Derivatives of ln x and logₐ(x): Worked examples

4.1.20 · D3 · Maths › Calculus I — Limits & Derivatives › Derivatives of ln x and logₐ(x)

Scenario matrix

Kuch bhi work karne se pehle, har case-class list karte hain jo is topic se aa sakti hai. Har row ek "cell" hai. Neeche ke worked examples us cell se tagged hain jise woh cover karte hain, taaki milke poora grid fill ho jaye.

#	Cell (case class)	Kya alag hai isme	Example
C1	Plain $ln$ of a polynomial	pure chain rule $u^{'} / u$	(a)
C2	Non- $e$ base $lo g_{a}$	extra $\frac{1}{ln a}$ factor	(b)
C3	Pehle simplify karo (log laws)	chain rule ki zaroorat hi nahi	(c)
C4	Negative / absolute-value input	domain, $ln ∣ x ∣$	(d)
C5	Product/quotient inside	log laws use karke sum ban jaata hai	(e)
C6	Variable exponent mein	logarithmic differentiation	(f)
C7	Real-world rate (word problem)	units, interpretation	(g)
C8	Exam twist / degenerate	limiting value, hidden constant	(h)

Hum aath examples se aathon cells cover karenge.

Pehle picture padho

Har example ke peeche ek hi fact hai — $ln x$ ki shape aur uska slope $1/ x$ . Figure dekho: height badhti hai lekin dheere dheere, aur slope triangles right jaane par flatten hote jaate hain.

Figure — Derivatives of ln x and logₐ(x)

Worked examples

(a) Cell C1 — plain $ln$ of a polynomial

(b) Cell C2 — non- $e$ base

(c) Cell C3 — pehle log laws se simplify karo

Worked example

$h (x) = ln \frac{x ^{3}}{x + 1}$ differentiate karo Forecast: yeh ek quotient-plus-chain nightmare lagta hai. Log laws ise do-term sum bana dete hain. Guess karo answer mein kitne terms honge.

Log laws se expand karo: $h (x) = ln x^{3} - ln (x + 1)^{1/2} = 3 ln x - \frac{1}{2} ln (x + 1)$ . Yeh step kyun? $ln \frac{A}{B} = ln A - ln B$ aur $ln A^{p} = p ln A$ ek mushkil log ko do aasaan logon mein badal dete hain (Logarithm Laws).
Term by term differentiate karo: $h^{'} (x) = 3 \cdot \frac{1}{x} - \frac{1}{2} \cdot \frac{1}{x + 1} .$ Yeh step kyun? Ab har term plain $ln$ hai; doosra $\frac{d}{d x} ln (x + 1) = \frac{1}{x + 1}$ use karta hai.
Tidy karo: $h^{'} (x) = \frac{3}{x} - \frac{1}{2 ( x + 1 )} .$ Yeh step kyun? Sirf cosmetic — sum ke roop mein chhodo, yeh already clean hai.

Verify: original ka domain $x > 0$ hai (chahiye $x^{3} > 0$ aur $x + 1 > 0$ ). $x = 1$ par: $h^{'} (1) = 3 - \frac{1}{4} = \frac{11}{4} = 2.75$ . Reasonable aur positive — function wahan increasing hai. ✓

(d) Cell C4 — negative input aur $ln ∣ x ∣$

Worked example

$p (x) = ln ∣ x ∣$ differentiate karo, aur $p^{'} (- 2)$ evaluate karo Forecast: negative number ka $ln$ undefined hai — lekin $ln ∣ x ∣$ $x < 0$ ke liye theek hai. Kya tumhare khayaal mein slope formula left side par sign flip karta hai?

$x > 0$ ke liye: $∣ x ∣ = x$ , toh $p (x) = ln x$ aur $p^{'} (x) = \frac{1}{x}$ . Yeh step kyun? Positive side par koi absolute value ki chinta nahi.
$x < 0$ ke liye: $∣ x ∣ = - x$ , toh $p (x) = ln (- x)$ . Chain rule $u = - x$ , $u^{'} = - 1$ ke saath: $p^{'} (x) = \frac{u ^{'}}{u} = \frac{- 1}{- x} = \frac{1}{x} .$ Yeh step kyun? Do minus signs cancel ho jaate hain — slope dono sides par $1/ x$ hai.
Combine karo: $p^{'} (x) = \frac{1}{x}$ sabhi $x \neq = 0$ ke liye. Yeh step kyun? Ek tidy rule poora domain cover karta hai minus forbidden point $x = 0$ .

Verify: $p^{'} (- 2) = \frac{1}{- 2} = - 0.5$ . Neeche figure mein mirror-image branch dikhti hai; left side par curve $x$ ke $0$ ki taraf badhne par girta hai, toh negative slope sahi hai. ✓

(e) Cell C5 — log ke andar product

(f) Cell C6 — variable exponent mein (logarithmic differentiation)

Worked example

$x > 0$ ke liye $y = (2 x)^{x}$ differentiate karo Forecast: exponent variable hai, toh na power rule na $a^{x}$ rule akele kaam karta hai. Kaun sa tool exponent ko neeche kheenchta hai? Logs. Dekho Logarithmic Differentiation.

Dono sides ka $ln$ lo: $ln y = x ln (2 x)$ . Yeh step kyun? $ln A^{p} = p ln A$ variable exponent $x$ ko ek aisi product mein laata hai jise hum differentiate kar sakte hain.
Right side simplify karo: $ln (2 x) = ln 2 + ln x$ , toh $ln y = x ln 2 + x ln x$ . Yeh step kyun? Pehle split karna product rule ko aasaan banata hai.
Dono sides differentiate karo (left par Implicit Differentiation): $\frac{y ^{'}}{y} = ln 2 + (ln x + x \cdot \frac{1}{x}) = ln 2 + ln x + 1.$ Yeh step kyun? Left side: $\frac{d}{d x} ln y = \frac{y ^{'}}{y}$ (chain rule). Right: $x ln x$ par product rule.
$y^{'}$ ke liye solve karo: $y^{'} = (2 x)^{x} (ln 2 + ln x + 1) .$ Yeh step kyun? $y = (2 x)^{x}$ se multiply karo wapas.

Verify: note karo $ln 2 + ln x = ln (2 x)$ , toh $y^{'} = (2 x)^{x} (ln (2 x) + 1)$ — same shape jaise parent ka $x^{x} (ln x + 1)$ , log ke andar $x \to 2 x$ ke saath. $x = 1$ par: $y = 2^{1} = 2$ aur $y^{'} = 2 (ln 2 + 0 + 1) = 2 (ln 2 + 1) \approx 3.386$ . ✓

(g) Cell C7 — real-world rate (word problem)

Worked example Ek quiet zone ki loudness decibels mein

$L (I) = 10 lo g_{10} (\frac{I}{I _{0}})$ hai, jahan $I$ sound intensity hai aur $I_{0}$ ek fixed reference hai. $L$ ka $I$ ke saath rate of change kitna hai? $I = 100 I_{0}$ par evaluate karo. Forecast: zyada intensity matlab zyada loudness — lekin log ka matlab hai har extra unit of $I$ kam add karta hai. Guess karo: kya $\frac{d L}{d I}$ $I$ badhne par grow karega ya shrink karega?

Rewrite karo: $L = 10 (lo g_{10} I - lo g_{10} I_{0})$ ; doosra term constant hai. Yeh step kyun? $lo g_{10} I_{0}$ ka derivative zero hai — sirf $lo g_{10} I$ vary karta hai.
Master rule $\frac{d}{d I} lo g_{10} I = \frac{1}{I l n 10}$ use karke differentiate karo: $\frac{d L}{d I} = 10 \cdot \frac{1}{I l n 10} = \frac{10}{I l n 10} .$ Yeh step kyun? Base-10 log $\frac{1}{l n 10}$ factor laata hai (Cell C2 ka lesson).
$I = 100 I_{0}$ par evaluate karo: $\frac{d L}{d I} = \frac{10}{100 I _{0} ln 10} = \frac{1}{10 I _{0} ln 10} .$ Yeh step kyun? Value plug karo aur $I_{0}$ ko symbolic rakhho (yeh intensity ke units ke saath ek fixed constant hai).

Verify: units hain $decibels per unit intensity$ , aur answer $1/ I$ ki tarah shrink karta hai — exactly figure s01 se wahi flattening. Louder sounds ko same decibel gain ke liye intensity ka bada jump chahiye, isliye decibels logarithmic hain. Numerically $\frac{10}{100 l n 10} \approx 0.04343$ dB per unit ( $I_{0} = 1$ leke). ✓

(h) Cell C8 — exam twist / degenerate case

Worked example

$\frac{d}{d x} lo g_{a} x$ find karo aur batao kya hota hai jab base $a \to 1^{+}$ . Forecast: formula hai $\frac{1}{x l n a}$ . Jab $a$ $1$ ki taraf slide karta hai toh $ln a$ kya karta hai? Yahi trap hai.

Master formula: $\frac{d}{d x} lo g_{a} x = \frac{1}{x ln a}$ . Yeh step kyun? Seedha change of base se; fixed $a$ ke liye $ln a$ ek constant hai.
Degenerate limit: jab $a \to 1^{+}$ , $ln a \to 0^{+}$ , toh $\frac{1}{x ln a} \to + \infty$ . Yeh step kyun? $0$ ki taraf shrink hoti cheez se divide karne par slope blow up ho jaata hai.
Interpret karo: $lo g_{1} x$ undefined hai — isliye parent note demand karta hai $a \neq = 1$ . Derivative ka diverge karna forbidden base ko signal karta hai. Yeh step kyun? Calculus boundary par warn karta hai: infinite slope matlab function khud collapse ho gaya.

Verify: doosra special base $a = e$ sanity check karo: $ln e = 1$ deta hai $\frac{1}{x \cdot 1} = \frac{1}{x}$ , plain $ln$ recover ho gaya (parent §3). Aur $a = 2$ par $x = 1$ par $\frac{1}{1 \cdot l n 2} \approx 1.4427$ milta hai. Dono finite aur correct; sirf $a = 1$ misbehave karta hai. ✓

Active recall

Recall Har problem kaun sa cell hit karta hai? (right side cover karo)

$\frac{d}{d x} ln (x^{2} + 4)$ ::: C1 — plain chain rule, answer $\frac{2 x}{x ^{2} + 4}$ . $\frac{d}{d x} lo g_{10} (7 x)$ ::: C2 — extra $\frac{1}{l n 10}$ , answer $\frac{1}{x l n 10}$ . $\frac{d}{d x} ln \frac{x ^{3}}{x + 1}$ ::: C3 — pehle split karo, answer $\frac{3}{x} - \frac{1}{2 ( x + 1 )}$ . $\frac{d}{d x} ln ∣ x ∣$ ::: C4 — $\frac{1}{x}$ sabhi $x \neq = 0$ ke liye. $\frac{d}{d x} ln (x^{2} e^{3 x})$ ::: C5 — $\frac{2}{x} + 3$ . $\frac{d}{d x} (2 x)^{x}$ ::: C6 — $(2 x)^{x} (ln (2 x) + 1)$ . $lo g_{a} x$ $a \to 1$ par kyun break karta hai? ::: C8 — $ln a \to 0$ , slope $\to \infty$ , base $1$ forbidden.

Connections

Chain Rule — C1, C4 ka engine.
Logarithm Laws — C3, C5 ko sums mein badalta hai.
Logarithmic Differentiation — C6 ka eklauta raasta.
Implicit Differentiation — C6 mein $\frac{d}{d x} ln y$ ke andar use hota hai.
Derivative of e^x and a^x — C5 mein $ln e^{3 x}$ simplify karne wala inverse fact.
Standard Limit (1+t)^{1/t} → e — woh foundation jis par parent ka $1/ x$ tika hai.
Parent topic — woh master formulas jo yeh page exercise karta hai.