=\lim_{h\to0}\frac1h\ln\!\Big(\frac{x+h}{x}\Big).$$
*Why this step?* Quotient rule for logs: $\ln A-\ln B=\ln\frac{A}{B}$.
Let $t=h/x$ so $h=xt$ and $t\to0$:
$$=\lim_{t\to0}\frac{1}{xt}\ln(1+t)=\frac1x\lim_{t\to0}\frac{\ln(1+t)}{t}=\frac1x\cdot 1=\frac1x.$$
*Why this step?* The standard limit $\lim_{t\to0}\frac{\ln(1+t)}{t}=1$ (it's the derivative of $\ln$ at $1$, or equivalently $\ln e =1$ since $(1+t)^{1/t}\to e$).
> [!formula] Master results
> $$\frac{d}{dx}\ln x=\frac1x \qquad
> \frac{d}{dx}\log_a x=\frac{1}{x\ln a}$$
> Chain-rule versions: $\displaystyle \frac{d}{dx}\ln u=\frac{u'}{u},\qquad \frac{d}{dx}\log_a u=\frac{u'}{u\ln a}.$
---
## 3. Deriving $\dfrac{d}{dx}\log_a x$
**HOW:** don't memorise — convert to $\ln$ using change of base.
$$\log_a x=\frac{\ln x}{\ln a}.$$
*Why this step?* $\ln a$ is a **constant** (just a number), so we differentiate $\ln x$ and divide.
$$\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:** set $a=e$. Then $\ln a=\ln e=1$, giving $\frac{1}{x\cdot1}=\frac1x$. ✓ Consistent with $\ln x$.
![[4.1.20-Derivatives-of-ln-x-and-loga(x).png]]
---
## 4. Worked examples
> [!example] (a) $f(x)=\ln(3x^2+1)$
> Use $\frac{d}{dx}\ln u=\frac{u'}{u}$ with $u=3x^2+1$.
> $u'=6x$. So $f'(x)=\dfrac{6x}{3x^2+1}.$
> *Why this step?* The chain rule: differentiate the inside, divide by the inside.
> [!example] (b) $g(x)=\log_2(5x)$
> $g'(x)=\dfrac{(5x)'}{(5x)\ln 2}=\dfrac{5}{5x\ln 2}=\dfrac{1}{x\ln 2}.$
> *Why this step?* The constant $5$ cancels — note $\log_2(5x)=\log_2 5+\log_2 x$, and $\log_2 5$ is constant, so only $\log_2 x$ contributes. Same answer, faster.
> [!example] (c) $h(x)=\ln\sqrt{x}=\tfrac12\ln x$
> Simplify **first**: $h(x)=\frac12\ln x \Rightarrow h'(x)=\dfrac{1}{2x}.$
> *Why this step?* Log laws ($\ln x^{1/2}=\frac12\ln x$) turn a chain-rule problem into a one-liner.
> [!example] (d) Logarithmic differentiation: $y=x^x$ ($x>0$)
> Take $\ln$: $\ln y = x\ln x$. *Why?* Exponent has a variable in it; logs pull it down.
> Differentiate: $\frac{y'}{y}=\ln x + x\cdot\frac1x=\ln x+1$.
> So $y'=x^x(\ln x+1).$
---
## 5. Common mistakes (Steel-man + fix)
> [!mistake] "$\dfrac{d}{dx}\log_a x=\dfrac1x$"
> **Why it feels right:** it looks just like $\ln$. **The truth:** only $\ln$ (base $e$) gives clean $1/x$. For other bases you pay an extra factor $\dfrac{1}{\ln a}$. **Fix:** always write $\frac{1}{x\ln a}$ and sanity-check with $a=e\Rightarrow\ln a=1$.
> [!mistake] "$\dfrac{d}{dx}\ln(3x)=\dfrac{1}{3x}\cdot 3=\dfrac3{3x}$... so it's different from $\ln x$?"
> **Why it feels right:** chain rule says multiply by $u'=3$. **The truth:** $\frac{3}{3x}=\frac1x$ — same as $\ln x$! Because $\ln(3x)=\ln 3+\ln x$ and the constant dies. **Fix:** both methods agree; trust the algebra.
> [!mistake] Forgetting the domain.
> **Why it feels right:** $1/x$ is defined for negative $x$. **The truth:** $\ln x$ needs $x>0$. For $\ln|x|$, the derivative is $\frac1x$ on *both* sides $x\neq0$. **Fix:** state the domain.
---
## 6. Active recall
> [!recall]- Quick self-test (cover the answers)
> - $\frac{d}{dx}\ln x = ?$ → $\frac1x$
> - $\frac{d}{dx}\log_a x = ?$ → $\frac1{x\ln a}$
> - Why is base $e$ special? → $\ln e=1$, killing the $\ln a$ factor.
> - $\frac{d}{dx}\ln(\cos x)=?$ → $\frac{-\sin x}{\cos x}=-\tan x$.
> [!recall]- Feynman: explain to a 12-year-old
> Imagine climbing a hill whose height is $\ln x$ "log-steps." When you're far to the right (big $x$), each extra step barely lifts you — the hill is nearly flat, slope tiny like $1/x$. Near $x=1$ the slope is exactly $1$ — steep. The slope is always "one over how far right you are." For a different base log it's the same hill but **stretched** by the number $\ln a$, so the slope shrinks by that factor.
> [!mnemonic] Memory hook
> **"Log down, base in the basement."** $\log_a$ → answer is **one over** $x$ times $\ln a$ sitting **down** in the denominator (the "basement"). For $\ln$, the basement is empty ($\ln e=1$).
---
## 7. Connections
- [[Derivative of e^x and a^x]] — these are the inverses; slopes are reciprocals.
- [[Chain Rule]] — powers every $\ln u$ derivative.
- [[Implicit Differentiation]] — Method A above.
- [[Logarithm Laws]] — simplify before differentiating.
- [[Logarithmic Differentiation]] — for $x^x$, products of many factors.
- [[Standard Limit (1+t)^{1/t} → e]] — underpins Method B.
---
#flashcards/maths
Derivative of $\ln x$ ::: $\dfrac1x$ for $x>0$.
Derivative of $\log_a x$ ::: $\dfrac{1}{x\ln a}$.
Why does base $e$ give the simplest derivative? ::: Because $\ln e=1$, so the factor $\ln a$ disappears.
Derive $\frac{d}{dx}\ln x$ in one line ::: $e^y=x\Rightarrow e^y y'=1\Rightarrow y'=1/e^y=1/x$.
$\frac{d}{dx}\ln u = ?$ (chain rule) ::: $\dfrac{u'}{u}$.
$\frac{d}{dx}\ln(3x)=?$ ::: $\dfrac1x$ (the constant $3$ cancels).
$\frac{d}{dx}\ln|x|=?$ ::: $\dfrac1x$ for all $x\neq0$.
Derivative of $x^x$ ::: $x^x(\ln x+1)$, via $\ln y=x\ln x$.
Standard limit used in the first-principles proof ::: $\lim_{t\to0}\frac{\ln(1+t)}{t}=1$.
## 🖼️ 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]
```
## 🔊 Hinglish (regional understanding)
> [!intuition]- Hinglish mein samjho
> Dekho, idea bilkul simple hai. $\ln x$ jo hai woh $e^x$ ka **inverse** hai. Aur $e^x$ ki khaas baat yeh hai ki uska derivative khud $e^x$ hota hai. Isi wajah se jab hum inverse ka slope nikaalte hain, woh super clean aa jaata hai: $\frac{d}{dx}\ln x = \frac1x$. Proof yaad rakhne ke liye implicit method best hai — $y=\ln x$ ko $e^y=x$ likho, dono taraf differentiate karo, $e^y\,y'=1$, aur $e^y=x$ daal do, ho gaya $y'=1/x$.
>
> Ab $\log_a x$ ke liye alag formula ratne ki zaroorat nahi. Bas **change of base** lagao: $\log_a x = \frac{\ln x}{\ln a}$. Yahan $\ln a$ to ek constant number hai, toh derivative ban jaata hai $\frac{1}{x\ln a}$. Check karo — agar $a=e$, toh $\ln e=1$, aur answer wapas $\frac1x$ aa jaata hai. Yahi reason hai ki base $e$ ko "natural" kehte hain: extra factor gayab ho jaata hai.
>
> Sabse common galti: students $\log_a x$ ka derivative bhi $\frac1x$ likh dete hain. Galat — $\ln a$ wala factor mat bhoolna. Doosri baat, chain rule: $\frac{d}{dx}\ln u = \frac{u'}{u}$ — andar ko differentiate karo, andar se divide karo. Aur ek pro tip: differentiate karne se pehle log laws se simplify kar lo (jaise $\ln\sqrt x=\frac12\ln x$), problem aadhi aasaan ho jaati hai. Exam mein 80/20 rule: yeh do formula aur change-of-base — bas itne se hi poora subtopic cover ho jaata hai.
![[audio/4.1.20-Derivatives-of-ln-x-and-loga(x).mp3]]