Chain rule — proof, composite function derivatives: Question bank

4.1.16 · D5 · Maths › Calculus I — Limits & Derivatives › Chain rule — proof, composite function derivatives

Traps se pehle, ek refresher taaki kuch unexplained na rahe:

True or false — justify

TF1. "

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

true hai kyunki

d u

's fractions ki tarah cancel ho jaate hain."

False — equation sahi hai, lekin reason nahi.

d y / d u

aur

d u / d x

single limits hain, shared factor wale fractions nahi; cancellation sirf ek memory aid hai, jo theek se error-function argument se prove hoti hai.

TF2. "Agar

f

aur

g

dono har jagah differentiable hain, toh

f \circ g

bhi har jagah differentiable hai."

True — chain rule ki hypotheses (

g

x

par differentiable,

f

g (x)

par differentiable) har

x

par poori hoti hain, isliye composite wahan differentiable hai product formula ke saath.

TF3. "Chain rule fail ho sakta hai even if answer

f^{'} (g (x)) g^{'} (x)

ek number ke roop mein exist karta hai."

True —

f

ki differentiability point $g (x)$ par honi chahiye, sirf kahi par nahi. Agar

f

us particular

u

par differentiable nahi hai, toh formula ki value meaningless hai, chahe wo compute ho bhi jaye.

TF4. "

\frac{d}{d x} sin (x^{2}) = cos (2 x)

False — inside ko intact rakhna aur uski rate se multiply karna zaroori hai:

cos (x^{2}) \cdot 2 x

cos (2 x)

galat tarike se angle ko double karta hai

x^{2}

rakhne aur

g^{'} = 2 x

se multiply karne ki jagah.

TF5. "Rigorous proof mein, error function

E

Δ u

se divide karti hai, isliye yeh tab bhi tootti hai jab

Δ u = 0

False —

E

defined hai ki

s = 0

par

0

ke barabar hai, aur key identity

f (u + s) - f (u) = (f^{'} (u) + E (s)) s

s

se multiply karti hai. Kabhi bhi

Δ u

se divide nahi hota.

TF6. "

g

ki differentiability sirf

g^{'} (x)

appear karaane ke liye chahiye;

g

ki continuity ka koi role nahi."

False —

g

ki continuity hi guarantee karti hai ki

Δ u \to 0

jab

Δ x \to 0

, aur yahi cheez

E (Δ u) \to E (0) = 0

hone deti hai. Iske bina proof ruk jaata hai.

TF7. "Constant inner function

g (x) = c

ke liye, naïve

Δ u

-cancellation proof theek kaam karta hai."

False — yahan har

Δ x

ke liye

Δ u = 0

hai, isliye

Δ u

se divide karna hamesha undefined hai; sirf error-function proof survive karta hai, aur sahi tarike se

h^{'} (x) = f^{'} (c) \cdot 0 = 0

deta hai (maante hue ki

f

c

par differentiable hai).

TF8. "Agar

h = f \circ g

x

par differentiable hai, toh

f

aur

g

dono wahan differentiable hone chahiye."

False — ek composite smooth ho sakta hai chahe ek piece na ho; jaise

g (x) = ∣ x ∣

(

0

par differentiable nahi) ko

f (u) = u^{2}

mein feed karo toh

h (x) = x^{2}

milta hai, jo

0

par differentiable hai. Chain rule sufficient hai, necessary nahi.

Why questions

WH1. Do rates multiply kyun karte hain instead of add?

Kyunki wiggles gears ki tarah sequence mein act karte hain:

g

Δ x

Δ u \approx g^{'} Δ x

mein scale karta hai, phir

f

use

Δ y \approx f^{'} Δ u

mein scale karta hai, isliye

Δ y \approx f^{'} g^{'} Δ x

— compounding hota hai, isliye product.

WH2. Proof Leibniz form ki jagah derivative ki limit definition se kyun start karta hai?

Leibniz "cancellation" sirf ek mnemonic hai; ek genuine derivation mein

h^{'}

ko difference-quotient limit tak reduce karna padta hai, jo derivative ki actual definition hai.

WH3. Error function

E

0

par continuous kyun banaya jaata hai (

E (0) = 0

set karke)?

Taaki single identity

f (u + s) - f (u) = (f^{'} (u) + E (s)) s

s = 0

samete sabhi

s

ke liye hold kare, aur taaki

E (Δ u) \to E (0) = 0

jab

Δ u \to 0

— har division-by-zero ki chinta khatam.

WH4.

f^{'}

x

par nahi balki

g (x)

par kyun evaluate karte hain?

Outer machine kabhi bhi sirf number

u = g (x)

dekhti hai,

x

khud nahi. Uski slope wahan read honi chahiye jo actually usmein enter ho raha hai, yaani

g (x)

WH5. "

g

differentiable

\Rightarrow

g

continuous" "

g

differentiable" se zyada kyun matter karta hai?

Product formula ko

g^{'} (x)

(differentiability) aur

Δ u \to 0

(continuity) dono chahiye taaki error term khatam ho sake. Continuity wo bridge hai jo

E (Δ u)

ki limit ko

0

pe collapse karati hai.

WH6. Implicit differentiation "bas kaam" kyun karta hai bina kisi naye rule ke?

Kyunki yeh chain rule hi hai:

y

y (x)

maanke,

y

mein kisi bhi expression ko differentiate karne par exactly chain rule ki demand ke mutabiq

\frac{d y}{d x}

ka factor aata hai (dekho Implicit differentiation).

WH7. Yahi trick Inverse function derivative prove karne ke liye kyun kaam karti hai?

Identity

f (f^{- 1} (x)) = x

par chain rule lagao: milta hai

f^{'} (f^{- 1} (x)) \cdot (f^{- 1})^{'} (x) = 1

, isliye

(f^{- 1})^{'} (x) = 1/ f^{'} (f^{- 1} (x))

— inverse-derivative formula seedha nikal aata hai.

Edge cases

EC1.

h^{'} (x)

kya hoga jab inner function constant hai,

g (x) = c

(

f

c

par differentiable hai)?

h (x) = f (c)

khud constant hai, isliye

h^{'} (x) = 0

; formula bhi agree karta hai kyunki

g^{'} (x) = 0

deta hai

f^{'} (c) \cdot 0 = 0

. (Agar

f

c

par differentiable na hota, toh formula simply apply nahi hota — lekin

h

phir bhi constant hai, isliye

h^{'} = 0

directly.)

EC2. Us point par kya hoga jahan

g^{'} (x) = 0

lekin

f^{'} (g (x)) \neq = 0

Composite ka wahan zero slope hoga:

h^{'} (x) = f^{'} (g (x)) \cdot 0 = 0

. Ek stationary inner wiggle koi output wiggle produce nahi karta, chahe

f

kitni bhi steep ho.

EC3. Kya

g^{'} (x) = 0

hote hue

h^{'} (x) \neq = 0

ho sakta hai?

Chain rule ke zariye nahi — jab bhi uski hypotheses hold hoti hain, formula force karta hai

h^{'} = f^{'} \cdot 0 = 0

. Agar

h^{'} \neq = 0

observe ho, toh koi hypothesis (differentiability of

f

g (x)

) fail hui hogi, isliye formula kabhi apply hua hi nahi.

EC4. Kya chain rule

g (x) = x^{2} sin (1/ x)

(

g (0) = 0

ke saath) par

x = 0

par apply hoti hai?

Haan agar

g

0

par differentiable hai aur

f

g (0)

par — aur yahi woh pathological case hai jahan infinitely many

Δ x

ke liye

Δ u = 0

hota hai, jo exactly dikhata hai kyun error-function proof (naïve cancellation nahi) zaroori hai.

EC5. Triple nest

f (g (p (x)))

ke liye, kitne factors hain aur har derivative kahan evaluate hota hai?

Teen factors:

f^{'} (g (p (x))) \cdot g^{'} (p (x)) \cdot p^{'} (x)

— har outer derivative us value par read hota hai jo actually usmein feed hoti hai, sab multiply kiye jaate hain (bahar se andar peelo). (Innermost function ko hum $p$ naam dete hain taaki error-function argument se confuse na ho.)

EC6. Agar

g

x_{0}

par corner hai (differentiable nahi) lekin

f

smooth hai, toh kya

f \circ g

necessarily

x_{0}

par non-differentiable hai?

Nahi — smoothing ho sakti hai.

f (u) = u^{2}

aur

g (u) = ∣ u ∣

ke saath, composite

x^{2}

0

par bilkul differentiable hai; chain rule simply apply nahi hoti, lekin composite direct check se phir bhi differentiable ho sakti hai.

Connections

Chain rule — proof, composite function derivatives — parent; yahan ke har trap us proof ke ek step ko target karta hai.
Derivative — limit definition — WH2 aur SE3 difference-quotient definition par hinge karte hain.
Continuity implies differentiability fails converse — WH5, EC6 "differentiable ⇒ continuous" use karte hain.
Implicit differentiation — WH6 ise chain rule ke roop mein recast karta hai.
Inverse function derivative — WH7 ise $f (f^{- 1} (x)) = x$ par chain-rule karke derive karta hai.
Product rule, Quotient rule — sibling rules jo aksar chain rule ke saath combine hoti hain.

Question bank — Chain rule — proof, composite function derivatives