Exercises — Derivatives of sin x, cos x — proofs from first principles

Definition. Derivative wo limit hai jo average slope ki hoti hai jab interval shrink hokar kuch nahi reh jaata: $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}.$ Yahan $f(x+h)-f(x)$ rise hai aur $h$ run hai, toh fraction ek slope hai; limit $x$ par tangent line ki slope pick karti hai.

Ingredients. ${\lim }_{h\to 0}\frac{\sin h}{h}=1,{\lim }_{h\to 0}\frac{\cos h-1}{h}=0.$

$\sin (x+h)=\sin x\cos h+\cos x\sin h,$ $\cos (x+h)=\cos x\cos h-\sin x\sin h.$ Ye Trigonometric addition formulas se aate hain. Proof mein inका kaam hai moving part $h$ ko fixed part $x$ se alag karna, taaki $h$-limits sirf $h$ wale pieces par act kar sakein.

(a) $1$. (b) $0$. (c) $-\sin x$.

Idea: ingredient chahta hai ki wahi angle upar aur neeche ho. $\frac{5}{5}$ se multiply karke force karo: $\frac{\sin (5h)}{h}=\frac{\sin (5h)}{5h}\cdot 5.$ Jab $h\to 0$, inner angle $5h\to 0$ bhi, toh $\frac{\sin (5h)}{5h}\to 1$. Isliye ${\lim }_{h\to 0}\frac{\sin (5h)}{h}=1\cdot 5=\boxed{5}.$

Har sine ko apne angle se match karo taaki dono ingredient ban jaayein: $\frac{\sin (3h)}{\sin (7h)}=\frac{\sin (3h)}{3h}\cdot \frac{7h}{\sin (7h)}\cdot \frac{3h}{7h}.$ Jab $h\to 0$: pehla factor $\to 1$, doosra factor $\to 1$ (ingredient ka reciprocal), teesra factor constant $\frac{3}{7}$ hai. Toh ${\lim }_{h\to 0}\frac{\sin (3h)}{\sin (7h)}=1\cdot 1\cdot \frac{3}{7}=\boxed{\frac{3}{7}}.$

$f^{\prime }\left(\frac{\pi }{2}\right)={\lim }_{h\to 0}\frac{\sin (\frac{\pi }{2}+h)-\sin \frac{\pi }{2}}{h}.$ Kyunki $\sin \frac{\pi }{2}=1$ aur $\cos \frac{\pi }{2}=0$, addition formula deta hai $\sin (\frac{\pi }{2}+h)=\sin \frac{\pi }{2}\cos h+\cos \frac{\pi }{2}\sin h=\cos h.$ Toh $f^{\prime }\left(\frac{\pi }{2}\right)={\lim }_{h\to 0}\frac{\cos h-1}{h}=0.$ Aur $\cos \frac{\pi }{2}=0$. ✔ Dono agree karte hain. (Geometrically: $\sin x$ apni flat peak par hai $x=\frac{\pi }{2}$ par, toh slope $0$.)

Setup. $\frac{d}{dx}\cos x={\lim }_{h\to 0}\frac{\cos (x+h)-\cos x}{h}.$ Step 1 — $x$ ko $h$ se alag karo addition formula se (ye ek hi algebraic tool hai jo sum-angle ko separate karta hai): $={\lim }_{h\to 0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}.$ Step 2 — $\cos x$ terms group karo taaki har ingredient appear ho sake: $={\lim }_{h\to 0}\left[\cos x\cdot \frac{\cos h-1}{h}-\sin x\cdot \frac{\sin h}{h}\right].$ Step 3 — $\cos x$ aur $\sin x$, $h$ ke respect mein constants hain, inhe bahar nikalo aur ingredients ($0$ aur $1$) substitute karo: $=\cos x\cdot 0-\sin x\cdot 1=\boxed{-\sin x}.$ Kaun sa ingredient deta hai? Surviving $-\sin x$ $\lim \frac{\sin h}{h}=1$ se aata hai ($\frac{\cos h-1}{h}$ wala term $0$ ho jaata hai). Minus sign cosine addition formula ke andar ke $-$ se aata hai.

Neeche diya figure dekho.

Unit circle (radius $1$) par, centre par angle $h$ mark karo.

• Inner triangle (blue): base $1$ horizontal par, height $=\sin h$ (point ki vertical drop). Area $=\frac{1}{2}\cdot 1\cdot \sin h=\frac{1}{2}\sin h$.
• Sector (green wedge): disc ka ek slice angle $h$ wala. Radius $1$ wali poori disc ka area $\pi$ hai; slice ka fraction $\frac{h}{2\pi }$ hai, jo deta hai $\frac{h}{2\pi }\cdot \pi =\frac{1}{2}h$. Is step ko radians chahiye — fraction $\frac{h}{2\pi }$ tabhi kaam karta hai jab $h$ radians mein ho.
• Outer triangle (orange): iski vertical side $(1,0)$ par circle ki tangent line hai, height $\tan h$. Kyun $\tan$? Kyunki $\tan h=\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{vertical side}}{1}$, toh tangent segment exactly measure karta hai ki outer triangle kitna upar jaata hai jab uska base radius $1$ ho. Iska area $=\frac{1}{2}\cdot 1\cdot \tan h=\frac{1}{2}\tan h$.

Geometrically inner triangle sector ke andar baitha hai, jo outer triangle ke andar baitha hai, toh areas nest karte hain: $\frac{1}{2}\sin h\le \frac{1}{2}h\le \frac{1}{2}\tan h.$ $2$ se multiply karo aur $\sin h>0$ se divide karo (small $h>0$ ke liye valid): $1\le \frac{h}{\sin h}\le \frac{1}{\cos h},$ kyunki $\frac{\tan h}{\sin h}=\frac{1}{\cos h}$. Jab $h\to 0$, $\cos h\to 1$, toh dono ends $\to 1$ aur Squeeze Theorem $\frac{h}{\sin h}\to 1$ trap karta hai, isliye $\frac{\sin h}{h}\to 1$.

Conjugate trick use karo — $\frac{1+\cos h}{1+\cos h}$ se multiply karo taaki cosine sine mein convert ho ($1-{\cos }^{2}h={\sin }^{2}h$ se): $\frac{1-\cos h}{h^2}\cdot \frac{1+\cos h}{1+\cos h}=\frac{1-{\cos }^{2}h}{h^2(1+\cos h)}=\frac{{\sin }^{2}h}{h^2(1+\cos h)}.$ Ab ingredient pieces mein split karo: $={\left(\frac{\sin h}{h}\right)}^2\cdot \frac{1}{1+\cos h}.$ Jab $h\to 0$: ${\left(\frac{\sin h}{h}\right)}^2\to 1^2=1$, aur $\frac{1}{1+\cos h}\to \frac{1}{1+1}=\frac{1}{2}$. Toh ${\lim }_{h\to 0}\frac{1-\cos h}{h^2}=\boxed{\frac{1}{2}}.$ Ingredient 2 se farq: ingredient 2 mein denominator $h$ hai (ek power) aur equals $0$ hai; is wale mein $h^2$ hai aur equals $\frac{1}{2}$ hai. $h$ ki extra power hi $\frac{1}{2}$ ko rescue karti hai.

$g^{\prime }(x)={\lim }_{h\to 0}\frac{\sin (2(x+h))-\sin (2x)}{h}={\lim }_{h\to 0}\frac{\sin (2x+2h)-\sin (2x)}{h}.$ Step 1 — addition formula moving angle $2h$ ke saath: $\sin (2x+2h)=\sin (2x)\cos (2h)+\cos (2x)\sin (2h).$ Step 2 — subtract aur group karo: $g^{\prime }(x)={\lim }_{h\to 0}\left[\sin (2x)\cdot \frac{\cos (2h)-1}{h}+\cos (2x)\cdot \frac{\sin (2h)}{h}\right].$ Step 3 — har limit ko ingredient form mein repair karo (angle $2h$ upar aur neeche match hona chahiye): $\frac{\cos (2h)-1}{h}=\frac{\cos (2h)-1}{2h}\cdot 2\to 0\cdot 2=0,\frac{\sin (2h)}{h}=\frac{\sin (2h)}{2h}\cdot 2\to 1\cdot 2=2.$ Step 4 — substitute karo: $g^{\prime }(x)=\sin (2x)\cdot 0+\cos (2x)\cdot 2=\boxed{2\cos (2x)}.$ $2$ kahan paida hota hai: Step 3 mein, $\frac{\sin (2h)}{h}\to 2$ se — inner angle ki rate se. Ye Chain Rule ka haath se bana preview hai.

$\frac{d}{dx}\sin x=\cos x$ aur $\frac{d}{dx}\cos x=-\sin x$ baar baar apply karo: $f^{\prime }(x)=\cos x,f^{\prime \prime }(x)=-\sin x,f^{\prime \prime \prime }(x)=-\cos x,f^{(4)}(x)=\sin x.$ Pattern: har differentiation wave ko $90^{\circ }=\frac{\pi }{2}$ aage advance karta hai: $\sin \to \cos \to -\sin \to -\cos \to \sin ,$ ek 4-step cycle. Equivalently $\frac{d}{dx}\sin x=\sin \left(x+\frac{\pi }{2}\right)$. Toh $f^{(n)}(x)=\sin \left(x+\frac{n\pi }{2}\right)$, aur derivatives period $4$ se repeat karte hain. Ye Taylor series of sin and cos se juda hai.

$\frac{d}{dx}\tan x={\lim }_{h\to 0}\frac{\tan (x+h)-\tan x}{h}={\lim }_{h\to 0}\frac{1}{h}\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right].$ Step 1 — common denominator: $={\lim }_{h\to 0}\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{h\cos (x+h)\cos x}.$ Step 2 — numerator ulta addition formula hai: $\sin (x+h)\cos x-\cos (x+h)\sin x=\sin ((x+h)-x)=\sin h.$ (Ye Trigonometric addition formulas ka subtraction formula hai — choose isliye kiya kyunki ye poore numerator ko ek single $\sin h$ mein collapse kar deta hai.) $={\lim }_{h\to 0}\frac{\sin h}{h\cos (x+h)\cos x}.$ Step 3 — ek ingredient aur baaki mein split karo: $={\lim }_{h\to 0}\frac{\sin h}{h}\cdot \frac{1}{\cos (x+h)\cos x}=1\cdot \frac{1}{\cos x\cdot \cos x}=\frac{1}{{\cos }^{2}x}=\boxed{{\sec }^{2}x}.$ Ye bilkul wahi result hai jo Derivatives of tan, sec, csc, cot mein hai, yahan raw ingredients se build kiya gaya.

Har sine ko apne angle se match karo: $\frac{\sin (ah)}{\sin (bh)}=\frac{\sin (ah)}{ah}\cdot \frac{bh}{\sin (bh)}\cdot \frac{ah}{bh}.$ Jab $h\to 0$, dono inner angles $\to 0$, toh $\frac{\sin (ah)}{ah}\to 1$ aur $\frac{bh}{\sin (bh)}\to 1$. Aakhri factor constant $\frac{a}{b}$ hai. Isliye ${\lim }_{h\to 0}\frac{\sin (ah)}{\sin (bh)}=\frac{a}{b}.$ Degenerate case $b=0$: tab denominator $\sin (0)=0$ hai har $h$ ke liye, toh expression $\frac{\sin (ah)}{0}$ undefined hai — limit exist nahi karta (division by zero). Formula $\frac{a}{b}$ bhi toot jaata hai ($b$ denominator mein hai), consistently failure signal karta hai.

Kyun invalid hai — circularity. L'Hôpital $\frac{\sin h}{h}$ ko $\frac{(\sin h{)}^{\prime }}{(h{)}^{\prime }}=\frac{\cos h}{1}$ se replace karta hai. Lekin $(\sin h{)}^{\prime }=\cos h$ jaanna matlab hai sine ka derivative — bilkul wahi theorem jiska proof $\frac{\sin h}{h}\to 1$ par depend karta hai. Ise apni hi foundation prove karne ke liye use karna ek loop hai: tumne conclusion assume kar liya. Sabse chhota valid route. Unit-circle area squeeze (Exercise 3.2): $\frac{1}{2}\sin h\le \frac{1}{2}h\le \frac{1}{2}\tan h$ deta hai $\cos h\le \frac{\sin h}{h}\le 1$, aur kyunki $\cos h\to 1$ toh Squeeze Theorem $\frac{\sin h}{h}\to 1$ force karta hai — sine ka koi bhi derivative use kiye bina.