Visual walkthrough — Derivatives of ln x and logₐ(x)

4.1.20 · D2 · Maths › Calculus I — Limits & Derivatives › Derivatives of ln x and logₐ(x)

Shuru karne se pehle, ek word jis par hum baar baar jaayenge: slope matlab hai "curve kitna steep hai" — height kitni change hoti hai jab tum thodi si right side chalte ho. Slope $2$ ka matlab hai "ek step right, do step climb." Slope $\frac{1}{2}$ ka matlab hai "ek step right, sirf aadha step climb." Yahi pura game hai: hum chahte hain curve $y=\ln x$ ka slope har point par.

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Step 1 — $\ln x$ asal mein kya hai? (ek picture, formula nahi)

KYA HAI. $\ln x$ ko function $e^x$ ka inverse define kiya jaata hai. "$y=\ln x$" kehna bilkul wahi baat hai jo "$e^y=x$" kehna hai. Yeh ek hi fact ko likhne ke do tarike hain.

KYUN. Abhi hum koi slope formula use nahi kar sakte — hamare paas hai nahi. Toh hume $\ln$ ko kisi aisi cheez se anchor karna hoga jis par hum trust karte hain: $e^x$, jiska slope law hum Derivative of e^x and a^x se lete hain (woh apna khud ka slope hai). Neeche ka har result uss ek relation $e^y=x$ se squeeze kiya jaata hai.

PICTURE. Figure dekho. Purple curve hai $e^x$. Coral curve hai $\ln x$. Yeh dono dashed grey line $y=x$ ke across mirror images hain — "inverse" aisa hi dikhta hai: horizontal aur vertical axes swap karo aur ek curve doosra ban jaata hai.

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Step 2 — Mirror se slope padhna

KYA HAI. $y=x$ ke across mirror "horizontal step" aur "vertical step" ke roles swap kar deta hai. Toh yeh rise aur run swap karta hai. Aur slope hai $\frac{\text{rise}}{\text{run}}$ — toh mirror slope ko ulta kar deta hai (uska reciprocal le leta hai).

KYUN. Yeh poore page ka geometric dil hai. Hum $e^x$ ka slope jaante hain; hum chahte hain uske mirror image $\ln x$ ka slope. Agar mirroring slope flip karta hai, toh hum bina kisi algebra ke ek se doosra nikal sakte hain.

PICTURE. Us point par jahan $e^x$ ka ek chhota right-triangle "step" hai run $1$, rise $m$ (slope $m$), $\ln x$ par mirrored point par wahi same triangle hai lekin legs swapped hain: run $m$, rise $1$ — slope $\frac{1}{m}$. Lal triangle aur uski reflected teal copy yeh leg-swap dikhati hai.

Ab hum is picture ko ek equation mein badlenge. Yahi Step 3 hai.

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Step 3 — $e^y=x$ ke dono sides differentiate karo

KYA HAI. Relation $e^y=x$ lo aur poochho "jab $x$ thoda right nudge kare toh har side kitni change hoti hai?" Hum same operation $\frac{d}{dx}$ (="$x$ ke respect mein rate of change") dono sides par apply karte hain.

KYUN. Ek equation tab bhi true rehti hai jab tum dono sides par same cheez karo. Right side $x$ rate $1$ par change hoti hai (thoda right jaao, $x$ exactly utna hi badhta hai). Left side $e^y$ bhi change hoti hai — lekin $y$ khud bhi silently slide kar raha hai jab $x$ move karta hai. Yahi implicit idea hai.

$\underset{\text{left side}}{\underbrace{\frac{d}{dx}\left(e^y\right)}}=\underset{=1}{\underbrace{\frac{d}{dx}(x)}}$

Yahan $\frac{d}{dx}(x)=1$ kyunki $x$ ka rate of change khud ke respect mein $1$ hota hai.

PICTURE. Figure do curves ko ek chain ki tarah stack karke dikhata hai: $x\to y\to e^y$. Ek tiny push $dx$ bottom par travel karta hai $y$ ke through aur $e^y$ se baahir nikalti hai. Hume track karna hai ki push har stage par kitni badi hai — yahi Chain Rule agle step mein karta hai.

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Step 4 — Chain rule left side ko unpack karta hai

KYA HAI. Left side $e^y$ ek function hai $y$ ka, aur $y$ ek function hai $x$ ka. Chain rule kehta hai: total rate = (rate of $e^y$ per unit $y$) $\times$ (rate of $y$ per unit $x$).

$\frac{d}{dx}\left(e^y\right)=\underset{\begin{array}{c}\text{how fast }{e}^y\\ \text{grows per unit }y\end{array}}{\underbrace{e^y}}\cdot \underset{\begin{array}{c}\text{how fast }y\\ \text{grows per unit }x\end{array}}{\underbrace{\frac{dy}{dx}}}$

KYUN. Hume exponential ke liye sirf ek slope law pata hai: $e^y$, $e^y$ ke rate par grow karta hai per unit of $y$. Lekin $x$ woh variable hai jo push ho raha hai, aur $y$ sirf $\frac{dy}{dx}$ fraction utna fast move karta hai. Dono speed fractions ko saath multiply karo — yahi chain rule hai. Iske bina hum galat units mein speed measure kar rahe hote.

PICTURE. Do gears meshed hain: $x$-gear ghoomana $y$-gear ko ratio $\frac{dy}{dx}$ par ghoomata hai, jo $e^y$-gear ko ratio $e^y$ par ghoomata hai. Overall gearing multiply hoti hai. Step 3 ki equation ab padhi jaati hai:

$e^y\cdot \frac{dy}{dx}=1.$

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Step 5 — Solve karo, aur $x$ wapas daalo

KYA HAI. Hamare paas hai $e^y\cdot \frac{dy}{dx}=1$. Dono sides ko $e^y$ se divide karo:

$\frac{dy}{dx}=\frac{1}{e^y}.$

Ab Step 1 ka relation yaad karo: $e^y=x$. Substitute karo:

$\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}.$

KYUN. Hum answer wahan express karna chahte hain jahan hum hain, $x$ — height $y$ ke terms mein nahi. Relation $e^y=x$ woh bridge hai jo $e^y$ ko plain number $x$ se swap karta hai. Yeh exactly wahi reciprocal hai jo Step 2 ke mirror se mili: $\frac{dy}{dx}=\frac{1}{e^y}$ matlab hai "one over slope of $e^x$", ab $x$ use karke likha gaya.

PICTURE. $\ln x$ ki tangent line teen jagahon par — $x=\frac{1}{2}$, $x=1$, $x=4$ — apna slope print karke. Slopes hain $2$, $1$, $\frac{1}{4}$: exactly $\frac{1}{x}$. Dekho tangent kaise flat hoti jaati hai jab $x$ badhta hai.

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Step 6 — Wahi answer first principles se (ek sanity check)

KYA HAI. Slope officially ek limit ki tarah define hota hai: step $h\to 0$ shrink karo aur dekho average slope kahan settle hoti hai.

$\frac{d}{dx}\ln x={\lim }_{h\to 0}\underset{\text{ek tiny chord ka rise over run}}{\underbrace{\frac{\ln (x+h)-\ln x}{h}}}$

Log law $\ln A-\ln B=\ln \frac{A}{B}$ use karke Logarithm Laws se, aur substitution $t=h/x$:

$=\frac{1}{x}{\lim }_{t\to 0}\frac{\ln (1+t)}{t}=\frac{1}{x}\cdot 1=\frac{1}{x}.$

KYUN. Pure geometric mirror argument (Steps 1–5) beautiful hai lekin $e^x$ ke slope par lean karta hai. Yeh doosra raasta instead lean karta hai standard limit $\frac{\ln (1+t)}{t}\to 1$ par — ek bilkul independent foundation. Jab do independent roads $\frac{1}{x}$ par pahunch jaayein, hum trust karte hain.

PICTURE. Chord (ek straight line jo $\ln x$ par do nearby points ko join karta hai) tilt hoti hai aur tangent par settle hoti hai jab $h$ shrink hota hai. Uska printed slope $\frac{1}{x}$ ki taraf creep karta hai.

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Step 7 — Koi bhi base, aur $\ln$ ka empty basement

KYA HAI. General base $a$ ke liye (jahan $a>0,a\ne 1$), base change karo: ${\log }_{a}x=\frac{\ln x}{\ln a}$. Kyunki $\ln a$ sirf ek fixed number hai, hum slope ko uss se divide karte hain:

$\frac{d}{dx}{\log }_{a}x=\frac{1}{\ln a}\cdot \frac{1}{x}=\frac{1}{x\ln a}.$

KYUN. Hum doosra rule memorise karne se mana karte hain. Har $\log$ asal mein $\ln$ ki ek rescaled copy hai: poore curve ki height ko constant $\ln a$ se divide karo toh har slope bhi $\ln a$ se divide ho jaata hai. $a=e$ set karo toh $\ln e=1$ milta hai, toh factor gayab ho jaata hai aur hum wapas $\frac{1}{x}$ par — "basement is empty."

PICTURE. Teen logs — $\ln x$, ${\log }_{2}x$, ${\log }_{10}x$ — same shape hain lekin vertically $\ln a$ se squashed hain. $x=1$ par unke slopes hain $1$, $\frac{1}{\ln 2}\approx 1.44$, $\frac{1}{\ln 10}\approx 0.43$: bada base matlab flatter, gentler curve.

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Step 8 — Degenerate & edge cases (koi gap mat chhodna)

KYA & KYUN. Teen boundaries jo smooth story quietly rely karti hai:

• $x\to 0^{+}$ (domain ka left edge). Slope $\frac{1}{x}\to +\infty$: curve vertical tangent ke saath $-\infty$ tak dive karti hai. Yeh kabhi $x=0$ nahi pahunchti; $y$-axis ek wall hai (ek asymptote). $\ln x$ simply exist nahi karta $x\le 0$ ke liye — yeh match karta hai $e^y=x$ ke saath jo kabhi non-positive $x$ produce nahi karta.
• $x=1$. $\frac{1}{1}=1$: woh ek point jahan log-hill $45^{\circ }$ ramp jitna steep hai. Yahan $\ln 1=0$, toh curve axis cross karti hai.
• Negative $x$ via $\ln |x|$. $x<0$ ke liye hum phir bhi $\ln |x|$ le sakte hain. Uska slope phir bhi $\frac{1}{x}$ hai (ab $x<0$ ke liye ek negative number) — formula cleanly extend ho jaata hai har $x\ne 0$ tak, lekin sirf absolute value ke through.

PICTURE. Poora curve jisme $x=0$ wall marked hai, $x=1$ par $45^{\circ }$ crossing flagged hai, aur $x<0$ ke liye mirrored branch $\ln |x|$ faint draw ki gayi hai.

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Ek-picture summary

Sab kuch ek saath: $e^x$ aur $\ln x$ $y=x$ ke across mirrored; step-triangles ka matched pair jiske legs swapped hain (slope $m$ ban jaata hai $\frac{1}{m}$); $\ln x$ ki tangent ek general $x$ par slope $\frac{1}{x}$ ke saath labeled; aur $e^y=x$ bridge ki tarah printed.

Recall Feynman retelling — poora walk simple words mein

Logarithm sirf exponential ka reflection hai ek $45^{\circ }$ par tilted mirror mein. Mirror left–right steps ko up–down steps se swap karta hai, aur slope hai up-over-across — toh mirror har slope ko ulta kar deta hai. Exponential ka slope height $x$ wale point par $x$ hi hota hai (yeh self-copying function hai). Use ulta karo aur milta hai $\frac{1}{x}$ — $\ln$ ka slope. Safe rehne ke liye maine yeh doosre, bilkul alag tarike se bhi kiya: tiny chords liye, do logs ko ek mein fold karne ke liye log law use ki, aur standard limit $\frac{\ln (1+t)}{t}\to 1$ par ride kiya finish tak — wahi $\frac{1}{x}$. Doosre bases wahi hill hain lekin fixed number $\ln a$ se thodi chhoti squashed, toh har slope shrink ho jaata hai $\frac{1}{x\ln a}$ tak; base $e$ basement ko empty chhodta hai kyunki $\ln e=1$. Aur fine print: hill sirf zero ke right exist karti hai, $x=1$ par yeh perfect $45^{\circ }$ ramp hai, aur $x=0$ approach karte hi yeh vertical wall ki taraf dive karti hai.

Recall Rapid self-check

$x=4$ par $\ln x$ ka slope? ::: $\frac{1}{4}$. $x=1$ par $\ln x$ ka slope? ::: $1$ (ek $45^{\circ }$ crossing). $y=x$ ke across mirror slopes ko kyun invert karta hai? ::: Yeh rise aur run swap karta hai, aur slope hai rise/run. $x=1$ par ${\log }_{2}x$ ka slope? ::: $\frac{1}{\ln 2}\approx 1.44$. $\ln x$ ka vertical tangent kahan hota hai? ::: Jab $x\to 0^{+}$, jahan $\frac{1}{x}\to \infty$.

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Connections

• Derivative of e^x and a^x — woh self-copying slope jise hum mirror karte hain.
• Implicit Differentiation — Steps 3–5.
• Chain Rule — Step 4 ke meshed gears.
• Logarithm Laws — Step 6 mein chord fold karta hai.
• Standard Limit (1+t)^{1/t} → e — Step 6 ka independent foundation.
• Logarithmic Differentiation — jahan yeh slope law kaam mein aata hai.