Quotient rule — proof — Hinglish notes

4.1.15 · Maths › Calculus I — Limits & Derivatives

The statement

Derivation from first principles

Hum limit definition use karte hain: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h} .$

Step 1 — difference quotient likho. $f (x + h) - f (x) = \frac{u ( x + h )}{v ( x + h )} - \frac{u ( x )}{v ( x )} .$ Yeh step kyun? Yeh literally definition hai; baaki sab algebra hai taaki limit computable ho sake.

Step 2 — common denominator par combine karo. $= \frac{u ( x + h ) v ( x ) - u ( x ) v ( x + h )}{v ( x + h ) v ( x )} .$ Yeh step kyun? Ek single fraction manipulate karna aasaan hai, aur yeh denominator $v (x + h) v (x)$ ko isolate karta hai jo limit mein $v^{2}$ ban jaayega.

Step 3 — "add and subtract zero" trick. Numerator mein $- u (x) v (x) + u (x) v (x) = 0$ insert karo: $u (x + h) v (x) - u (x) v (x + h) = change in u [u (x + h) v (x) - u (x) v (x)] - change in v [u (x) v (x + h) - u (x) v (x)] .$ Yeh step kyun? Hum $u$ aur $v$ ke liye alag-alag difference quotients engineer karte hain. Factor karo: $= v (x) [u (x + h) - u (x)] - u (x) [v (x + h) - v (x)] .$

Step 4 — $h$ se divide karo. Yaad karo $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ , toh poori cheez ko $h$ se divide karo: $\frac{f ( x + h ) - f ( x )}{h} = \frac{1}{v ( x + h ) v ( x )} [v (x) \frac{u ( x + h ) - u ( x )}{h} - u (x) \frac{v ( x + h ) - v ( x )}{h}] .$ Yeh step kyun? Ab do pehchaane jaane wale difference quotients dikhte hain: yeh $u^{'} (x)$ aur $v^{'} (x)$ ki definitions hain.

Step 5 — limit $h \to 0$ lo. Use karo ki $v$ differentiable hai (isliye continuous bhi hai), toh $v (x + h) \to v (x)$ : $f^{'} (x) = \frac{1}{v ( x ) v ( x )} [v (x) u^{'} (x) - u (x) v^{'} (x)] = \frac{u ^{'} ( x ) v ( x ) - u ( x ) v ^{'} ( x )}{( v ( x ) ) ^{2}} . ■$ Continuity kyun matter karti hai: bina $v (x + h) \to v (x)$ ke, denominator $v (x + h) v (x)$ $v (x)^{2}$ ki taraf tend nahi karta aur poora argument collapse ho jaata hai.

Worked examples

Common mistakes (steel-manned)

Flashcards

State the quotient rule for

(u / v)^{'}

\frac{u ^{'} v - u v ^{'}}{v ^{2}}

Proof mein, kaun sa algebraic trick do difference quotients create karta hai?

Numerator mein

u (x) v (x)

add aur subtract karo ("add zero").

Proof ke last step ke liye

v

continuous kyun hona chahiye?

Taaki

v (x + h) \to v (x)

, aur denominator

v (x)^{2}

ki taraf tend kare.

Quotient rule mein minus sign kyun hai?

Kyunki denominator ka badhna fraction ko chhhota karta hai;

1/ v

differentiate karne se

- v^{'} / v^{2}

milta hai.

Quotient rule se

\frac{d}{d x} tan x

derive karo.

\frac{c o s ^{2} x + s i n ^{2} x}{c o s ^{2} x} = sec^{2} x

u = 1

se kaun sa special case milta hai?

Reciprocal rule

\frac{d}{d x} (1/ v) = - v^{'} / v^{2}

u^{'} / v^{'}

" ko kill karne wala quick test: kaun sa function?

f = x^{2} / x = x

mein

f^{'} = 1

hai, lekin

u^{'} / v^{'} = 2 x

— contradiction.

Recall Feynman: ek 12-saal ke bachhe ko explain karo

Pizza share karne ki imagine karo: top = tumhare paas kitne slices hain, bottom = kitne dost share kar rahe hain. Agar zyada slices milti hain (top badhta hai), toh sabka share badhta hai — yeh wala $+ u^{'} v$ part hai. Agar zyada dost aa jaate hain (bottom badhta hai), toh sabka share chhhota ho jaata hai — yeh wala $- u v^{'}$ part hai (minus!). Aur pehle se bheed jitni badi hai, utna hi kam fark padta hai ek extra dost se — isliye hum crowd size squared, $v^{2}$ , se divide karte hain.

Connections

Product rule — proof — wahi "add-zero" trick; quotient rule = $u$ aur $1/ v$ ka product rule.
Chain rule — $1/ v$ ko $v^{- 1}$ dekhne ke liye zaroori hai jo $- v^{'} / v^{2}$ deta hai.
Limit definition of derivative — is poore proof ka engine.
Continuity — $v (x + h) \to v (x)$ justify karta hai.
Derivatives of trig functions — $tan, cot, sec, csc$ sab is rule se aate hain.