Derivatives of eˣ and aˣ — proofs: Question bank

4.1.19 · D5 · Maths › Calculus I — Limits & Derivatives › Derivatives of eˣ and aˣ — proofs

Shuru karne se pehle, teen tools assumed hain. Agar koi shaky lage, toh use dobara dekho:

Derivative from first principles — slope as a limit — from Limits — first principles definition of derivative.
Chain rule — ek function ke andar ek function differentiate karna — from Chain Rule.
Natural log $ln$ — $e^{(\cdot)}$ ka inverse — from Natural Logarithm ln x.

Har trap ke peeche ek hi picture hai

Is page ke almost har trap ka daroga ek hi number hai. Main use yahan build karta hoon taaki tumhe page chhodna na pade.

Yeh hai kyun woh ek number poori derivative control karta hai. Exponent law $a^{x + h} = a^{x} a^{h}$ use karke,

\frac{d}{d x} a^{x} = h \to 0 lim \frac{a ^{x} a ^{h} - a ^{x}}{h} = a^{x} M (a) h \to 0 lim \frac{a ^{h} - 1}{h} = M (a) a^{x} .

$a^{x}$ bahar nikal aata hai kyunki usme koi $h$ nahi hai. Toh $a^{x}$ ki derivative bas $a^{x}$ hai jise uske slope-at-zero se scale kiya gaya hai. Aur woh scaling number exactly natural log nikalta hai:

\frac{d}{d x} a^{x} = a^{x} ln a, M (a) = ln a

kyunki $a = e^{l n a}$ likhne se $a^{x} = e^{(l n a) x}$ milta hai, jiski derivative (chain rule) $e^{(l n a) x} \cdot ln a = a^{x} ln a$ hai.

Neeche ka figure dikhata hai ki $M (a) = ln a$ geometrically kya matlab rakhta hai: badi bases ke liye $(0, 1)$ par khinchi gayi tangent line zyada tilt karti hai.

Figure — Derivatives of eˣ and aˣ — proofs

Doosra figure curvature dikhata hai — kyun decay curves ( $0 < a < 1$ ) phir bhi upar ki taraf bend karti hain.

True or false — justify karo

Har line: true ya false decide karo, phir reason do. Sirf "true/false" se kuch nahi milega.

e^{x}

akela function hai jo apni khud ki derivative ke barabar hai.

False. Koi bhi constant multiple

C e^{x}

bhi satisfy karta hai

\frac{d}{d x} (C e^{x}) = C e^{x}

; family

C e^{x}

(

C = 0

sameta) poori set hai, lekin

e^{x}

akela nahi hai.

Agar

\frac{d}{d x} a^{x} = a^{x} ln a

hai, toh ek badi base hamesha har jagah steeper curve deti hai.

True.

ln a

a

ke saath badhta hai aur

a^{x} > 0

hamesha hota hai, toh ek bada base har

x

par slope factor bada karta hai — poori curve steeper hoti hai.

0 < a < 1

ke liye function

a^{x}

increasing hai.

False. Tab

ln a < 0

hota hai, toh

\frac{d}{d x} a^{x} = a^{x} ln a < 0

har jagah (kyunki

a^{x} > 0

): curve girती hai — yeh decay hai.

x = 0

par

a^{x}

ka slope wahan uski height ke barabar hai.

Generally false.

x = 0

par height hai

a^{0} = 1

, lekin slope hai

M (a) = ln a

(upar build kiya gaya slope constant); woh tab equal hote hain jab

ln a = 1

, yaani

a = e

1^{x}

ki derivative

1^{x} ln 1 = 0

hai.

True.

1^{x} = 1

ek flat horizontal line hai, aur

ln 1 = 0

hai, toh iska slope har jagah

0

hai — bilkul consistent.

Constant

M (a) = lim_{h \to 0} \frac{a ^{h} - 1}{h}

dono

a

aur

x

par depend karta hai.

False. Variable

x

is limit mein kabhi enter nahi karta;

M (a)

ek pure number hai jo sirf base

a

se fix hota hai (yeh

ln a

nikalta hai).

\frac{d}{d x} e^{- x} = e^{- x}

False. Inner function

- x

hai jiska derivative

- 1

hai, toh chain rule se

\frac{d}{d x} e^{- x} = - e^{- x}

— ek decreasing curve.

e^{x}

bade negative

x

ke liye negative ho sakta hai.

False.

e^{x} > 0

har real

x

ke liye; jaise

x \to - \infty

, yeh

0

ki taraf upar se approach karta hai lekin kabhi usse touch ya cross nahi karta.

Error pakdo

Har line ek "proof" ya claim batati hai jisme ek hidden flaw hai. Flaw ko ek sentence mein naam do.

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

, power rule se."

Power rule

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

ko fixed exponent aur varying base chahiye; yahan base fixed hai aur exponent vary karta hai, ulti situation, toh yeh apply nahi hota — answer

a^{x} ln a

hai.

\frac{d}{d x} e^{5 x} = e^{5 x}

kyunki

e^{kuch bhi}

apni khud ki derivative hoti hai."

Self-derivative rule sirf tab hold karta hai jab exponent exactly

x

ho; yahan inner function

5 x

hai, toh chain rule ek factor

5

add karta hai, jo

5 e^{5 x}

deta hai.

"Kyunki

a = e^{l n a}

hai, hume milta hai

M (a) = lo g_{10} a

Rewrite natural log use karta hai (

a = e^{l n a}

sirf

ln

se true hai), toh constant

ln a

hona chahiye,

lo g_{10} a

nahi.

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

, bas

a^{x} ln a

apply karke."

Exponent yahan

x^{2}

hai,

x

nahi, toh yeh plain rule nahi hai;

e^{x^{2} l n 2}

mein convert karke aur chain rule use karke ek extra inner derivative

2 x ln 2

milti hai, toh answer

2^{x^{2}} (2 x) ln 2

hai.

e^{x}

apni khud ki derivative hai, toh iska second derivative kuch naya hona chahiye."

e^{x}

ko dobara differentiate karne par bas

e^{x}

wapas milta hai;

e^{x}

ki har derivative

e^{x}

hai, toh kuch bhi naya kabhi nahi aata.

M (a) = lim_{h \to 0} \frac{a ^{h} - 1}{h}

0

hai kyunki numerator

\to 0

Denominator

h \to 0

bhi hai, toh yeh

\frac{0}{0}

indeterminate form hai; ratio

ln a

par settle hota hai,

0

par nahi — tum sirf numerator se yeh nahi padh sakte.

"Kyunki

\frac{d}{d x} a^{x} = a^{x} ln a

hai,

a^{x}

ko do baar differentiate karne par phir

a^{x} (ln a)

milta hai."

Har differentiation ek aur factor

ln a

multiply karta hai, toh second derivative

a^{x} (ln a)^{2}

hai,

a^{x} ln a

nahi.

a^{x} = (e^{l n a})^{x} = e^{l n a} \cdot x

likhke rewrite karte hain."

Powers products mein nahi badte;

(e^{l n a})^{x} = e^{(l n a) x}

exponents multiply karke milta hai,

e^{l n a} \cdot x

se nahi.

Why questions

Har ek ko genuine reasoning ki ek ya do sentences mein jawab do.

a^{x}

difference-quotient limit se bahar kyun slide karta hai jabki

a^{h} - 1

andar rehta hai?

Kyunki

a^{x + h} = a^{x} a^{h}

hai, aur

a^{x}

mein koi

h

nahi hota — woh

h \to 0

ke liye constant hai — toh woh factor out ho jaata hai, sirf

h

-dependent piece

\frac{a ^{h} - 1}{h}

limit ke andar rehta hai.

e

ko natural base kyun kehte hain, say,

10

ki jagah?

Kyunki

e

woh unique base hai jiska slope-at-height-1 constant

M (a)

exactly

1

ke barabar hota hai, jisse

e^{x}

apni khud ki derivative banti hai bina kisi extra factor ke — sabse clean possible growth law.

a^{x}

ko differentiate karne se pehle base

e

mein convert kyun karna padta hai?

Chain rule hume clean derivatives sirf known rule

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

ke through deta hai;

a^{x} = e^{(l n a) x}

likhna expression ko uss form mein daalta hai taaki derivative machinery apply ho sake.

e

se thoda bada base ek slope constant

1

se thoda bada kyun deta hai?

Kyunki

M (a) = ln a

hai aur

ln

increasing hai

ln e = 1

ke saath;

e

se thoda upar ka base

ln a

thoda upar

1

se rakhta hai, toh curve

x = 0

par thodi steeper hoti hai.

Power rule aur exponential rule

2^{x}

ke liye kabhi agree kyun nahi kar sakte?

Power rule

x 2^{x - 1}

deta, jo

2^{x}

ke relative without bound badhta, jabki true derivative

2^{x} ln 2

function ka fixed multiple rehta hai — yeh fundamentally different behaviours describe karte hain.

e

ki limit definition (

M (e) = 1

) compound-interest definition

e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}

se match kyun karti hai?

M (e) = 1

ka matlab hai

e^{h} \approx 1 + h

chote

h

ke liye, toh

e \approx (1 + h)^{1/ h}

;

h = 1/ n

set karke aur

n \to \infty

let karne par exactly compound-interest limit wapas milti hai. Dekho The number e — definitions.

Constant

M (a)

"slope of

a^{x}

x = 0

" ke same kyun hai?

General derivative

\frac{d}{d x} a^{x} = M (a) a^{x}

mein

x = 0

plug karne par

M (a) \cdot a^{0} = M (a) \cdot 1 = M (a)

milta hai, toh

M (a)

literally origin point

(0, 1)

par slope hi hai.

Edge cases

Degenerate aur boundary scenarios jinhe rules survive karne chahiye.

\frac{d}{d x} a^{x}

kya hai jab

a = 1

1^{x} = 1

constant hai, aur

ln 1 = 0

hai, toh derivative

1^{x} \cdot 0 = 0

hai — flat line se match karta hai, koi contradiction nahi.

Kya

\frac{d}{d x} a^{x} = a^{x} ln a

a = 0

ke liye valid hai?

Nahi;

0^{x}

ek well-defined positive exponential nahi hai (

x \leq 0

par undefined hai aur

ln 0

- \infty

tak diverge karta hai), toh rule ko

a > 0

chahiye.

\frac{d}{d x} a^{x}

ka kya hota hai jaise

a \to e

dono sides se?

Slope factor

ln a \to ln e = 1

continuously, toh derivative smoothly

e^{x}

ki taraf approach karti hai —

e

single balance point hai jahan

M = 1

hota hai.

0

aur

1

ke beech

a

ke liye,

a^{x}

ka second derivative positive hai ya negative?

Positive:

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

, aur

ln a

ko square karna use positive bana deta hai chahe

ln a < 0

ho, toh decay curves convex hain (woh flatten out hoti hain, upar ki taraf curve karti hain).

x = 0

par,

e^{- x}

(1/2)^{x}

mein se kaun steeper hai?

e^{- x}

ka slope

ln (e^{- 1}) = - 1

hai;

(1/2)^{x}

ka slope

ln (1/2) \approx - 0.693

hai. Toh

e^{- x}

zyada fast girta hai (

- 1

more negative hai

- 0.693

se), yaani woh origin par neeche ki taraf steeper hai.

Jaise

x \to - \infty

e^{x}

ka slope kya hai aur kya curve kabhi completely flatten hoti hai?

Slope height

e^{x}

ke barabar hoti hai, jo

\to 0^{+}

karta hai; curve horizontal ki taraf flatten hoti hai lekin kisi bhi finite

x

par zero slope kabhi nahi reach karta.

Kya

\frac{d}{d x} a^{x} = a^{x} ln a

kabhi

a = e

ke through ke alawa kisi aur point par exactly

1

slope deta hai?

Haan; kisi bhi base

a > 1

ke liye,

a^{x} ln a = 1

set karo aur solve karo — koi

x

ise satisfy karta hai, kyunki

a^{x} ln a

saare positive values cover karta hai, lekin sirf

a = e

isse $x = 0$ par hone deta hai.

Connections

Derivatives of eˣ and aˣ — proofs — woh parent proof jin par ye traps stress-test kar rahe hain.
Chain Rule — har "inner derivative bhool gaye" trap ke peeche ka tool.
Natural Logarithm ln x — kyun $M (a)$ $ln$ hona chahiye, $lo g_{10}$ nahi.
The number e — definitions — "why" section mein cross-check ki gayi do definitions.
Exponential Growth and Decay — context mein sign-of- $ln a$ edge cases.
Power Rule — woh rule jis par ye traps baar baar galat apply na karne ki warning dete hain.