Exercises — Derivatives of ln x and logₐ(x)

Constant $5$ ka slope $0$ hai. Isliye $f^{\prime }(x)=\frac{1}{x}$. Kyun: Ek constant add karne se graph upar shift hota hai lekin kabhi tilt nahi hota — steepness unchanged rehti hai.

Yahan base $a=10$ hai, isliye $\ln a=\ln 10$. $g^{\prime }(x)=\frac{1}{x\ln 10}.$ Kyun $1/x$ nahi? Sirf base $e$ clean $1/x$ deta hai, kyunki $\ln e=1$. Kisi bhi aur base ko $\frac{1}{\ln a}$ ka factor pay karna padta hai.

Constant $7$ ek multiplier ki tarah saath rehta hai: $h^{\prime }(x)=7\cdot \frac{1}{x}=\frac{7}{x}.$

Chain rule: andar $u=2x$ hai, isliye $u^{\prime }=2$, jo deta hai $\frac{u^{\prime }}{u}=\frac{2}{2x}=\frac{1}{x}$. Isliye (ii) aur (iii) ek hi cheez hain — dono sahi hain; (i) galat hai. Gehra reason: $\ln (2x)=\ln 2+\ln x$, aur $\ln 2$ ek constant hai jo differentiate hone par khatam ho jaata hai. Log ke andar ka multiplier kabhi derivative nahi badalta.

Andar $u=3x^2+1$ hai, isliye $u^{\prime }=6x$. $f^{\prime }(x)=\frac{u^{\prime }}{u}=\frac{6x}{3x^2+1}.$

Andar $u=\sin x$ hai, isliye $u^{\prime }=\cos x$. $g^{\prime }(x)=\frac{\cos x}{\sin x}=\cot x.$ Kyun: "andar ka derivative, andar se divide." Chain Rule saara kaam karta hai.

$h^{\prime }(x)=\frac{(5x{)}^{\prime }}{(5x)\ln 2}=\frac{5}{5x\ln 2}=\frac{1}{x\ln 2}.$ Kyun $5$ gayab hota hai: ${\log }_2(5x)={\log }_{2}5+{\log }_{2}x$; constant ${\log }_{2}5$ ka slope $0$ hai.

Pehle simplify karo Logarithm Laws use karke: $\ln (x^2+1{)}^3=3\ln (x^2+1)$. $k^{\prime }(x)=3\cdot \frac{2x}{x^2+1}=\frac{6x}{x^2+1}.$ Kyun simplify karein: power law $\ln A^n=n\ln A$ ek scary chain ko one-liner mein badal deta hai.

Quotient law se split karo: $\ln \frac{x}{x+1}=\ln x-\ln (x+1)$. $f^{\prime }(x)=\frac{1}{x}-\frac{1}{x+1}=\frac{(x+1)-x}{x(x+1)}=\frac{1}{x(x+1)}.$ Kyun split karein: do simple logs differentiate karna, chain rule ke andar dabbe mein band quotient rule se behtar hai.

$x>0$ ke liye: $\ln |x|=\ln x$, derivative $\frac{1}{x}$. $x<0$ ke liye: $\ln |x|=\ln (-x)$; andar $u=-x$, $u^{\prime }=-1$, isliye $\frac{-1}{-x}=\frac{1}{x}$. Dono branches ek hi answer dete hain. Figure dekho: do curve pieces mirror images hain, aur unke slopes single formula $1/x$ se match karte hain, sirf forbidden point $x=0$ chhodkar. $\boxed{\frac{d}{dx}\ln |x|=\frac{1}{x},x\ne 0.}$

Andar $u=\cos x$, $u^{\prime }=-\sin x$. $g^{\prime }(x)=\frac{-\sin x}{\cos x}=-\tan x.$

Domain: $x^2-4>0\Rightarrow |x|>2$ chahiye, yaani $x<-2$ ya $x>2$. Derivative: andar $u=x^2-4$, $u^{\prime }=2x$: $\frac{2x}{x^2-4}.$ Note karo yeh $(-\infty ,-2)$ par negative aur $(2,\infty )$ par positive hai — graph pehle girta hai phir chadta hai, do alag branches se match karta hai.

Variable exponent mein hai, isliye koi plain power rule apply nahi hoga. Dono sides ka $\ln$ lo: $\ln y=x\ln x.$ Differentiate karo (left side mein Implicit Differentiation kyunki $y$, $x$ par depend karta hai): $\frac{y^{\prime }}{y}=\ln x+x\cdot \frac{1}{x}=\ln x+1.$ $y=x^x$ se wapas multiply karo: $y^{\prime }=x^x(\ln x+1).$

$\ln$ lo aur log laws se expand karo (products → sums, powers → multipliers): $\ln y=2\ln (x+1)+\frac{1}{2}\ln (x-1)-4\ln (x+3).$ Term by term differentiate karo: $\frac{y^{\prime }}{y}=\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{4}{x+3}.$ Isliye $y^{\prime }=\frac{(x+1{)}^2\sqrt{x-1}}{(x+3{)}^4}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{4}{x+3}\right].$ Kyun yeh quotient+product+chain se behtar hai: logs ek ulajhe hue fraction ko simple pieces ki clean sum mein badal dete hain.

Variable base aur variable exponent — logarithmic differentiation zaroori hai. $\ln y=\sin x\cdot \ln x.$ $\frac{y^{\prime }}{y}=\cos x\cdot \ln x+\sin x\cdot \frac{1}{x}.$ $y^{\prime }=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).$

Change of base karo jahan naya base $x$ hai: $y={\log }_{x}2=\frac{\ln 2}{\ln x}=\ln 2\cdot (\ln x{)}^{-1}.$ $\ln 2$ ek constant hai. Power differentiate karo: $y^{\prime }=\ln 2\cdot (-1)(\ln x{)}^{-2}\cdot \frac{1}{x}=-\frac{\ln 2}{x(\ln x{)}^2}.$ Kyun yeh subtle hai: variable log ke base mein hai, isliye tum $\frac{1}{x\ln a}$ directly use nahi kar sakte — woh formula constant base maanta hai. Pehle rewrite karo.

$\frac{d}{dx}\ln x={\lim }_{h\to 0}\frac{\ln (x+h)-\ln x}{h}={\lim }_{h\to 0}\frac{1}{h}\ln \left(\frac{x+h}{x}\right).$ Kyun: quotient law $\ln A-\ln B=\ln \frac{A}{B}$ difference ko collapse kar deta hai. $t=h/x$ substitute karo (isliye $h=xt$, aur $t\to 0$ jab $h\to 0$): $={\lim }_{t\to 0}\frac{1}{xt}\ln (1+t)=\frac{1}{x}{\lim }_{t\to 0}\frac{\ln (1+t)}{t}=\frac{1}{x}\cdot 1=\frac{1}{x}.$ Engine hai standard limit $\underset{t\to 0}{\lim }\frac{\ln (1+t)}{t}=1$, jo khud $(1+t{)}^{1/t}\to e$ aur $\ln e=1$ se aata hai.

Point: $x=e$ par, $y=\ln e=1$, isliye $(e,1)$. Slope: $y^{\prime }=\frac{1}{x}$, $x=e$ par $\frac{1}{e}$ deta hai. Tangent: $y-1=\frac{1}{e}(x-e)\Rightarrow y=\frac{x}{e}-1+1=\frac{x}{e}$. $\boxed{y=\frac{x}{e}.}$ $x=e$ par tangent origin se guzarti hai — ek clean, yaadgaar fact. Figure dekho.

$y^{\prime }=\frac{1}{x}=x^{-1}$, isliye $y^{\prime \prime }=-x^{-2}=-\frac{1}{x^2}$. Kisi bhi real $x\ne 0$ ke liye, $x^2>0$, isliye sab $x>0$ ke liye $y^{\prime \prime }<0$. Interpretation: negative second derivative ka matlab hai curve har jagah concave down hai — slope $1/x$ $x$ badhne ke saath ghatta rehta hai. $\ln x$ ki hill kabhi flattening band nahi karti.

$\ln$ lo: $\ln y=v\ln u$. Dono sides differentiate karo (right par product rule, left par implicit): $\frac{y^{\prime }}{y}=v^{\prime }\ln u+v\cdot \frac{u^{\prime }}{u}.$ $y=u^v$ se multiply karo: $\boxed{y^{\prime }=u^v\left(v^{\prime }\ln u+\frac{v{u}^{\prime }}{u}\right).}$ L4.1 check karo ($u=x,v=x$): $u^{\prime }=1,v^{\prime }=1$, jo deta hai $x^x(\ln x+\frac{x}{x})=x^x(\ln x+1)$. ✓ L4.3 check karo ($u=x,v=\sin x$): $u^{\prime }=1,v^{\prime }=\cos x$, jo deta hai $x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})$. ✓

${\log }_{10}x$ ka derivative ::: $\frac{1}{x\ln 10}$ $\frac{d}{dx}\ln (\sin x)$ ::: $\cot x$ $\frac{d}{dx}\ln |x|$ aur kahan ::: $\frac{1}{x}$ sab $x\ne 0$ ke liye $\ln x$ ki $x=e$ par tangent ::: $y=x/e$ (origin se guzarti hai) $\ln x$ ka second derivative ::: $-\frac{1}{x^2}$ ($x>0$ ke liye hamesha concave down) $u^v$ ke liye master rule ::: $u^v\left(v^{\prime }\ln u+\frac{v{u}^{\prime }}{u}\right)$