Derivatives of eˣ and aˣ — proofs: Worked examples

4.1.19 · D3 · Maths › Calculus I — Limits & Derivatives › Derivatives of eˣ and aˣ — proofs

Yeh page parent proofs ki drill-hall hai. Parent ne bataya kyun $\frac{d}{d x} e^{x} = e^{x}$ aur $\frac{d}{d x} a^{x} = a^{x} ln a$ hota hai. Yahan hum har tarah ke exponential jo ek problem mein aa sakte hain ko dhundh ke solve karte hain — including woh sneaky degenerate cases (base $= 1$ , decaying bases, negative exponents) jo log bhool jaate hain.

Scenario matrix

Har exponential-derivative problem inhi boxes mein se kisi ek mein aati hai. Hum har row ka kam se kam ek example solve karenge.

#	Case class	Isme kya tricky hai	Covered by
1	Base $= e$ , exponent $= x$	pure self-derivative	Ex 1
2	Base $a > 1$ , exponent $= x$	ek $ln a$ factor aa jaata hai	Ex 2
3	Base $0 < a < 1$ (decay)	$ln a < 0$ → slope negative hoti hai	Ex 3
4	Degenerate base $a = 1$	$ln 1 = 0$ → derivative zero hai	Ex 4
5	Exponent ek linear inside $k x$ hai	chain rule → extra constant $k$	Ex 5
6	Exponent negative / non-linear hai	$u^{'} (x)$ ka sign aur shape	Ex 6
7	General base with a function exponent	pehle base $e$ mein convert karo	Ex 7
8	Slope at a point (geometry)	derivative evaluate karo, tangent padho	Ex 8
9	Real-world word problem (growth rate)	units lagao, number interpret karo	Ex 9
10	Exam twist: base ke liye solve karo	derivative rule ko ulta chalao	Ex 10

Case 1 — pure self-derivative

Figure — Derivatives of eˣ and aˣ — proofs

Case 2 — bada base $ln a$ ka coat pehanta hai

Case 3 — shrinking base se NEGATIVE slope milti hai

Worked example Ex 3 (cell 3,

decay). $h (x) = (\frac{1}{2})^{x}$ ko differentiate karo aur $x = 0$ pe slope nikalo. Forecast: yeh curve $x$ badhne ke saath girta hai. Toh slope… negative honi chahiye. Confirm karte hain ki sign sahi aata hai.

$h^{'} (x) = (\frac{1}{2})^{x} ln \frac{1}{2}$ . Yeh step kyun? Same rule $a^{x} ln a$ ; isme kuch bhi require nahi kiya ki $a > 1$ ho — sirf $a > 0$ , aur $\frac{1}{2} > 0$ . ✓
Note: $ln \frac{1}{2} = - ln 2 \approx - 0.6931$ . Yeh step kyun? $1$ se chhote number ka $ln$ negative hota hai — wahi negative puri derivative ko negative banata hai, encoding "curve neeche jaati hai."
$x = 0$ pe: $h^{'} (0) = 1 \cdot (- ln 2) = - ln 2 \approx - 0.6931$ . Yeh step kyun? $(\frac{1}{2})^{0} = 1$ ; $x = 0$ pe slope personality number ke barabar hai, yahan negative.

Verify: function har jagah decrease karti hai, toh negative slope sahi hai. Dhyan do ki iska size $0.6931$ , $2^{x}$ ki slope ke barabar hai — mirror bases $2$ aur $\frac{1}{2}$ ke opposite-sign personality numbers hain kyunki $ln \frac{1}{2} = - ln 2$ . ✓ Decay curve aur uska downward tangent neeche dikha hai. Dekho Exponential Growth and Decay.

Case 4 — degenerate base $a = 1$

Case 5 — linear inside function

Worked example Ex 5 (cell 5,

chain rule / linear inside). $e^{5 x}$ ko differentiate karo. Forecast: answer $e^{5 x}$ nahi hoga. Kaun sa extra factor aa jayega? (Andar ka slope, $5$ .)

Inside $u = 5 x$ identify karo, toh $u^{'} = 5$ . Yeh step kyun? Exponent bare $x$ nahi hai; Chain Rule ko andar ke slope ki zaroorat hai.
$e^{u}$ ka outside derivative $e^{u}$ hai; $u^{'}$ se multiply karo: $\frac{d}{d x} e^{5 x} = e^{5 x} \cdot 5 = 5 e^{5 x}$ . Yeh step kyun? Chain rule = (outside slope) $\times$ (inside slope). Yahan outside self-derivative hai, toh sirf $5$ add hota hai.

Verify (limit sanity): first-principles definition se, derivative hai $lim_{h \to 0} \frac{e ^{5 (x + h)} - e ^{5 x}}{h} = lim_{h \to 0} \frac{e ^{5 x} ( e ^{5 h} - 1 )}{h}$ . Yahan hum linearization $e^{t} \approx 1 + t$ use karte hain chhote $t$ ke liye (yeh exactly parent ki defining fact hai $lim_{t \to 0} \frac{e ^{t} - 1}{t} = 1$ , yaani $e^{t} - 1 \approx t$ ). $t = 5 h$ ke saath: $e^{5 h} - 1 \approx 5 h$ , toh quotient $\approx \frac{e ^{5 x} \cdot 5 h}{h} = 5 e^{5 x}$ . ✓ Match karta hai.

Case 6 — negative / non-linear inside

Worked example Ex 6 (cell 6).

$q (x) = e^{- x^{2}}$ (bell-curve shape) ko differentiate karo. Forecast: inside $- x^{2}$ ki slope $- 2 x$ hai. Toh answer mein $- 2 x$ ka factor aana chahiye — matlab slope $x = 0$ pe $0$ hogi aur $x > 0$ ke liye negative. Compute karne se pehle sign predict karo.

Inside $u = - x^{2}$ , toh $u^{'} = - 2 x$ . Yeh step kyun? Non-linear exponent → uski derivative khud $x$ ka function hai, constant nahi. $- x^{2}$ pe Power Rule use karo.
Chain rule: $q^{'} (x) = e^{- x^{2}} \cdot (- 2 x) = - 2 x e^{- x^{2}}$ . Yeh step kyun? Outside self-derivative $e^{u}$ times inside slope $- 2 x$ .
$x = 1$ pe: $q^{'} (1) = - 2 (1) e^{- 1} = - 2/ e \approx - 0.7358$ . Yeh step kyun? Point plug karo actual tangent slope paane ke liye.

Verify: $x = 0$ pe, $q^{'} (0) = - 2 (0) e^{0} = 0$ — bell ka peak, jahan tangent flat hoti hai. ✓ $x > 0$ ke liye factor $- 2 x$ negative hai, toh curve right pe neeche jaati hai — exactly bell ka right side, neeche draw kiya hai.

Case 7 — general base with a function exponent

Worked example Ex 7 (cell 7).

$r (x) = 2^{x^{2}}$ ko differentiate karo. Forecast: yahan do cheezein ek saath hain — base $e$ nahi hai aur exponent non-linear bhi hai. Safe move kya hai? Pehle base $e$ mein convert karo, phir chain rule.

Rewrite karo: $2^{x^{2}} = e^{(l n 2) x^{2}}$ . Yeh step kyun? Clean chain rule base $e$ mein rehti hai; koi bhi positive base ban jaata hai $a = e^{l n a}$ , toh $2^{x^{2}} = e^{x^{2} l n 2}$ .
Inside $u = (ln 2) x^{2}$ , slope $u^{'} = 2 x ln 2$ . Yeh step kyun? $ln 2$ ek constant multiplier hai; power rule se $x^{2}$ differentiate karo → $2 x$ , constant rakho.
$r^{'} (x) = e^{x^{2} l n 2} \cdot 2 x ln 2 = 2^{x^{2}} (2 x) ln 2$ . Yeh step kyun? Yeh phir chain rule hai — outside derivative times inside derivative: outside $e^{u}$ differentiate hokar khud ban jaata hai $e^{x^{2} l n 2}$ , aur hum andar ke slope $u^{'} = 2 x ln 2$ se multiply karte hain step 2 se. Phir $e^{x^{2} l n 2}$ ko wapas $2^{x^{2}}$ likha clean answer ke liye.
$x = 1$ pe: $r^{'} (1) = 2^{1} \cdot 2 \cdot ln 2 = 4 ln 2 \approx 2.7726$ . Yeh step kyun? Ek concrete check ke liye number mein evaluate karo.

Verify: $x = 0$ pe factor $2 x$ zero hai, toh $r^{'} (0) = 0$ — aur $2^{x^{2}}$ ka $x = 0$ pe flat minimum hai (uska lowest point, value $1$ ). ✓

Case 8 — slope-at-a-point, geometrically padhna

Figure (Ex 8): blue curve $y = e^{x}$ aur uski orange tangent $y = x + 1$ red point $(0, 1)$ pe milti hain. Tangent ki slope ( $1$ ) curve ki height ke barabar hai wahan ( $e^{0} = 1$ ) — self-derivative property visible hai. $x = 0$ se door straight line alag ho jaati hai, kyunki actual curve steepen hoti rehti hai.

Case 9 — real-world growth rate (units ke saath!)

Worked example Ex 9 (cell 9,

word problem). Ek bacteria population $P (t) = 100 e^{0.4 t}$ hai, jahan $t$ hours mein hai. $t = 3$ hours pe yeh kitni tez badh rahi hai? Forecast: "kitni tez" matlab ek derivative. Answer ki units houngi bacteria per hour, bacteria nahi. Guess: kya rate current count se badi hogi ya choti? (Rate $= 0.4 \times$ count, toh choti.)

$P^{'} (t) = 100 e^{0.4 t} \cdot 0.4 = 40 e^{0.4 t}$ . Yeh step kyun? Inside $u = 0.4 t$ ki slope $0.4$ hai; chain rule self-derivative ko $0.4$ se multiply karta hai. Constant $100$ untouched rehta hai.
$t = 3$ pe: $P^{'} (3) = 40 e^{1.2} \approx 40 \times 3.3201 \approx 132.8$ . Yeh step kyun? Maange gaye time pe evaluate karo instantaneous rate paane ke liye.

Verify (units + structure): $P^{'} (3) \approx 132.8$ bacteria per hour. Cross-check: current count $P (3) = 100 e^{1.2} \approx 332.0$ hai, aur $P^{'} (3) = 0.4 \times P (3) = 0.4 \times 332.0 = 132.8$ . ✓ Kisi bhi self-scaling exponential ke liye, growth rate $=$ (growth constant) $\times$ (current size) — yeh Exponential Growth and Decay ki pehchaan hai.

Case 10 — exam twist: rule ulta chalao

Recall Exams ke liye case-checklist

Jab bhi exponential differentiate karni ho, is order mein poocho: Is the base $e$ ya koi aur positive number? ::: Agar $e$ nahi hai, toh personality number $ln a$ nikalo (ya pehle $e^{(l n a) u}$ mein convert karo). Kya exponent bare $x$ hai ya andar ka function $u (x)$ ? ::: Agar andar ka function hai, toh $u^{'} (x)$ se multiply karo — chain rule. Kya $0 < a < 1$ hai? ::: Toh $ln a < 0$ , yaani slope negative hai — yeh decay hai. Kya $a = 1$ hai? ::: Toh $ln 1 = 0$ , yaani derivative $0$ hai (yeh bas constant hai). Kya unhe ek number chahiye (slope at a point)? ::: Pehle differentiate karo, TAB point substitute karo.

Connections

Derivatives of eˣ and aˣ — proofs — woh parent jisko yeh page drill karta hai.
Chain Rule — cases 5, 6, 7, 9 mein use hota hai.
Power Rule — andar ke exponents jaise $- x^{2}$ , $x^{2}$ differentiate karne ke liye use hota hai.
Natural Logarithm ln x — Ex 10 mein $ln a = 2$ invert karne ke liye.
Exponential Growth and Decay — cells 3 aur 9 mein.
The number e — definitions — Ex 8 ka $e^{x} \approx 1 + x$ tangent.
Limits — first principles definition of derivative — Ex 5 mein sanity check.