4.1.13
Differentiation ek linear operation hai. Ye ek akela word is note ki poori baat pack kar leta hai: ek combination of functions ka derivative sirf unke derivatives ka wahi combination hota hai. Aapko kabhi wapas limit definition pe jaane ki zaroorat nahi hai kisi sum jaise x 2 + sin x
WHY does this matter? Lagbhag har function jo aap dekhte ho woh simpler pieces ko add karne, subtract karne, aur scale karne se bana hota hai. Agar derivatives is building process ko respect karti hain, toh calculus modular ban jaata hai. Karti hain — aur ye teen rules iska proof hain.
Definition The three linearity rulesMaano f g x c
Sum rule: d x d [ f ( x ) + g ( x ) ] === f ′ ( x ) + g ′ ( x ) == Difference rule: d x d [ f ( x ) − g ( x ) ] === f ′ ( x ) − g ′ ( x ) == Constant multiple rule: d x d [ c f ( x ) ] === c f ′ ( x ) ==
Ek linearity statement mein combine karke:
d x d [ a f ( x ) + b g ( x ) ] = a f ′ ( x ) + b g ′ ( x )
WHAT we start from. Sirf ek cheez assume karne ki permission hai — derivative ki definition:
h ′ ( x ) = lim Δ x → 0 Δ x h ( x + Δ x ) − h ( x ) limit laws : sum ka limit, limits ka sum hota hai; aur constants limits se bahar aa jaate hain.
Maano h ( x ) = f ( x ) + g ( x )
h ′ ( x ) = lim Δ x → 0 Δ x [ f ( x + Δ x ) + g ( x + Δ x ) ] − [ f ( x ) + g ( x ) ]
Why this step? Maine sirf h = f + g
Ab numerator ko regroup karo, f f
= lim Δ x → 0 [ Δ x f ( x + Δ x ) − f ( x ) + Δ x g ( x + Δ x ) − g ( x ) ]
Why this step? Addition commutative hoti hai, toh main charon terms ko do difference quotients mein shuffle kar sakta hoon. Yehi poora trick hai.
Sum law for limits apply karo (allowed kyunki dono limits assumption ke hisaab se exist karti hain):
= f ′ ( x ) Δ x → 0 lim Δ x f ( x + Δ x ) − f ( x ) + g ′ ( x ) Δ x → 0 lim Δ x g ( x + Δ x ) − g ( x ) = f ′ ( x ) + g ′ ( x ) ■
Maano h ( x ) = c f ( x )
h ′ ( x ) = lim Δ x → 0 Δ x c f ( x + Δ x ) − c f ( x ) = lim Δ x → 0 c ⋅ Δ x f ( x + Δ x ) − f ( x )
Why this step? Numerator se c
= c lim Δ x → 0 Δ x f ( x + Δ x ) − f ( x ) = c f ′ ( x ) ■
Why this step? Ek constant factor limit ke through pass ho jaata hai — woh Δ x
f − g = f + ( − 1 ) ⋅ g Sum rule apply karo, phir constant multiple rule c = − 1 d x d [ f − g ] = f ′ + ( − 1 ) g ′ = f ′ − g ′ ■
Intuition Why difference is "free"Subtraction bas ek negative add karna hai. Jab ek baar linearity aa jaaye, difference alag se prove nahi karna padta — woh khud nikal aata hai. Yahi 80/20 lesson hai: sum + constant rules master karo; baaki sab recombination hai.
Worked example Example 1 — a polynomialy = 3 x 4 − 5 x 2 + 7
y ′ = d x d ( 3 x 4 ) − d x d ( 5 x 2 ) + d x d ( 7 ) Why? Sum/difference rule teen terms ko split karta hai.
= 3 ⋅ 4 x 3 − 5 ⋅ 2 x = 12 x 3 − 10 x Why? Constant multiple 3 5 x n 7 0
Worked example Example 2 — mixing function typesf ( x ) = 2 sin x + 4 e x − 3 1 ln x
f ′ ( x ) = 2 cos x + 4 e x − 3 1 ⋅ x 1 Why? Har term ek constant times ek known derivative hai (sin ′ = cos ( e x ) ′ = e x ln ′ = 1/ x
Worked example Example 3 — rewrite first, then applyg ( x ) = 2 x 2 + 6 x
Step 1 — powers ke constant multiples ki tarah rewrite karo:
g ( x ) = 2 1 x 2 + 3 x 1/2 Why? 2 2 1 x = x 1/2 Step 2 — term by term differentiate karo:
g ′ ( x ) = 2 1 ⋅ 2 x + 3 ⋅ 2 1 x − 1/2 = x + 2 x 3
Common mistake "Product ka derivative, derivatives ka product hota hai — jaise sum rule."Why it feels right: Sum rule cleanly split hua, toh brain over-generalize karta hai: surely multiplication bhi split hogi? Symmetry seductive hoti hai.
The fix: Linearity sirf + − × product rule chahiye: ( f g ) ′ = f ′ g + f g ′ d x d ( x ⋅ x ) = 2 x x ′ ⋅ x ′ = 1 ⋅ 1 = 1 = 2 x
Common mistake Bhool jaana ki constant ka derivative0 Why it feels right: Constants "terms jaisi" lagti hain, toh students 7 7 1 The fix: c f ( x ) f ( x ) = 1 c ⋅ 0 = 0 7 0 7
Common mistake Multi-term second function ke saath subtraction mein sign galat handle karna.f − ( x 2 − 3 x ) f ′ − 2 x − 3 Galat — minus distribute hota hai: f ′ − ( 2 x − 3 ) = f ′ − 2 x + 3 The fix: − ( g ) ( − 1 ) ⋅ g g
Recall Compute karne se pehle predict karoy = 10 x 3 − 5 x + cos x y ′ forecast karo, phir expand karo.
✅ y ′ = 30 x 2 − 5 1 − sin x 5 x = 5 1 x 5 1 cos ′ = − sin
Sum rule for derivatives d x d [ f + g ] = f ′ + g ′
Difference rule d x d [ f − g ] = f ′ − g ′
Constant multiple rule d x d [ c f ] = c f ′
Kaun sa limit law sum rule ko kaam karwaata hai? Limit of a sum equals sum of the limits (jab dono exist karti hain).
Difference rule alag se prove kyun nahi kiya jaata? Kyunki f − g = f + ( − 1 ) g
Ek constant c 0
Kya linearity products par apply hoti hai? Nahi — products ko product rule chahiye ( f g ) ′ = f ′ g + f g ′
Ek-line "linearity" statement d x d [ a f + b g ] = a f ′ + b g ′
d x d ( 3 x 4 − 5 x 2 + 7 ) 12 x 3 − 10 x
Recall Feynman: explain to a 12-year-oldSocho tum do runners ki speed track kar rahe ho. Agar tum unki positions ko milaake ek "combined" position banaao, toh us combined cheez ki speed sirf dono speeds ka sum hogi. Aur agar ek runner triple speed set treadmill par daude, toh uski recorded speed sirf tegun normal hogi. Yehi sab ye rules kehte hain: adding added rehti hai, scaling scaled rehti hai. Lekin tum "runner A ki distance times runner B ki distance" ke liye yeh nahi kar sakte — multiply karna unhe mix karta hai, aur uske liye alag rule chahiye.
"Slopes add, constants ride along."
Sums → slopes add karo. c c ride karta hai . Subtraction = negative add karna.
Limit laws (sum, scalar, product of limits) — in proofs ke peeche ka engine.Definition of the derivative (limit of difference quotient) — jahan se humne shuru kiya.Power rule — linearity ke saath mila ke kisi bhi polynomial ko differentiate karo.Product rule — non-linear sibling; dhyaan se contrast karo.Linearity of integration — bilkul isi idea ka integral version.Linear operators — abstract structure: d x d
Derivative limit definition
Limit laws: sum splits, constants pull out
Constant multiple rule c f'
Linearity of differentiation
Modular calculus of built-up functions
