4.1.20 · Maths › Calculus I — Limits & Derivatives
WHAT hum chahte hain: y=lnx ki slope.
Defining relation se shuru karo, kisi formula se nahi:
y=lnx⟹ey=x.
Dono sides ko x ke respect mein differentiate karo:
dxd(ey)=dxd(x).
Yeh step kyun? Left side y ka function hai, aur y, x par depend karta hai, isliye hum chain rule use karte hain.
ey⋅dxdy=1.
Yeh step kyun? dxdey=eydxdy — chain rule, kyunki hum sirf dydey=ey jaante hain.
dxdy ke liye solve karo aur ey=x substitute karo:
dxdy=ey1=x1.
dxdlnx=x1,x>0.
=\lim_{h\to0}\frac1h\ln\!\Big(\frac{x+h}{x}\Big).$$
*Yeh step kyun?* Logs ke liye quotient rule: $\ln A-\ln B=\ln\frac{A}{B}$.
$t=h/x$ let karo toh $h=xt$ aur $t\to0$:
$$=\lim_{t\to0}\frac{1}{xt}\ln(1+t)=\frac1x\lim_{t\to0}\frac{\ln(1+t)}{t}=\frac1x\cdot 1=\frac1x.$$
*Yeh step kyun?* Standard limit $\lim_{t\to0}\frac{\ln(1+t)}{t}=1$ (yeh $1$ par $\ln$ ka derivative hai, ya equivalently $\ln e =1$ kyunki $(1+t)^{1/t}\to e$).
> [!formula] Master results
> $$\frac{d}{dx}\ln x=\frac1x \qquad
> \frac{d}{dx}\log_a x=\frac{1}{x\ln a}$$
> Chain-rule versions: $\displaystyle \frac{d}{dx}\ln u=\frac{u'}{u},\qquad \frac{d}{dx}\log_a u=\frac{u'}{u\ln a}.$
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## 3. $\dfrac{d}{dx}\log_a x$ derive karna
**HOW:** memorise mat karo — change of base use karke $\ln$ mein convert karo.
$$\log_a x=\frac{\ln x}{\ln a}.$$
*Yeh step kyun?* $\ln a$ ek **constant** hai (bas ek number), isliye hum $\ln x$ differentiate karte hain aur divide karte hain.
$$\frac{d}{dx}\log_a x=\frac{1}{\ln a}\cdot\frac{d}{dx}\ln x=\frac{1}{\ln a}\cdot\frac1x=\boxed{\frac{1}{x\ln a}.}$$
**Check:** $a=e$ set karo. Tab $\ln a=\ln e=1$, jo $\frac{1}{x\cdot1}=\frac1x$ deta hai. ✓ $\ln x$ ke saath consistent hai.
![[4.1.20-Derivatives-of-ln-x-and-loga(x).png]]
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## 4. Worked examples
> [!example] (a) $f(x)=\ln(3x^2+1)$
> $\frac{d}{dx}\ln u=\frac{u'}{u}$ use karo jahan $u=3x^2+1$.
> $u'=6x$. Toh $f'(x)=\dfrac{6x}{3x^2+1}.$
> *Yeh step kyun?* Chain rule: andar differentiate karo, andar se divide karo.
> [!example] (b) $g(x)=\log_2(5x)$
> $g'(x)=\dfrac{(5x)'}{(5x)\ln 2}=\dfrac{5}{5x\ln 2}=\dfrac{1}{x\ln 2}.$
> *Yeh step kyun?* Constant $5$ cancel ho jaata hai — note karo $\log_2(5x)=\log_2 5+\log_2 x$, aur $\log_2 5$ constant hai, isliye sirf $\log_2 x$ contribute karta hai. Same answer, faster.
> [!example] (c) $h(x)=\ln\sqrt{x}=\tfrac12\ln x$
> Pehle **simplify** karo: $h(x)=\frac12\ln x \Rightarrow h'(x)=\dfrac{1}{2x}.$
> *Yeh step kyun?* Log laws ($\ln x^{1/2}=\frac12\ln x$) ek chain-rule problem ko one-liner mein convert kar dete hain.
> [!example] (d) Logarithmic differentiation: $y=x^x$ ($x>0$)
> $\ln$ lo: $\ln y = x\ln x$. *Kyun?* Exponent mein variable hai; logs usse neeche kheench lete hain.
> Differentiate karo: $\frac{y'}{y}=\ln x + x\cdot\frac1x=\ln x+1$.
> Toh $y'=x^x(\ln x+1).$
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## 5. Common mistakes (Steel-man + fix)
> [!mistake] "$\dfrac{d}{dx}\log_a x=\dfrac1x$"
> **Kyun sahi lagta hai:** yeh bilkul $\ln$ jaisa dikhta hai. **Sachai:** sirf $\ln$ (base $e$) clean $1/x$ deta hai. Doosre bases ke liye ek extra factor $\dfrac{1}{\ln a}$ pay karna padta hai. **Fix:** hamesha $\frac{1}{x\ln a}$ likho aur $a=e\Rightarrow\ln a=1$ se sanity-check karo.
> [!mistake] "$\dfrac{d}{dx}\ln(3x)=\dfrac{1}{3x}\cdot 3=\dfrac3{3x}$... toh yeh $\ln x$ se alag hai?"
> **Kyun sahi lagta hai:** chain rule kehta hai $u'=3$ se multiply karo. **Sachai:** $\frac{3}{3x}=\frac1x$ — $\ln x$ jaisa hi! Kyunki $\ln(3x)=\ln 3+\ln x$ aur constant mar jaata hai. **Fix:** dono methods agree karte hain; algebra par trust karo.
> [!mistake] Domain bhool jaana.
> **Kyun sahi lagta hai:** $1/x$ negative $x$ ke liye defined hai. **Sachai:** $\ln x$ ko $x>0$ chahiye. $\ln|x|$ ke liye, derivative $\frac1x$ hai *dono* sides $x\neq0$ par. **Fix:** domain state karo.
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## 6. Active recall
> [!recall]- Quick self-test (answers cover karo)
> - $\frac{d}{dx}\ln x = ?$ → $\frac1x$
> - $\frac{d}{dx}\log_a x = ?$ → $\frac1{x\ln a}$
> - Base $e$ special kyun hai? → $\ln e=1$, jo $\ln a$ factor ko khatam kar deta hai.
> - $\frac{d}{dx}\ln(\cos x)=?$ → $\frac{-\sin x}{\cos x}=-\tan x$.
> [!recall]- Feynman: ek 12-saal ke bachche ko explain karo
> Socho ek pahaad par chadh rahe ho jiska height $\ln x$ "log-steps" hai. Jab tum kaafi door right mein ho (bada $x$), har extra step tumhe mushkil se upar uthaata hai — pahaad almost flat hai, slope $1/x$ ki tarah chhoti. $x=1$ ke paas slope exactly $1$ hai — steep. Slope hamesha "kitne door right mein ho, uska ulta" hota hai. Doosre base ke log ke liye wahi pahaad hai lekin $\ln a$ number se **stretched** hai, isliye slope us factor se chhoti ho jaati hai.
> [!mnemonic] Memory hook
> **"Log neeche, base basement mein."** $\log_a$ → answer hai **one over** $x$ times $\ln a$ jo **neeche** denominator mein baitha hai ("basement" mein). $\ln$ ke liye, basement khaali hai ($\ln e=1$).
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## 7. Connections
- [[Derivative of e^x and a^x|e^x aur a^x ka Derivative]] — yeh inverses hain; slopes reciprocals hain.
- [[Chain Rule]] — har $\ln u$ derivative ko power deta hai.
- [[Implicit Differentiation]] — upar wala Method A.
- [[Logarithm Laws]] — differentiate karne se pehle simplify karo.
- [[Logarithmic Differentiation]] — $x^x$, kai factors ke products ke liye.
- [[Standard Limit (1+t)^{1/t} → e]] — Method B ko underpin karta hai.
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#flashcards/maths
Derivative of $\ln x$ ::: $\dfrac1x$ for $x>0$.
Derivative of $\log_a x$ ::: $\dfrac{1}{x\ln a}$.
Base $e$ sabse simple derivative kyun deta hai? ::: Kyunki $\ln e=1$, isliye $\ln a$ factor gayab ho jaata hai.
Ek line mein $\frac{d}{dx}\ln x$ derive karo ::: $e^y=x\Rightarrow e^y y'=1\Rightarrow y'=1/e^y=1/x$.
$\frac{d}{dx}\ln u = ?$ (chain rule) ::: $\dfrac{u'}{u}$.
$\frac{d}{dx}\ln(3x)=?$ ::: $\dfrac1x$ (constant $3$ cancel ho jaata hai).
$\frac{d}{dx}\ln|x|=?$ ::: $\dfrac1x$ sabhi $x\neq0$ ke liye.
$x^x$ ka derivative ::: $x^x(\ln x+1)$, $\ln y=x\ln x$ ke through.
First-principles proof mein use hone wali standard limit ::: $\lim_{t\to0}\frac{\ln(1+t)}{t}=1$.
## 🖼️ Concept Map
```mermaid
flowchart TD
EXP[e^x is its own derivative] -->|inverse of| LN[ln x]
LN -->|defined by| REL[e^y = x]
REL -->|implicit diff| SLOPE[dy/dx = 1/e^y]
SLOPE -->|sub e^y=x| D1[d/dx ln x = 1/x]
LIM[Limit definition] -->|uses lim ln 1+t /t = 1| D1
D1 -->|chain rule| CHAIN[d/dx ln u = u'/u]
LN -->|change of base| CB[log_a x = ln x / ln a]
CB -->|ln a is constant| D2[d/dx log_a x = 1/ x ln a]
D1 -->|derives| D2
D2 -->|set a=e| CHK[recovers 1/x]
CHAIN -->|example| EX[ln 3x^2+1 gives 6x/ 3x^2+1]
```