4.1.22 · Maths › Calculus I — Limits & Derivatives
Kabhi kabhi y ek equation ke andar phansa hota hai aur bahar nahi aata. Try karo x 2 + y 2 = 25 ko y ke liye solve karna: milta hai y = ± 25 − x 2 — do gandhi branches. Try karo sin ( x y ) = x + y : y ko closed form mein bilkul bhi solve nahi kar sakte.
Trick yeh hai: hume y ko azaad karne ki zaroorat nahi . Hum bas declare karte hain ki "y koi function hai x ka" aur poori equation ko as-is differentiate karte hain, chain rule apply karte hain har baar jab y aata hai. Phir result ko algebraically d x d y ke liye solve karte hain.
Definition Explicit vs Implicit
Ek explicit function mein y akela ek side pe hota hai: y = f ( x ) .
Ek implicit equation mein x aur y milke aate hain: F ( x , y ) = 0 , jaise x 2 + y 2 = 25 .
Implicit differentiation ka matlab hai ki implicit equation ke dono sides ko x ke respect mein differentiate karo, y ko x ka function maante hue , phir d x d y ke liye solve karo.
Poora mechanism ek assumption + ek rule pe tika hai.
Assumption. Kisi point ke paas, curve F ( x , y ) = 0 define karta hai y ko kisi differentiable function y ( x ) ke roop mein, chahe hum usse likh na sakein. (Implicit Function Theorem guarantee karta hai yeh jahan ∂ F / ∂ y = 0 ho.)
Rule. Chain rule. Agar u = g ( y ) aur y = y ( x ) , toh
d x d g ( y ) = g ′ ( y ) ⋅ d x d y .
Toh jab bhi tum kisi y -wale term ko differentiate karte ho, ek extra factor d x d y nikal aata hai. Yahi ordinary differentiation se ek maatra farq hai.
Intuition Kyun aata hai extra
d x d y factor?
d x d ( x 3 ) = 3 x 2 kyunki andar wala (x ) rate 1 se change karta hai x ke saath.
d x d ( y 3 ) = 3 y 2 ⋅ d x d y kyunki andar wala (y ) rate d x d y se change karta hai x ke saath. Chain rule ek "toll" leta hai y ′ ka, y se guzarne ke liye.
F ( x , y ) = 0 ko x ke w.r.t. multivariable chain rule se differentiate karo:
F x + F y d x d y = 0 ⟹ d x d y = − F y F x
jahan F x = ∂ F / ∂ x , F y = ∂ F / ∂ y . Yeh ek shortcut hai — lekin term-by-term method jo neeche hai, wahi actually use hota hai.
Worked example Example 1 — Circle
x 2 + y 2 = 25
Dono sides differentiate karo:
d x d ( x 2 ) + d x d ( y 2 ) = d x d ( 25 )
2 x + 2 y d x d y = 0
Yeh step kyun? d x d ( y 2 ) = 2 y ⋅ y ′ chain rule se — y square ke andar chhupi hai.
Solve karo: d x d y = − 2 y 2 x = − y x .
Explicit se check: upper half hai y = 25 − x 2 , toh y ′ = 25 − x 2 − x = − y x . ✓ Same answer, bahut kam kaam — aur yeh dono branches ek saath cover karta hai.
Worked example Example 2 — Product term
x 3 + y 3 = 6 x y (folium of Descartes)
3 x 2 + 3 y 2 d x d y = 6 ( y + x d x d y )
Yeh step kyun? RHS 6 x y ek product hai, toh d x d ( 6 x y ) = 6 ( y ⋅ 1 + x ⋅ y ′ ) product rule se — aur x -factor ka derivative sirf 1 hai.
y ′ terms collect karo:
3 y 2 y ′ − 6 x y ′ = 6 y − 3 x 2 ⇒ y ′ ( 3 y 2 − 6 x ) = 6 y − 3 x 2
d x d y = 3 y 2 − 6 x 6 y − 3 x 2 = y 2 − 2 x 2 y − x 2 .
Point ( 3 , 3 ) pe: y ′ = 9 − 6 6 − 9 = 3 − 3 = − 1 . Wahan ki tangent ka slope − 1 hai.
Worked example Example 3 — Trig/transcendental:
sin ( x y ) = x + y
cos ( x y ) ⋅ d x d ( x y ) = 1 + d x d y
cos ( x y ) ( y + x d x d y ) = 1 + d x d y
Yeh step kyun? sin pe outer chain rule, phir x y pe product rule. Yahan y ke liye pehle solve karna literally impossible hai — implicit diff hi ek maatra raasta hai.
Expand aur collect karo:
y cos ( x y ) + x cos ( x y ) y ′ = 1 + y ′
y ′ ( x cos ( x y ) − 1 ) = 1 − y cos ( x y )
d x d y = x c o s ( x y ) − 1 1 − y c o s ( x y ) .
( a , b ) pe slope hai m = d x d y ( a , b ) ; tangent y − b = m ( x − a ) , normal ka slope = − 1/ m .
d x d ( arcsin x ) derive karna
Maano y = arcsin x ⇒ sin y = x . Implicitly differentiate karo:
cos y ⋅ d x d y = 1 ⇒ d x d y = c o s y 1 .
Kyunki sin y = x aur cos y = 1 − sin 2 y = 1 − x 2 (range pe positive):
d x d arcsin x = 1 − x 2 1 .
Har inverse-function derivative isi tarah nikali jaati hai.
y = x x differentiate karo
ln lo: ln y = x ln x . Implicitly differentiate karo:
y 1 d x d y = ln x + 1 ⇒ d x d y = x x ( ln x + 1 ) .
Kyun? Power rule nahi use kar sakte (exponent variable hai) aur exponential rule bhi nahi (base variable hai). Log lene se product/power pehle sums mein badal jaata hai.
Quantities jo ek equation se judi hain, dono time t mein change ho rahi hain — t ke w.r.t. implicitly differentiate karo. (Same chain-rule logic; x aur y dono t ke functions hain.)
d x 2 d 2 y for x 2 + y 2 = 25
Humne nikala tha y ′ = − x / y . Dobara differentiate karo (quotient rule), yaad rakhte hue y ′ = − x / y :
y ′′ = − y 2 ( 1 ) y − x y ′ = − y 2 y − x ( − x / y ) = − y 2 y + x 2 / y = − y 3 y 2 + x 2 = − y 3 25 .
Sundar: circle pe curvature info collapse ho jaata hai ek constant mein jo y 3 par divide hoti hai.
d x d y factor bhool jaana
Galat: d x d ( y 2 ) = 2 y .
Kyun sahi lagta hai: yeh bilkul d x d ( x 2 ) = 2 x jaisa dikhta hai, jo humara sabse drilled rule hai.
Fix: y variable nahi hai — yeh x ka function hai. Chain rule ek toll leta hai: d x d ( y 2 ) = 2 y d x d y .
x y pe product rule ki jagah power rule use karna
Galat: d x d ( x y ) = y ′ ya = 1 .
Kyun sahi lagta hai: x y "ek cheez" jaisa lagta hai.
Fix: Yeh do functions ka product hai: d x d ( x y ) = 1 ⋅ y + x ⋅ y ′ = y + x y ′ .
Common mistake Point plug karne se pehle back substitute karna bhool jaana
Jab tum y ′′ ke liye dobara differentiate karte ho, y ′ ki expression wapas replace karni hi padegi — warna tumhara "answer" abhi bhi y ′ contain karega.
d x d ( 25 ) ko nonzero maanna
Constants differentiate hokar 0 dete hain. Pura point yahi hai: ek balanced equation d x d ke under balanced rehti hai.
Recall Feynman: 12-year-old ko samjhao
Socho x aur y do dost hain jo ek rassi se bandhe hain (equation). Agar tum x ko thoda hilao, y ko bhi hilna padta hai — dono bandhe hain. Humein parwah nahi ki y exactly kya hai; hum sirf poochhte hain "jab x thoda sa move kare, y kitna move karta hai? " Woh ratio hai d x d y . Har baar jab hum math mein kisi y ko touch karte hain, hum multiply karte hain "y kitni tezi se move karta hai" se, kyunki y x ke saath dance kar raha hai, woh apni jagah khada nahi hai.
"y ko touch karo, toll y ′ chukao."
Normally differentiate karo; har baar jab y se guzro, ek d x d y lagao. Phir C.F.D. : C ollect, F actor, D ivide.
Implicit differentiation y ke baare mein kya assume karta hai? Ki y x ka (differentiable) function hai, chahe hum explicitly solve na kar sakein.
d x d ( y 3 ) ko extra factor kyun milta hai?Chain rule — y andar hai, toh d x d ( y 3 ) = 3 y 2 d x d y .
d x d ( x y ) = ? y + x d x d y (product rule).
x 2 + y 2 = r 2 ke liye, d x d y = ? − y x .
F ( x , y ) = 0 ke slope ka general formula?d x d y = − F y F x .
d x d arcsin x derive karne ka methodsin y = x set karo, implicitly differentiate karo:
cos y y ′ = 1 , toh
y ′ = 1/ 1 − x 2 .
d x d ( x x ) = ? aur technique?Logarithmic differentiation; x x ( ln x + 1 ) .
Technique ke 4 steps? Dono sides differentiate karo; har y -term mein y ′ lagao; y ′ terms collect karo; factor karo aur divide karo.
Implicit differentiation only option kab hota hai? Jab equation ko y ke liye closed form mein solve nahi kar sakte (jaise sin ( x y ) = x + y ).
Second-derivative ka trap? y ′ ki expression wapas substitute karni hi padegi; final answer mein y ′ nahi hona chahiye.
Chain Rule — har d x d y toll ke peeche ka engine.
Product Rule — mixed x y terms ke liye zaroor hai.
Derivatives of Inverse Functions — implicit differentiation se derive hote hain.
Logarithmic Differentiation — variable bases/exponents.
Related Rates — time t mein implicit differentiation.
Implicit Function Theorem — guarantee karta hai ki y ( x ) exist karta hai jahan F y = 0 .
Tangent and Normal Lines — main geometric application.
guarantees where Fy not 0
Assume y is function of x
Implicit Function Theorem
Product rule for xy terms