Exercises — Quotient rule — proof

Top hai $u=3x+2$; bottom hai $v=x-5$.

• $u^{\prime }=3$ (seedhi line $3x+2$ ki slope).
• $v^{\prime }=1$ ($x-5$ ki slope). Rule mein place karne par: $f^{\prime }(x)=\frac{(3)(x-5)-(3x+2)(1)}{(x-5{)}^2}.$ WHAT humne kiya: har piece ko ek slot se match kiya. WHY: quotient rule ki har galti mis-labelling se shuru hoti hai — yeh sahi karo toh algebra mechanical ho jaata hai. Domain: sirf $x\ne 5$ ke liye valid, kyunki wahan bottom $v=x-5$ zero ho jaata hai aur $f$ undefined hai.

$u=\cos x,v=x^2$. Toh $u^{\prime }=-\sin x$ (Derivatives of trig functions se) aur $v^{\prime }=2x$. $g^{\prime }(x)=\frac{(-\sin x)(x^2)-(\cos x)(2x)}{(x^2{)}^2}.$ WHAT humne kiya: ek trig top ko polynomial bottom ke upar slot kiya. WHY it matters: $u,v$ ko pehchanna bilkul same kaam karta hai chahe pieces trig hon, polynomial hon, ya exponential — function ka type kabhi nahi badalta ki use kaun se slot mein jaana hai, sirf differentiate karne ka tarika badalta hai. Domain: sirf $x\ne 0$ ke liye valid, kyunki wahan bottom $v=x^2$ zero ho jaata hai aur $g$ undefined hai.

$u=x^2-1,v=x^2+1$; $u^{\prime }=2x,v^{\prime }=2x$. $f^{\prime }(x)=\frac{(2x)(x^2+1)-(x^2-1)(2x)}{(x^2+1{)}^2}.$ WHAT aage: numerator dhyan se expand karo. $=\frac{2x^3+2x-2x^3+2x}{(x^2+1{)}^2}=\frac{4x}{(x^2+1{)}^2}.$ WHY cancellation acchi hai: $2x^3$ terms ek doosre ko khatam kar dete hain, ek clean $4x$ bacha dete hain. Toh $f^{\prime }=0$ sirf $x=0$ par. Domain: saare real $x$ ke liye — yahan $v=x^2+1\ge 1$ kabhi zero nahi hota, toh koi point exclude nahi hota.

$u=e^x,v=x$; $u^{\prime }=e^x,v^{\prime }=1$. $h^{\prime }(x)=\frac{e^x\cdot x-e^x\cdot 1}{x^2}=\frac{e^x(x-1)}{x^2}.$ WHY $e^x$ factor karo: yeh dikhata hai ki $h^{\prime }$ ka sign $(x-1)$ se control hota hai, kyunki $e^x>0$ aur $x^2>0$ hamesha. Toh $h$ $x<1$ ke liye decrease karta hai, $x>1$ ke liye increase karta hai. Domain: sirf $x\ne 0$ ke liye valid, kyunki wahan bottom $v=x$ zero ho jaata hai aur $h$ undefined hai.

Reciprocal special case use karo (parent ka Example 3): $u=1,u^{\prime }=0$, toh ${\left(\frac{1}{v}\right)}^{\prime }=-\frac{v^{\prime }}{v^2}$. Yahan $v=x^2+4,v^{\prime }=2x$: $r^{\prime }(x)=-\frac{2x}{(x^2+4{)}^2}.$ WHY yeh shortcut legal hai: $u=1$ set karne par full quotient rule ke andar $u^{\prime }v$ term khatam ho jaata hai, sirf $-v^{\prime }/v^2$ bachta hai. Domain: saare real $x$ ke liye — yahan $v=x^2+4\ge 4$ kabhi zero nahi hota, toh koi point exclude nahi hota.

$u=x,v=x^2+1$; $u^{\prime }=1,v^{\prime }=2x$. $f^{\prime }(x)=\frac{(1)(x^2+1)-(x)(2x)}{(x^2+1{)}^2}=\frac{x^2+1-2x^2}{(x^2+1{)}^2}=\frac{1-x^2}{(x^2+1{)}^2}.$ Horizontal tangent ka matlab slope $=0$ hai, yaani numerator $=0$: $1-x^2=0\Rightarrow x=\pm 1$. $f^{\prime }$ ka sign check karo: $|x|<1$ ke liye numerator $>0$ (rising), $|x|>1$ ke liye numerator $<0$ (falling). Toh $x=1$ ek maximum hai, $x=-1$ ek minimum hai. Domain: saare real $x$ ke liye — yahan $v=x^2+1\ge 1$ kabhi zero nahi hota.

Neeche figure mein, solid curve $f$ hai; pink dashed line $x=1$ par horizontal tangent mark karta hai aur blue dashed line $x=-1$ par. Notice karo ki dono dashed lines bilkul flat hain (slope $0$), aur curve unke beech rise karta hai aur bahar fall karta hai — exactly sign analysis se match karta hai.

Parent ne nikala tha $f^{\prime }(x)=\frac{x(x+2)}{(x+1{)}^2}$. $x=1$ par evaluate karo: $f^{\prime }(1)=\frac{1\cdot 3}{2^2}=\frac{3}{4}.$ Yeh positive hai, toh $f$ $x=1$ par increasing hai slope $\frac{3}{4}$ ke saath. WHY it matters: parent ki "backwards subtraction" mistake $-\frac{3}{4}$ deti — same size, galat sign. Kisi known-increasing function ka sign check karna woh error turant pakad leta hai. Domain: sirf $x\ne -1$ ke liye valid, jahan bottom $v=x+1$ zero hai.

$u=\sin x,v=1+\cos x$; $u^{\prime }=\cos x,v^{\prime }=-\sin x$. $y^{\prime }=\frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x{)}^2}=\frac{\cos x+{\cos }^{2}x+{\sin }^{2}x}{(1+\cos x{)}^2}.$ ${\cos }^{2}x+{\sin }^{2}x=1$ use karo: $=\frac{\cos x+1}{(1+\cos x{)}^2}=\frac{1}{1+\cos x}.$ WHY collapse hota hai: numerator exactly denominator ka ek factor ban jaata hai, toh ek $(1+\cos x)$ cancel ho jaata hai. Domain: wahan valid hai jahan $1+\cos x\ne 0$, yaani $x\ne \pi +2\pi k$ (jahan $\cos x=-1$).

$u=(2x+1{)}^3$. $u$ par chain rule: outer power $3(\cdot {)}^2$ times inner derivative $2$, toh $u^{\prime }=3(2x+1{)}^2\cdot 2=6(2x+1{)}^2$. $v=x,v^{\prime }=1$. $F^{\prime }(x)=\frac{6(2x+1{)}^2\cdot x-(2x+1{)}^3\cdot 1}{x^2}.$ WHAT aage: numerator se common factor $(2x+1{)}^2$ bahar nikalo: $=\frac{(2x+1{)}^2[6x-(2x+1)]}{x^2}=\frac{(2x+1{)}^2(4x-1)}{x^2}.$ WHY expand karne se pehle factor karo: yeh bina brute-force algebra ke slope zeros $x=-\frac{1}{2}$ (double) aur $x=\frac{1}{4}$ par reveal karta hai. Domain: sirf $x\ne 0$ ke liye valid, jahan bottom $v=x$ zero hai.

$f=u\cdot v^{-1}$ likho. Product rule: $f^{\prime }=u^{\prime }\cdot v^{-1}+u\cdot (v^{-1}{)}^{\prime }$. Chain rule se, $(v^{-1}{)}^{\prime }=-v^{-2}{v}^{\prime }=-\frac{v^{\prime }}{v^2}$. Toh $f^{\prime }=\frac{u^{\prime }}{v}-\frac{u{v}^{\prime }}{v^2}.$ Common denominator $v^2$ par rakho: $=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}.$ Yeh exactly quotient rule hai. WHY satisfying hai: quotient rule koi nayi law nahi hai — yeh product rule plus "$v$ ka ulta" hai. Domain: jahan bhi $v\ne 0$ ho wahan valid, kyunki $v^{-1}$ ke liye yeh zaroori hai.

$h$ yahan kya hai: $h$ $x$ mein ek chhota sa step hai; "$h\to 0$" ka matlab hai hum us step ko kuch nahi ki taraf sikudte hain, toh point $x+h$ wapis $x$ par slide karta hai. Woh property jo force karti hai $v(x+h)\to v(x)$ jab $x+h\to x$ hai continuity of $v$; differentiability isse imply karti hai. Yeh denominator $v(x+h)v(x)\to v(x{)}^2$ banata hai. Agar $v$ $x$ par discontinuous hoti: maano $v$ jump kare, toh $v(x+h)\to L\ne v(x)$. Toh denominator $Lv(x)$ ki taraf jaata hai, $v(x{)}^2$ ki taraf nahi, aur "clean" formula wahan simply galat hai. Example: $v(x)=\{\begin{array}{ll}1,& x<0\\ 2,& x\ge 0\end{array}$ $0$ par discontinuous hai — wahan koi derivative exist nahi karta, toh quotient rule apply nahi ho sakta.

$u=\cos x,v=\sin x$; $u^{\prime }=-\sin x,v^{\prime }=\cos x$. $\frac{d}{dx}\cot x=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{{\sin }^{2}x}=\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}.$ Numerator $=-({\sin }^{2}x+{\cos }^{2}x)=-1$: $=\frac{-1}{{\sin }^{2}x}=-{\csc }^{2}x.$ Undefined jahan $\sin x=0$, yaani $x=k\pi$ (wahan $\cot x$ khud blow up karta hai).

General: $f^{\prime }(a)=\frac{u^{\prime }(a)v(a)-u(a)v^{\prime }(a)}{v(a{)}^2}$. Kyunki $u(a)=0$ hai, term $u(a)v^{\prime }(a)=0$ vanish ho jaata hai: $f^{\prime }(a)=\frac{u^{\prime }(a)v(a)}{v(a{)}^2}=\frac{u^{\prime }(a)}{v(a)}.$ Geometric meaning: top ke ek root par, fraction zero cross karta hai, aur wahan ke paas denominator ek constant $v(a)$ ki tarah act karta hai. Toh curve locally seedhi line $\frac{u^{\prime }(a)}{v(a)}(x-a)$ jaisi dikhti hai — bottom ka change $v^{\prime }$ us instant par matter nahi karta kyunki woh fraction ki (zero) height se multiply ho raha hai.

Neeche figure mein solid curve $f=(x-1)/(x^2+1)$ hai, jiska $u(a)=0$ $a=1$ par hai (marked). Dekho pink dashed tangent curve ko exactly wahan touch karta hai jahan woh $x$-axis cross karta hai: iska slope $u^{\prime }(1)/v(1)=1/2$ hai. Chahe denominator $x^2+1$ wahan change ho raha ho, woh change slope par koi fingerprint nahi chhodta — exactly kyunki $u(a)=0$ hai.

Sachchi baat yeh hai ki $f(x)=\frac{x^3}{x^2}=x$ (for $x\ne 0$), toh $f^{\prime }(x)=1$. Galat "term-wise" rule: $\frac{u^{\prime }}{v^{\prime }}=\frac{3x^2}{2x}=\frac{3x}{2}$, jo constant nahi hai — clearly galat. Sahi quotient rule: $\frac{(3x^2)(x^2)-(x^3)(2x)}{(x^2{)}^2}=\frac{3x^4-2x^4}{x^4}=\frac{x^4}{x^4}=1$. ✓