata hai, woh path khatam; sirf

\frac{\partial z}{\partial y} \frac{d y}{d t}

bachta hai. Formula khud sahi ho jaata hai — koi term haath se delete nahi karni padti.

Polar coordinates mein origin par (

r = 0

), kya

\frac{\partial z}{\partial r} = z_{x} cos θ + z_{y} sin θ

abhi bhi meaningful hai?

Formula theek hai, lekin

θ

r = 0

par undefined hai, isliye radial direction ambiguous hai; value is baat par depend kar sakti hai ki kis

θ

se approach kiya — yeh coordinate map ki ek sachchi geometric singularity signal karta hai.

Agar koi variable explicitly aur ek inner variable ke through dono jagah appear kare, aur uske dono contributions cancel ho jaayein, to iska kya matlab hai?

Iska matlab hai output us variable ke liye locally insensitive hai — explicit aur indirect effects offset ho gaye. Math sahi hai; zero ek real feature hai, jaise koi quantity jo motion ke along conserve ho.

Kya chain rule tab bhi apply ho sakta hai agar

f

ke partials exist karte hon lekin

f

us point par differentiable na ho?

Koi guarantee nahi. Partials ka exist karna differentiability se weaker condition hai; standard chain rule wahan fail ho sakta hai, isliye differentiability hypothesis decorative nahi hai.

J_{u \circ x} = J_{u} J_{x}

mein kya hoga jab

u

ek single number (scalar field) par map kare?

J_{u}

ek single row ban jaati hai (gradient

\nabla u

row vector ke roop mein), aur product

\nabla u J_{x}

composed partials ki ek row return karta hai — scalar case bas general matrix rule hai jisme ek output row hai.

Agar do alag inner variables ek jaisi functions hon, maano

x = y = t^{2}

, to kya hum phir bhi do path terms likhenge?

Haan. Tree structure, numerical coincidence nahi, paths dictate karta hai; tum

z_{x} \frac{d x}{d t} + z_{y} \frac{d y}{d t}

likhte ho aur tab equal values substitute karte ho — pehle collapse karne se ek term chhoot sakti hai.

Connections

4.4.07 Chain rule for multivariable functions — all cases — woh parent jise yeh bank stress-test karta hai.
Partial Derivatives — $d$ vs $\partial$ distinction yahan rehti hai.
Total Differential — "paths add" ka source.
Directional Derivative — polar edge case.
Gradient and Jacobian — matrix-order trap.
Implicit Differentiation (multivariable) — explicit-reappearance trap.
Single-variable Chain Rule — one-path baseline jisse yeh traps deviate karte hain.