Partial derivatives — notation, calculation, geometric meaning: Question bank

4.4.3 · D5 · Maths › Multivariable Calculus › Partial derivatives — notation, calculation, geometric meani

True or false — justify

Claim ya toh true hai ya false. Tumhara kaam hai kyun batana.

TF1.

\frac{\partial f}{\partial x}

aur

\frac{df}{d x}

ka matlab

f (x, y)

ke liye ek hi hai.

False. Seedha

d

demand karta hai ki

f

sirf

x

par depend kare; curly

\partial

signal deta hai ki doosre variables (

y

yahan) held fixed hain jabki

x

change hota hai.

d

use karne se ye chhup jaata hai ki ek poora variable freeze kiya gaya tha.

TF2. Agar

f_{x} (a, b) = 0

aur

f_{y} (a, b) = 0

, toh

f

ka wahan flat tangent plane hai.

True. Dono slice-slopes zero hain, toh tangent plane

z = f (a, b) + 0 \cdot (x - a) + 0 \cdot (y - b)

horizontal hai. (Lekin flat

\neq =

maximum — ye saddle bhi ho sakta hai.)

TF3. Koi function ek point par

f_{x}

exist karte hue bhi wahan discontinuous ho sakta hai.

True.

f_{x}

sirf us point ke through $x$ -axis line ke saath behaviour inspect karta hai; function doosre directions mein bura behave kar sakta hai aur phir bhi ek achha one-directional slope rakh sakta hai.

TF4. Agar kisi region par har jagah

f_{x} = f_{y} = 0

ho, toh

f

us region par constant hai.

True (ek connected region par). Dono

x

aur

y

directions mein zero slope hone par plane mein kahin bhi move karne par height change ki koi jagah nahi bachti.

TF5. Kisi bhi

y

ki akeli function

g

ke liye

\partial_{x} (g (y)) = 0

hoga.

True.

x

-derivative ke liye, sirf

y

se bani koi bhi cheez ek frozen constant hai, aur constants ka derivative zero hota hai — chahe

g

kitna bhi complicated kyun na lage.

TF6. Mixed partials hamesha

f_{x y} = f_{y x}

satisfy karte hain.

False. Sirf tab jab second partials continuous hon (Clairaut's / Schwarz condition). Aisi engineered functions hain jahan dono mixed partials ek single bad point par alag ho jaate hain.

TF7. Partial

f_{x}

khud generally dono

x

aur

y

ki function hoti hai.

True. Differentiate karne ke liye

y

ko freeze karna

y

ko delete nahi karta; jaise

f = x^{2} y

ke liye,

f_{x} = 2 x y

mein abhi bhi

y

hai. Tumne use operation ke dauraan freeze kiya, hamesha ke liye nahi.

TF8.

f_{y}

lene se pehle

f = \frac{x z}{y}

x z \cdot y^{- 1}

mein rewrite karna ek legal trick hai.

True. Ye wahi function hai, aur power rule

\frac{d}{d y} y^{- 1} = - y^{- 2}

(

x z

ke frozen hone ke saath) quotient rule se zyada clean hai. Rewriting se derivative kabhi nahi badlta.

Spot the error

Har ek ek galat worked line hai. Exact mistake batao.

SE1. "

f = x^{2} y \Rightarrow f_{x} = 2 x

Frozen

y

drop kar diya gaya. Ye ek constant multiplier hai, absent term nahi, isliye

f_{x} = 2 x y

hoga.

SE2. "

f = x^{3} - 7 y \Rightarrow f_{x} = 3 x^{2} - 7

- 7 y

term mein koi

x

nahi hai, isliye

x

ke w.r.t. ye ek constant hai aur iska derivative

0

hai,

- 7

nahi. Sahi:

f_{x} = 3 x^{2}

SE3. "

f = x^{2} y

ke liye

f_{x} (1, 2)

paane ke liye pehle plug in karo:

f (1, 2) = 2

, phir

2

ko differentiate karo to get

0

Differentiate karne se pehle substitute karna variable ko ek number mein badal deta hai aur uski slope destroy kar deta hai. Pehle differentiate karo (

f_{x} = 2 x y

), baad mein evaluate karo (

f_{x} (1, 2) = 4

SE4. "

f = sin (x y^{2}) \Rightarrow f_{x} = cos (x y^{2})

Chain rule ka inner derivative bhool gaye. Inner

x y^{2}

hai jiska

x

-derivative

y^{2}

hai, isliye

f_{x} = cos (x y^{2}) \cdot y^{2}

SE5. "

f = e^{x} cos y \Rightarrow f_{y} = e^{x} cos y

Unhone galat factor differentiate kiya.

e^{x}

ke frozen hone ke saath,

\frac{d}{d y} cos y = - sin y

, isliye

f_{y} = - e^{x} sin y

SE6. "Kyunki

f_{x y}

ka matlab hai '

x

ke w.r.t. phir

y

ke w.r.t. differentiate karo',

f = x^{2} y^{3}

ke liye hum paate hain

f_{x y} = \partial_{x} (\partial_{y} f)

Order/reading ki galti.

f_{x y}

ka matlab hai pehle

x

phir

y

\partial_{y} (\partial_{x} f)

. (Yahan ye Clairaut ki wajah se

f_{y x}

ke barabar hai, lekin definition phir bhi matter karti hai.)

SE7. "

f = \frac{x}{y}

y = 0

ke paas,

f_{y} = - \frac{x}{y ^{2}}

hai, jo har jagah theek hai."

Formula wahan sahi hai jahan $f$ defined hai, lekin

y = 0

par function khud undefined hai, isliye wahan partial evaluate nahi ki ja sakti — tum koi aisi value freeze nahi kar sakte jo function kabhi leta hi nahi.

Why questions

WHY1. Surface

z = f (x, y)

f (x)

ki tarah koi ek single "slope" kyun nahi hota?

Ek point par tum infinitely many directions mein chal sakte ho, har ek alag rate se upar jaata hai. Ek single number unhe sab describe nahi kar sakta, isliye hum separately measure karne ke liye special

x

aur

y

directions chunte hain.

WHY2. Partial derivative "ek saadha derivative hi disguise mein" kyun hai?

y = b

freeze karna

f (x, y)

ko ek single-variable function

g (x) = f (x, b)

mein badal deta hai; phir

f_{x} (a, b) = g^{'} (a)

, isliye Single-variable derivative ke saare rules unchanged apply hote hain.

WHY3. Poora tangent plane likhne ke liye dono partials

f_{x}, f_{y}

kaafi kyun hain?

Plane ko dono slice-slopes reproduce karne honge; uski

x

-slope ko

f_{x}

aur

y

-slope ko

f_{y}

banane par wo completely pin ho jaata hai, deta hai

z = f (a, b) + f_{x} (x - a) + f_{y} (y - b)

WHY4.

f_{x}, f_{y}

ko gradient

\nabla f

mein kyun bundle karte hain?

Kyunki do axis directions mein slope jaanna tumhe kisi bhi direction mein slope reconstruct karne deta hai (Directional derivative), aur gradient wo vector hai jo ye dot product se karta hai.

WHY5. Total differential

df = f_{x} d x + f_{y} d y

hai, sirf

df = f_{x} d x

kyun nahi?

Ek chhota move dono inputs ko change karta hai; har ek apni slope times apna step contribute karta hai, aur linear approximation in dono effects ko jod deta hai.

WHY6. Clairaut ki symmetry

f_{x y} = f_{y x}

geometrically believable kyun hai?

Dono surface ka wahi "twist" measure karte hain — ek direction mein slope kaise change hota hai jab tum doosre mein shift karte ho. Twist ko kisi bhi order se measure karna usi bend ko describe karta hai.

WHY7. Multivariable chain rule ko sabhi partials ki zaroorat kyun hai, sirf ek ki nahi?

Agar

x

aur

y

dono

t

par depend karte hain, toh

f

ka total change har variable ke pathway se ek contribution collect karta hai:

\frac{df}{d t} = f_{x} \frac{d x}{d t} + f_{y} \frac{d y}{d t}

WHY8.

\partial

("curly d") notation apne symbol ki zaroorat kyun hai?

Ye ek standing reminder hai ki doosre variables exist karte hain aur held constant hain. Ek plain

d

silently pretend karta ki

f

one-dimensional hai, bilkul wahi "main doosra variable bhool gaya" wali error ko invite karta.

Edge cases

EC1.

f (x, y) = x^{2} + y^{2}

ke liye

f_{z}

kya hai agar hum suddenly ek

z

-slot allow karein?

0

. Function mein koi

z

dependence nahi hai, isliye

z

wiggle karne se kuch nahi badlta — iska partial identically zero hai, ek absent variable ke liye honest answer.

EC2. Kya

f_{x} (a, b)

exist kar sakta hai jab

f_{y} (a, b)

exist nahi karta?

Haan. Dono directions independently inspect ki jaati hain;

x

ke saath slice smooth ho sakti hai jabki

y

ke saath slice usi point par corner ya jump rakh sakti hai.

EC3. Constant function

f (x, y) = 5

ke liye

f_{x}

kaisa dikhega?

f_{x} = 0

aur

f_{y} = 0

har jagah. Surface ek flat sheet hai; kisi bhi input mein koi wiggle height nahi badlta.

EC4.

(0, 0)

par

f (x, y) = ∣ x ∣ + y

ke liye, kya

f_{x}

exist karta hai?

Nahi.

x

-slice

z = ∣ x ∣

hai, jiska

x = 0

par ek corner hai — left slope

- 1

aur right slope

+ 1

disagree karte hain, isliye

f_{x}

define karne wali limit fail ho jaati hai. Lekin

f_{y} = 1

theek se exist karta hai.

EC5. Agar

f_{x} (a, b) > 0

hai, toh kya poori surface

(a, b)

ke paas rise kar rahi hai?

Zaroorat nahi. Ye sirf guarantee karta hai ki

+ x

direction mein step karne par rise hoga; surface saath hi

y

direction mein fall bhi kar sakti hai (ek saddle-like slope pattern).

EC6.

f = ln x

ki (

f (x, y) = ln x

ke roop mein dekha gaye)

x = 0

par

f_{x}

ka kya hoga?

Undefined.

f_{x} = 1/ x

x \to 0^{+}

par blow up karta hai aur

ln x

x \leq 0

ke liye defined bhi nahi hai, isliye wahan partial ki koi value nahi — ye ek genuine domain boundary hai.

EC7. Kya kisi isolated point par

f_{x}

computable hai jahan sirf

f (a, b)

pata ho?

Nahi. Limit definition ko chhote

h

ki poori range ke liye

f (a + h, b)

values chahiye; ek single point koi neighbouring heights compare karne ko nahi deta, isliye koi slope form nahi ho sakti.

Active Recall

Recall Ek trap jo sabko kam se kam ek baar pakad leta hai

Q: $f_{x}$ mein, constant term ka coefficient jaise $- 7 y$ rakhna galat kyun hai? A: $x$ ke relative, poora term $- 7 y$ mein koi $x$ nahi hai, isliye ye ek constant hai — aur kisi bhi constant ka derivative $0$ hota hai. $y$ ko freeze karna $- 7 y$ ko ek fixed number ki tarah behave karata hai.

Connections

Gradient vector — partials ko bundle karta hai; WHY4 dekho.
Directional derivative — WHY1 ka "any direction" answer.
Tangent plane and linear approximation — WHY3, TF2.
Chain rule (multivariable) — WHY7.
Total differential — WHY5.
Single-variable derivative — WHY2, frozen-variable reduction.
Clairaut's theorem — TF6, WHY6.