Quotient rule — proof: Visual walkthrough — Hinglish deep dive

4.1.15 · D2 · Maths › Calculus I — Limits & Derivatives › Quotient rule — proof

Shuru karne se pehle ek vaada: har letter pehli baar aane par define kiya jayega. Chalte hain wahan se jahan "function" aur "slope" ka matlab bhi clear ho.

Step 0 — Hum pooch kya rahe hain? (setup picture)

KYA. Hamare paas do machines hain. Machine $u$ ek number $x$ leta hai aur output deta hai $u (x)$ . Machine $v$ wahi $x$ leta hai aur output deta hai $v (x)$ . Hum ek ratio machine banate hain $f (x) = \frac{u ( x )}{v ( x )} .$ Yahan slash ka matlab ordinary division hai: " $u (x)$ ko $v (x)$ parts mein baanta". Zaroori hai ki $v (x) \neq = 0$ — zero parts mein nahi baant sakte.

KYUN. Hum $f$ ki slope chahte hain — yani jab input ko thoda sa badlate hain to output kitni tezi se badalta hai. Parent note mein ise derivative $f^{'} (x)$ kaha gaya hai.

DIKHTA KAISA HAI. Slope ek curve ki steepness hoti hai. Agar tum $f$ ke graph par khade ho aur thoda sa dahine kadam barhao, to slope hai kitna upar gaye divided by kitna kadam barhaya.

Step 1 — Rise ko do fractions ke difference ke roop mein likho

KYA. $f = u / v$ ko directly rise $f (x + h) - f (x)$ mein daalo: $f (x + h) - f (x) = nai value \frac{u ( x + h )}{v ( x + h )} - purani value \frac{u ( x )}{v ( x )} .$

KYUN. Yeh literally definition hai — abhi tak kuch bhi clever nahi kiya. Iske baad sab kuch sirf reshaping hai jab tak limit computable na ho jaye.

DIKHTA KAISA HAI. Do fractions hain jinka denominator alag-alag hai ( $v (x + h)$ vs $v (x)$ ). Alag bottoms matlab inhe abhi combine nahi kar sakte — jaise $\frac{1}{2} - \frac{1}{3}$ mein pehle common denominator chahiye.

Step 2 — Common denominator par combine karo

KYA. Har fraction ko multiply karo taaki dono ka bottom ek hi ho jaye $v (x + h) v (x)$ : $= \frac{u ( x + h ) v ( x ) naya top \times purana bottom - u ( x ) v ( x + h ) purana top \times naya bottom}{dono bottoms ka product v ( x + h ) v ( x )} .$

KYUN. Ek single fraction handle karna kaafi aasaan hai. Sabse important baat — denominator ab $v (x + h) v (x)$ hai — ise dhyan se dekho, kyunki jab $h \to 0$ hoga to do bottoms barabar ho jayenge aur yeh $v^{2}$ ban jayega. Yahan se hi squared denominator paida hota hai.

DIKHTA KAISA HAI. Ek rectangle-area picture: cross-multiply karne ke liye $\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$ , hum har fraction ko shared grid $b \times d$ par tile karte hain.

Step 3 — "Zero add karo" ka trick

KYA. Sirf numerator $u (x + h) v (x) - u (x) v (x + h)$ dekho. Ismein $- u (x) v (x) + u (x) v (x)$ slip in karo, jo $0$ ke barabar hai isliye kuch nahi badlega: $u (x + h) v (x) = 0, free mein add kiya - u (x) v (x) + u (x) v (x) - u (x) v (x + h) .$ Ab pieces group karo: $= u kitna badla [u (x + h) - u (x)] v (x) - u (x) v kitna badla [v (x + h) - v (x)] .$

KYUN. Hum chahte hain ki differences $u (x + h) - u (x)$ aur $v (x + h) - v (x)$ appear hon, kyunki yahi $u^{'}$ aur $v^{'}$ ke raw ingredients hain. Add-zero trick inhe kahin se bhi manufacture kar leta hai. (Yahi manoeuvre Product rule — proof mein bhi use hoti hai.)

DIKHTA KAISA HAI. Ek bar do coloured pieces mein split hoti hai: ek magenta piece jo carry karti hai " $u$ move hua, purane bottom $v (x)$ se weighted" aur ek violet piece jo carry karti hai " $v$ move hua, purane top $u (x)$ se weighted" — aur violet piece ke aage pehle se minus laga hua hai.

Step 4 — $h$ se divide karo taaki do slopes saamne aayen

KYA. Poori rise ab $v (x + h) v (x)$ ke upar hai; definition ke mutabik sab kuch $h$ se divide karo: $\frac{f ( x + h ) - f ( x )}{h} = \frac{1}{v ( x + h ) v ( x )} v (x) \to u^{'} (x) \frac{u ( x + h ) - u ( x )}{h} - u (x) \to v^{'} (x) \frac{v ( x + h ) - v ( x )}{h} .$

KYUN. Har bracketed difference ko $h$ se divide karne par woh ek difference quotient ban jaata hai — bilkul limit definition ki shape jaisi. To har bracket mein ek derivative chhupa hua hai jo reveal hone ka wait kar raha hai.

DIKHTA KAISA HAI. Step 3 ka har coloured piece ab ek chhota right-triangle slope ban jaata hai: rise $u (x + h) - u (x)$ over run $h$ (magenta), aur rise $v (x + h) - v (x)$ over run $h$ (violet).

Step 5 — Limit lo (jahan continuity apna kaam karti hai)

KYA. $h \to 0$ hone do. Chaar cheezein ek saath hoti hain: $\frac{u ( x + h ) - u ( x )}{h} \to u^{'} (x), \frac{v ( x + h ) - v ( x )}{h} \to v^{'} (x), v (x + h) \to v (x) .$ Teesri cheez yeh use karti hai ki $v$ differentiable hai, isliye continuous hai: $x$ mein hair-step se $v$ mein bhi sirf hair-step aata hai. To denominator $v (x + h) v (x) \to v (x) v (x) = (v (x))^{2}$ . Assemble karo: $f^{'} (x) = \frac{v ( x ) u ^{'} ( x ) - u ( x ) v ^{'} ( x )}{( v ( x ) ) ^{2}} = \frac{u ^{'} ( x ) v ( x ) - u ( x ) v ^{'} ( x )}{( v ( x ) ) ^{2}} . ■$

KYUN continuity zaroori hai. Agar $v$ jump karta (continuous nahi hota), to $v (x + h)$ kabhi $v (x)$ ke paas nahi aata, denominator $v^{2}$ nahi banta, aur proof collapse ho jaata.

DIKHTA KAISA HAI. Do shrinking triangles freeze hokar exact tangent slopes $u^{'} (x)$ aur $v^{'} (x)$ ban jaate hain, aur do nearby bottoms $v (x + h), v (x)$ slide hokar ek single value $v (x)$ ban jaate hain, jo squared hai.

Step 6 — Degenerate case: constant numerator ( $u = 1$ )

KYA. $u (x) = 1$ set karo, jo ek flat machine hai, to $u^{'} (x) = 0$ . Rule collapse ho jaata hai: $\frac{d}{d x} \frac{1}{v} = \frac{0 \cdot v u ^{'} v - 1 \cdot v ^{'}}{v ^{2}} = - \frac{v ^{'}}{v ^{2}} .$

KYUN. Yeh reciprocal rule hai jo andar chhup raha tha. Ek accha sanity check: quotient rule mein yeh hona hi chahiye. Yeh minus ko bhi confirm karta hai — top par kuch grow nahi kar sakta, to fraction ko move karne wali sirf ek hi cheez hai yani bottom, jo hamesha ulti direction mein push karti hai.

DIKHTA KAISA HAI. Flat top ke saath, $y = 1/ v$ tab girta hai jab $v$ badhta hai aur tab badhta hai jab $v$ girta hai — curve, $v$ ki motion ka mirror-flip hai.

Ek picture mein summary

Isse left se right padho: limit definition mein $f = u / v$ daala; common denominator ne future $v^{2}$ ko plant kiya; add-zero ne numerator ko ek magenta " $u$ -moved" piece aur ek violet " $v$ -moved" piece mein split kiya (aage minus hai); $h$ se divide karne par dono slopes ban gaye; limit ne inhe freeze karke $u^{'} v - u v^{'}$ over $v^{2}$ bana diya.

Recall Feynman retelling — poora walkthrough simple shabdon mein

Socho tum pizza kaat rahe ho. Top $u$ = tumhare paas slices, bottom $v$ = share karne wale dost. Tum jaanna chahte ho ki jab cheezein thodi si shift hoti hain to har insaan ka share kitna badlata hai. Pehle (Step 1–2) tum "naya share minus purana share" likhte ho aur dono ko ek common crowd-size ke upar rakhte ho. Phir (Step 3) magic hota hai: tum bilkul wahi amount add aur subtract karte ho — yeh zero add karna hai — free, harmless — lekin yeh tumhe "slices badle" aur "crowd badla" alag karne deta hai. Tiny time-step se divide karo (Step 4) aur har woh ek rate ban jaata hai: slices kitni tezi se badhti hain ( $u^{'}$ ), crowd kitni tezi se badhta hai ( $v^{'}$ ). Step shrink hone do (Step 5): ek pal mein crowd bahut kam hilta hai (yahi continuity hai), to do crowd sizes merge hokar ek ho jaati hain, squared. Final law: share-change = (crowd × slice-growth − slices × crowd-growth) ÷ crowd². Zyada dost (bottom badhna) sabka share chhota kar deta hai — yahi minus hai. Aur crowd jitna bada pehle se, ek aur dost utna kam matter karta hai — yahi squared bottom hai.

Connections

Limit definition of derivative — har step ka engine.
Product rule — proof — wahi add-zero trick.
Chain rule — $\frac{d}{d x} v^{- 1} = - v^{'} / v^{2}$ directly deta hai.
Continuity — Step 5 mein $v (x + h) \to v (x)$ possible banata hai.
Derivatives of trig functions — $tan x$ ka payoff.