4.4.6
Single-variable calculus mein, ek function a line ho. Multiple variables mein, differentiability ka matlab hai ki surface ka ek tangent plane ho jo graph ko genuinely hug kare — jab tum zoom in karo toh function linear dikhta hai.
Yahan ek trap hai: partial derivatives hona KAAFI NAHI hai. Ek point par ∂ f / ∂ x ∂ f / ∂ y stronger condition hai "partials exist" se.
Ek strict hierarchy hai. Har ek neeche wale ko imply karta hai, kabhi ulta nahi.
continuously diff. ( C 1 ) ⇒ differentiable ⇒ partials exist differentiable ⇒ continuous
Definition Partial derivativef : R 2 → R f x ( a , b ) = lim h → 0 h f ( a + h , b ) − f ( a , b ) , f y ( a , b ) = lim h → 0 h f ( a , b + h ) − f ( a , b ) . ek seedhi direction probe karta hai (axis ke saath). Yeh us line se hatke jo kuch bhi hota hai, usse andha hai.
Definition Differentiability (asli definition)f a = ( a , b ) differentiable hai agar ek linear map (gradient ∇ f ( a ) lim h → 0 ∥ h ∥ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h = 0. error ko distance ∥ h ∥ tez shrink karna chahiye jab tum har direction se ek saath approach karo.
Hum ek tangent plane L ( x ) = f ( a ) + m ⋅ ( x − a ) f E ( h ) = f ( a + h ) − plane [ f ( a ) + m ⋅ h ] .
Tangent plane ke liye hum sirf E → 0 E ∥ h ∥ negligible ho — ek higher-order error:
E ( h ) = o ( ∥ h ∥ ) ⟺ ∥ h ∥ E ( h ) → 0.
∥ h ∥ E → 0 ∥ h ∥ quadratically-fast maarne ke liye force karta hai. Yehi cheez ise tangent plane banati hai aur sirf koi bhi plane nahi jo point ko touch kare.
Kaun sa m x h = ( h , 0 ) ∥ h ∥ = ∣ h ∣ ∣ h ∣ f ( a + h , b ) − f ( a , b ) − m 1 h → 0 ⇒ m 1 = f x ( a , b ) . m 2 = f y ( a , b ) agar f ∇ f = ( f x , f y )
f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + o ( ∥ h ∥ )
Maano f a f ( a + h ) − f ( a ) = ∇ f ( a ) ⋅ h + E ( h ) , E = o ( ∥ h ∥ ) . h → 0 → 0 h E → 0 E /∥ h ∥ → 0 E → 0 f ( a + h ) → f ( a ) Continuous. ■
Worked example Example 1 — Aasaan tarike se differentiability confirm karof ( x , y ) = x 2 y + sin y ( 1 , 0 )
f x = 2 x y f y = x 2 + cos y Yeh step kyun? Partials compute karo dusre variable ko constant maanke.Dono sab jagah continuous hain (continuous functions ke products/sums). Kyun? C 1 ⇒
∇ f ( 1 , 0 ) = ( 0 , 1 + 1 ) = ( 0 , 2 ) z = 0 + 0 ( x − 1 ) + 2 ( y − 0 ) = 2 y
Worked example Example 2 — Partials exist karte hain par differentiable NAHIf ( x , y ) = ⎩ ⎨ ⎧ x 2 + y 2 x y 0 ( x , y ) = ( 0 , 0 ) ( 0 , 0 )
Origin par partials: y = 0 f ( x , 0 ) = 0 f x ( 0 , 0 ) = 0 f y ( 0 , 0 ) = 0 Kyun? y = 0 0 Par continuity fail karta hai: y = x f ( x , x ) = 2 x 2 x 2 = 2 1 = 0 = f ( 0 , 0 ) Yeh kyun khatam karta hai? Differentiable ⇒ continuous; yeh continuous nahi hai, toh differentiable nahi ho sakta — chahe dono partials exist karte ho!Moral: partials = 0 ek mirage of niceness hai. Diagonal direction sach dikhata hai.
Common mistake Steel-man: "Partials exist ⇒ differentiable"Sahi kyun lagta hai: 1D mein, "derivative exists" hi differentiability hai — toh hum same expect karte hain. Aur f x , f y Galat kyun hai: Partials sirf dono axis directions test karte hain. Ek function x y diagonal ke saath jagged ya discontinuous ho (Example 2). Differentiability har direction se ek saath achhe behaviour ki demand karta hai — ek 2D requirement.
Fix: Ya toh o ( ∥ h ∥ ) ya check karo ki partials continuous hain (C 1 C 1
Common mistake "Har direction mein directional derivatives exist karte hain" ko differentiability se confuse karnaChahe ek point se guzarne wali har line ke saath directional derivatives exist karte hon, f D u f = ∇ f ⋅ u
Recall Feynman: 12-saal ke bacche ko samjhaoEk ulhar-pulhar pahaad imagine karo. Ek jagah par khade hokar, tum slope check karte ho ekdum East chalte hue aur ekdum North chalte hue — woh "partial slopes" hain. Par iska North-East direction mein achanak ek cliff ke baare mein kuch pata nahi chalta! Ek pahaad "smooth" (differentiable) tabhi hai jab, chahe tum kisi bhi direction mein kadam rakho, zameen aise uthe jaise ek flat tilted board tumhare paairon ke neeche rakhi ho. Agar ek bhi diagonal direction tumhe dhoka de, toh pahaad us jagah se secretly toot hua hai — chahe East aur North bilkul theek lage hoon.
"Partials diagonals ke andhe hain; sirf plane sab dekh sakta hai."
Aur hierarchy: C ontinuously-diff → D ifferentiable → C ontinuous → (aur alag se) P artials.
"C ould D ucks C arry P lanes?" — left right ko imply karta hai, akele P ke liye kabhi ulta nahi.
f : R 2 → R a lim h → 0 ∥ h ∥ ∣ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h ∣ = 0
"Partials exist" differentiability ke liye kyun kaafi nahi? Partials sirf axis directions test karte hain; function diagonals ke saath discontinuous/jagged ho sakta hai (jaise x y / ( x 2 + y 2 )
Sufficient C 1 Agar f x , f y a f a
Differentiable hona kaun si weaker property imply karta hai? Continuity (aur sab directional derivatives ka exist karna).
Agar f Gradient ∇ f = ( f x , f y )
Definition mein error ko ∥ h ∥ Error ko higher-order (o ( ∥ h ∥ )
f ( x , y ) = x 2 + y 2 x y f x ( 0 , 0 ) , f y ( 0 , 0 ) f Dono = 0 f y = x 2 1
Estimation ke liye linearization formula f ( a + h ) ≈ f ( a ) + ∇ f ( a ) ⋅ h
Limit definition: error over norm h to 0
Gradient nabla f = fx, fy
Higher-order error o of norm h
Trap: partials alone insufficient
