Derivatives of ln x and logₐ(x): Question bank — Hinglish deep dive

4.1.20 · D5 · Maths › Calculus I — Limits & Derivatives › Derivatives of ln x and logₐ(x)

Intuition Is page ko kaise use karein

Neeche har ek item ek trap hai — ek aisi baat jo sahi lagti hai lekin kaunsa base, domain, chain rule, ya log ke slope ki limiting behaviour ke baare mein koi subtlety chupaati hai. Answer ko dhak lo, pehle apna verdict pakka karo, phir reveal karo. Agar tumhara reason hamaare se alag hai — chahe verdict match kare — toh tumne ek gap dhundh liya hai. Ye samajh test karte hain, arithmetic nahi — heavy computation ke liye doosre deep dives dekho.

Poore note mein, jab bhi log ke andar kuch $x$ se zyada complicated ho, hum us inner expression ko inside function kehte hain aur use $u = u (x)$ likhte hain; uski derivative $u^{'} = \frac{d u}{d x}$ hoti hai. $u$ ko pehle se naam dena hi hai jo neeche ke chain rule traps ko visible banata hai.

Woh ek fact jis par sab kuch tika hai: $\frac{d}{d x} ln x = \frac{1}{x}$ , aur kisi bhi doosre base ke liye tumhe $\frac{1}{ln a}$ ka toll dena padta hai. Poori derivation ke liye Derivatives of ln x and logₐ(x) dekho.

Har trap ke peeche teen pictures

Questions se pehle, teen images apne mind mein fix karo — zyaatar traps inhi teen mein se kisi ek picture ko misread karne se hote hain.

Picture 1 — slope $\frac{1}{x}$ hi $ln x$ ki steepness ki height hai. Jaise tum daayein slide karte ho, $ln x$ chadhta rehta hai lekin tangent line flat ki taraf jhukti jaati hai.

Figure — Derivatives of ln x and logₐ(x)

Picture 2 — base badalna curve ko sirf vertically stretch karta hai. $e$ se upar ka base curve ko squash karta hai (gentler slope); $1$ aur $e$ ke beech ka base usse stretch karta hai (steeper slope); $1$ se neeche ka base use ulta palat deta hai (negative slope).

Picture 3 — standard limit $lim_{t \to 0} \frac{l n ( 1 + t )}{t} = 1$ sirf $x = 1$ par $ln x$ ka slope hai. Point $(1, 0)$ ke paas log curve line $y = x - 1$ se chipki rehti hai, jiska slope exactly $1$ hai.

True ya False — justify karo

ln x

ka slope hamesha positive hota hai

True —

ln x

sirf

x > 0

ke liye defined hai, aur wahaan

\frac{1}{x} > 0

hota hai, toh curve hamesha chadhti rehti hai (kabhi flat nahi, kabhi girती nahi).

ln x

x \to \infty

ke saath tezi se steeper hoti jaati hai

False — slope

\frac{1}{x}

0

ki taraf shrink hota hai, toh curve flatten hoti hai; ye chadhti rehti hai lekin hamesha aur gently.

\frac{d}{d x} lo g_{a} x = \frac{1}{x}

har base

a

ke liye hota hai

False — sirf

a = e

ke liye. Generally ye

\frac{1}{x l n a}

hai; extra factor

\frac{1}{l n a}

sirf tab

1

hota hai jab

ln a = 1

, yaani

a = e

lo g_{a} x

aur

ln x

ek hi curve hain sirf ek vertical stretch tak

True —

lo g_{a} x = \frac{1}{l n a} ln x

, ek constant multiple, toh ye

ln x

ka vertically

\frac{1}{l n a}

se scaled version hai; isi liye unke slopes mein wahi same constant ka fark hota hai.

ln x

ka slope kisi point par exactly

1

ke barabar hota hai

True —

x = 1

par, kyunki

\frac{1}{1} = 1

. Ye wahi jagah hai jahan

ln x

4 5^{\circ}

par rise karta hai.

a > 1

ke liye

lo g_{a} x

ka slope har

x

par

ln x

ke slope se chhota hota hai

False — ye sirf tab hold hota hai jab

a > e

ho.

1 < a < e

ke liye

0 < ln a < 1

hai, toh

\frac{1}{x l n a} > \frac{1}{x}

: slope actually bada hota hai. Break-even base exactly

a = e

hai, jahaan

ln a = 1

\frac{d}{d x} ln (3 x)

\frac{d}{d x} ln x

se bada hota hai kyunki factor

3

hai

False —

ln (3 x) = ln 3 + ln x

, aur constant

ln 3

differentiate karne par vanish ho jaata hai, toh dono derivatives

\frac{1}{x}

ke barabar hain.

\frac{d}{d x} ln x = \frac{1}{x}

negative

x

ke liye bhi hold karta hai

False —

ln x

x \leq 0

ke liye undefined hai, toh uska derivative wahaan exist hi nahi karta. (Ye

ln ∣ x ∣

hai jo

x < 0

tak extend hota hai.)

\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}

sabhi

x \neq = 0

ke liye

True —

x < 0

par,

ln ∣ x ∣ = ln (- x)

, aur chain rule se

\frac{- 1}{- x} = \frac{1}{x}

;

x > 0

par ye clearly

\frac{1}{x}

hai. Toh ek clean formula dono sides cover karta hai.

ln x

ke domain mein kahin inflection point hai

False — iska second derivative

- \frac{1}{x ^{2}}

hai, jo har

x > 0

ke liye negative hai aur kabhi zero nahi hota. Concavity kabhi switch nahi hoti, toh koi inflection point nahi hai; curve poori tarah concave-down hai.

Error dhundho

\frac{d}{d x} lo g_{2} x = \frac{1}{x}

kyunki logs

\frac{1}{x}

par differentiate hote hain."

Base galat hai. Sirf base

e

\frac{1}{x}

milta hai; yahaan

\frac{d}{d x} lo g_{2} x = \frac{1}{x l n 2}

. Claim toll

\frac{1}{l n a}

bhool gaya.

\frac{d}{d x} ln (5 x) = \frac{5}{5 x} = \frac{5}{x}

Final cancellation galat ki gayi:

\frac{5}{5 x} = \frac{1}{x}

hota hai,

\frac{5}{x}

nahi. Yahaan inside function

u = 5 x

hai, toh

u^{'} = 5

andar ke

5

ko cancel kar deta hai.

\frac{d}{d x} ln (x^{2}) = \frac{1}{x ^{2}}

Chain rule drop ho gayi. Inside function

u = x^{2}

ke saath,

\frac{u ^{'}}{u} = \frac{2 x}{x ^{2}} = \frac{2}{x}

. (Equivalently

ln x^{2} = 2 ln x \Rightarrow \frac{2}{x}

"Kyunki

lo g_{a} x = \frac{l n x}{l n a}

, iska derivative

\frac{1/ x}{1/ a} = \frac{a}{x}

hai."

ln a

ek fixed number hai,

x

ka function nahi, toh ise

\frac{1}{a}

par differentiate nahi kiya jaata.

\frac{1}{l n a}

ko constant multiplier ki tarah treat karo: answer

\frac{1}{x l n a}

hai.

\frac{d}{d x} (ln x)^{2} = \frac{2}{x}

(ln x)^{2}

ln (x^{2})

se confuse kar raha hai.

ln x

ke square ke liye chain rule use karo:

2 ln x \cdot \frac{1}{x} = \frac{2 l n x}{x}

\frac{d}{d x} x^{x} = x \cdot x^{x - 1} = x^{x}

Power rule tab use nahi kar sakte jab exponent variable ho. Logarithmic differentiation use karo:

ln y = x ln x \Rightarrow y^{'} = x^{x} (ln x + 1)

. Dekho Logarithmic Differentiation.

\frac{d}{d x} ln (cos x) = \frac{1}{cos x}

Chain rule adhoori chhod di. Inside function

u = cos x

ke saath, uska derivative

u^{'} = - sin x

missing hai. Correct:

\frac{- s i n x}{c o s x} = - tan x

Why questions

Base

e

kyun sabse simple derivative deta hai?

Kyunki

ln e = 1

hai, toh toll factor

\frac{1}{l n a}

1

ban jaata hai aur disappear ho jaata hai, sirf bare

\frac{1}{x}

bachta hai. Dekho Derivative of e^x and a^x.

ln (3 x)

ke andar ka constant derivative ko affect kyun nahi karta?

Logarithm Laws se,

ln (3 x) = ln 3 + ln x

; ek constant ek flat vertical shift add karta hai, jiska slope zero hota hai, toh woh derivative mein kuch contribute nahi karta.

\frac{d}{d x} ln x

nikalne ke liye hum implicit differentiation kyun use karte hain?

Kyunki hum sirf

e^{x}

ka slope jaante hain (woh

e^{x}

hai), toh hum

e^{y} = x

se start karte hain aur Implicit Differentiation ko use karke woh known slope inverse ke slope mein transfer karte hain.

\frac{d}{d x} ln u = \frac{u ^{'}}{u}

kyun hota hai, sirf

\frac{1}{u}

kyun nahi?

Yahaan

u = u (x)

inside function hai aur

u^{'}

uski rate of change. Kyunki

u

khud

x

par depend karta hai, Chain Rule outer slope

\frac{1}{u}

ko inner slope

u^{'}

se multiply karta hai — ye account karta hai ki inside kitni tezi se move kar raha hai.

Standard limit

lim_{t \to 0} \frac{l n ( 1 + t )}{t} = 1

first-principles proof mein kyun aati hai?

x = 1

par

ln

ka derivative hi hai, aur ye

1

ke barabar hai kyunki

(1 + t)^{1/ t} \to e

aur

ln e = 1

. Dekho Standard Limit (1+t)^{1/t} → e.

Do alag-alag dikhne waale methods (algebra vs. chain rule)

ln (3 x)

ke liye same derivative kyun dete hain?

Woh same function ko do tareekon se describe karte hain; sahi maths mein agree karna zaroori hai. Chain rule

\frac{3}{3 x} = \frac{1}{x}

deta hai; log law

\frac{d}{d x} (ln 3 + ln x) = \frac{1}{x}

deta hai. Koi bhi disagreement ek algebra slip indicate karta hai.

lo g_{10} x

ka slope bade

x

ke liye itna tiny kyun hota hai?

Kyunki

ln 10 \approx 2.30

hai, toh slope

\frac{1}{x l n 10}

already

2.30

se divide ho chuka hai aur phir bade

x

se aur crush ho jaata hai — base-

10

curve

ln x

ka ek gentle, heavily-flattened version hai.

ln x

poori tarah concave down kyun hai?

Iska second derivative

- \frac{1}{x ^{2}}

hai, jo sabhi

x > 0

ke liye negative hai; negative second derivative ka matlab hai slope hamesha decreasing hai, toh tangent lines neeche ki taraf jhukti rehti hain aur curve poori jagah neeche bend karta hai.

Edge cases

x \to 0^{+}

par

\frac{d}{d x} ln x

kya hota hai?

Slope

\frac{1}{x} \to + \infty

: curve

0

ke just daayein almost vertically rise karta hai, ye reflect karta hai ki wahaan

ln x \to - \infty

kaise hota hai — domain edge par infinitely steep drop-off.

Kya koi

x

hai jahaan

ln x

ka slope zero ho?

Nahi —

\frac{1}{x}

finite

x > 0

ke liye kabhi

0

nahi hota. Slope sirf

x \to \infty

par

0

approach karta hai lekin kabhi reach nahi karta, toh

ln x

ka koi horizontal tangent nahi hai aur koi maximum nahi hai.

Kya

0 < a < 1

base ke liye

\frac{d}{d x} lo g_{a} x

ka sign change hota hai?

Haan — tab

ln a < 0

hota hai, toh

\frac{1}{x l n a} < 0

: log decreasing hai.

1

se neeche ka base curve ko flip kar deta hai, aur negative slope isi ko reflect karta hai.

a \to 1

par

\frac{1}{x l n a}

ka kya hota hai?

ln a \to 0

, toh derivative

\pm \infty

tak blow up ho jaata hai. Ye is rule ko mirror karta hai ki

lo g_{a}

a = 1

par undefined hai — base jitna

1

ke kareeb hoga, log ki steepness utni hi wild hogi.

x = 1

par, kya

ln x

aur

lo g_{a} x

ka slope same hota hai?

Nahi (jab tak

a = e

na ho). Dono point

(1, 0)

se guzarte hain, lekin wahaan unke slopes

1

aur

\frac{1}{l n a}

hain respectively — same height, alag steepness.

ln ∣ x ∣

ka slope

\frac{1}{x}

0

ke dono sides par kaise behave karta hai?

Daayein se (

x \to 0^{+}

) ye

+ \infty

par jaata hai, lekin bayein se (

x \to 0^{-}

) ye

- \infty

par jaata hai — dono one-sided slopes opposite infinities par jaate hain, toh woh

x = 0

par join nahi ho sakte.

Kya

ln ∣ x ∣

x = 0

par derivative hota hai?

Nahi —

ln ∣ x ∣

x = 0

par undefined hai (aur paas mein

- \infty

par blow karta hai), toh wahaan koi slope exist nahi karta. Formula

\frac{1}{x}

sirf

x \neq = 0

ke liye hold karta hai.

Kya

lo g_{a} x

ki concavity alag bases ke liye kabhi flip hoti hai?

a > 1

ke liye ye concave down hai (

- \frac{1}{x ^{2} l n a} < 0

);

0 < a < 1

ke liye,

ln a < 0

second derivative ko positive bana deta hai, toh flipped curve concave up hoti hai — lekin dono cases mein koi inflection point nahi hota, kyunki

x > 0

par sign kabhi change nahi hota.

Recall Ek-line self-check

Base $a$ ke liye toll? ::: Factor $\frac{1}{l n a}$ jo $\frac{1}{x}$ ko multiply karta hai.
Sirf woh base jisme koi toll nahi? ::: $a = e$ , kyunki $ln e = 1$ .
Base $a$ ki allowed values? ::: $a > 0$ aur $a \neq = 1$ .
$x \to \infty$ par $ln x$ ka slope? ::: $0$ tak shrink hota hai (flatten hota hai, negative nahi hota).
$ln x$ ka second derivative? ::: $- \frac{1}{x ^{2}}$ , poori tarah negative — concave down, koi inflection nahi.
$x < 0$ ke liye valid formula? ::: Sirf $ln ∣ x ∣$ , jo sabhi $x \neq = 0$ ke liye $\frac{1}{x}$ deta hai.
Woh base jo $lo g_{a} x$ ko $ln x$ se steeper banata hai? ::: Koi bhi $1 < a < e$ , kyunki tab $0 < ln a < 1$ .

Connections

Derivatives of ln x and logₐ(x) — parent note, full derivations ke saath.
Derivative of e^x and a^x — woh inverses jinke slopes yahaan sab kuch seed karte hain.
Chain Rule — woh tool jise in zyaatar traps ne abuse kiya ya bhool gaye.
Implicit Differentiation — kyun base $e$ $\frac{1}{x}$ deta hai.
Logarithm Laws — kyun log ke andar ke constants vanish ho jaate hain.
Logarithmic Differentiation — $x^{x}$ attack karne ka sahi tarika.
Standard Limit (1+t)^{1/t} → e — first-principles proof ke peeche.