4.1.16 · Maths › Calculus I — Limits & Derivatives
Intuition Core idea (WHY yeh exist karta hai)
Ek composite function f ( g ( x )) ek machine ke andar machine hai. Tum x ko thoda sa wiggle karte ho, inner machine g us wiggle ko apni rate g ′ ( x ) se amplify karti hai, phir outer machine f us wiggle ko apni rate f ′ ( g ( x )) se amplify karti hai. Total amplification = dono amplifications ka product.
Rates multiply hote hain kyunki yeh sequence mein act karte hain, jaise ek gearbox mein gear ratios hote hain.
Agar g , x par differentiable hai aur f , g ( x ) par differentiable hai, toh composite h ( x ) = f ( g ( x )) , x par differentiable hai aur
h ′ ( x ) = f ′ ( g ( x )) ⋅ g ′ ( x ) .
Leibniz notation mein, u = g ( x ) aur y = f ( u ) ke saath:
d x d y = d u d y ⋅ d x d u .
Leibniz form mein multiplication cancelling fractions jaisi lagti hai — yeh ek mnemonic hai, proof nahi (yeh literal fractions nahi hain).
Common mistake Steel-man: "
Δ u cancel karo — proof ek line mein ho jaata hai!"
Kyun sahi lagta hai: naïve cancellation se correct answer milta hai, aur d y / d x = ( d y / d u ) ( d u / d x ) fraction cancellation jaisa dikhta hai.
Kyun galat hai: Δ u bahut saare Δ x = 0 ke liye 0 ho sakta hai (jaise g ( x ) = x 2 sin ( 1/ x ) near 0 , ya koi bhi constant patch). Tab Δ u se divide karna undefined hai.
Fix: E ( k ) "error function" multiply karta hai — kabhi divide nahi karta — toh Δ u = 0 par bhi survive karta hai. Same answer milta hai, lekin har jagah legal hai.
h ( x ) = ( 3 x 2 + 1 ) 5
Outer f ( u ) = u 5 , inner g ( x ) = 3 x 2 + 1 .
f ′ ( u ) = 5 u 4 — kyun? outer shell par power rule.
g ′ ( x ) = 6 x — kyun? inside ki derivative.
h ′ ( x ) = 5 ( 3 x 2 + 1 ) 4 ⋅ 6 x = 30 x ( 3 x 2 + 1 ) 4 .
Yeh step kyun? u = g ( x ) ko f ′ ( u ) mein plug back karo, phir g ′ ( x ) se multiply karo.
h ( x ) = sin ( x 2 )
f ( u ) = sin u ⇒ f ′ ( u ) = cos u ; g ( x ) = x 2 ⇒ g ′ ( x ) = 2 x .
h ′ ( x ) = cos ( x 2 ) ⋅ 2 x = 2 x cos ( x 2 ) .
Yeh step kyun? Outer sin ko differentiate karo inside ko frozen rakhke, phir inside ki rate se multiply karo. Note karo cos ( x 2 ) = cos ( 2 x ) — inside ko intact rakho!
h ( x ) = e s i n x
f ( u ) = e u , f ′ ( u ) = e u ; g ( x ) = sin x , g ′ ( x ) = cos x .
h ′ ( x ) = e s i n x cos x .
Worked example 4. Triple nesting:
h ( x ) = cos ( 4 x ) = ( cos 4 x ) 1/2
Do baar chain karo: outermost power, phir cos , phir 4 x .
h ′ ( x ) = 2 1 ( cos 4 x ) − 1/2 ⋅ ( − sin 4 x ) ⋅ 4 = c o s 4 x − 2 s i n 4 x .
Yeh step kyun? Har layer ek factor contribute karti hai; saare layer-rates ko multiply karo (onion ko outside-in peel karo).
Worked example 5. Ek known rule ke against verify karo —
h ( x ) = ( x 2 ) 3 = x 6
Chain: 6 ( x 2 ) 2 ⋅ 2 x = 6 x 4 ⋅ 2 x = 12 x 5 . Direct: d x d x 6 = 6 x 5 .
Mismatch?! Re-check karo: outer f ( u ) = u 3 , f ′ = 3 u 2 , toh 3 ( x 2 ) 2 ⋅ 2 x = 3 x 4 ⋅ 2 x = 6 x 5 . ✓
Yeh step kyun? Forecast-then-verify ne meri arithmetic slip pakdi (maine f ′ ke liye 3 ki jagah 6 likha tha) — hamesha f ′ ko re-derive karo.
Recall Feynman: ek 12-saal ke bacche ko samjhao
Socho ek bike jisme do gears connected hain. Tum thoda pedal karte ho (yeh hai x ). Pehla gear teri pedaling se 3× tez ghoomta hai. Doosra gear pehle gear se 5× tez ghoomta hai. Toh pichla wheel teri feet se 5 × 3 = 15 times tez ghoomta hai. Chain rule bas yahi kehta hai: total speed-up paane ke liye, har gear ki speed-up ko multiply karo. f ′ ( g ( x )) doosre gear ka ratio hai, g ′ ( x ) pehle gear ka ratio hai.
Mnemonic "Outside-in, peel the onion"
D erivative of the O utside (inside ko akela chhodo), T imes derivative of the I nside.
DOTI : "Do the Outside, Then the Inside (aur multiply karo)."
Recall Active recall checkpoint
Answers cover karo. Kya tum (a) dono notations state kar sakte ho, (b) explain kar sakte ho kyun naïve cancellation fail hoti hai, (c) bina dekhke sin ( x 2 ) differentiate kar sakte ho?
What is the chain rule for h = f ( g ( x )) ? h ′ ( x ) = f ′ ( g ( x )) g ′ ( x ) .
Chain rule ko Leibniz notation mein state karo. d x d y = d u d y ⋅ d x d u with u = g ( x ) .
Proof mein hum simply Δ u se multiply/divide kyun nahi kar sakte? Δ u = g ( x + Δ x ) − g ( x ) , 0 ho sakta hai even jab Δ x = 0 ho, jisse division undefined ho jaati hai.
g ki kaun si property guarantee karti hai ki Δ x → 0 ke saath Δ u → 0 ?g ki differentiability implies g ki continuity.
Rigorous proof mein E ( k ) kya hai aur isko kyun use kiya jaata hai? Error E ( k ) = k f ( u + k ) − f ( u ) − f ′ ( u ) (with E ( 0 ) = 0 ); yeh humein divide karne ki jagah multiply karne deta hai, Δ u = 0 se bachata hai.
sin ( x 2 ) differentiate karo.2 x cos ( x 2 ) .
e s i n x differentiate karo.e s i n x cos x .
( 3 x 2 + 1 ) 5 differentiate karo.30 x ( 3 x 2 + 1 ) 4 .
Triple nesting mein kitne factors aate hain? Ek per layer; saari layer-derivatives ko multiply karo, har outer ko relevant inner value par evaluate karke.
sin ( x 2 ) differentiate karte waqt common slip kya hai?cos ( 2 x ) likhna — inside ko intact rakhna hoga: cos ( x 2 ) , phir 2 x se multiply karo.
Derivative — limit definition — chain rule directly difference-quotient limit se build hoti hai.
Continuity implies differentiability fails converse — hum g ke liye "differentiable ⇒ continuous" use karte hain.
Product rule aur Quotient rule — sibling differentiation rules; aksar chain rule ke saath combine hote hain.
Implicit differentiation — literally chain rule hai y ( x ) par apply ki hui.
Related rates — physical applications jahan d t d y = d x d y d t d x .
Inverse function derivative — f ( f − 1 ( x )) = x ko chain rule karke derive hoti hai.
wiggle amplified in sequence
multiply and divide by delta u
h' = f' of g x times g' x
dy/dx = dy/du times du/dx
Fraction cancelling mnemonic
Derivative definition limit