Derivatives of sin x, cos x — proofs from first principles: Question bank

4.1.17 · D5 · Maths › Calculus I — Limits & Derivatives › Derivatives of sin x, cos x — proofs from first principles

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

True or false — justify

lim_{h \to 0} \frac{s i n h}{h} = 1

chahe

h

degrees mein measure ho ya radians mein.

False. Degrees mein limit

\frac{π}{180} \approx 0.01745

hoti hai, kyunki sector-area bound

\frac{1}{2} h

(jo sirf unit circle, radius

= 1

, par valid hai, jahan arc length radian angle ke barabar hoti hai) ko

h

radians mein chahiye. Clean "

= 1

" sirf radians ka fact hai — Figure s02 mein unit-circle squeeze dekho.

Derivative

\frac{d}{d x} sin x = cos x

poori sine wave ko

9 0^{\circ}

left shift kar deta hai.

True.

cos x = sin (x + \frac{π}{2})

, isliye differentiate karne par wave ek quarter period aage badh jaati hai — slope ki graph sine graph hai jo

\frac{π}{2}

left slide ho gayi hai. Figure s03

sin x

aur uski slope-curve

cos x

ko overlay karta hai taaki tum phase shift seedha dekh sako.

Kyunki

cos h \to 1

jab

h \to 0

, hum conclude kar sakte hain ki

\frac{c o s h - 1}{h} \to \frac{0}{0}

, jo undefined hai.

False.

\frac{0}{0}

koi number nahi balki ek indeterminate form hai — ye harder kaam karne ka signal hai, final answer nahi. Conjugate trick isse definite value

0

par resolve kar deti hai.

\frac{d}{d x} cos x

x = 0

ke paas negative hai kyunki cosine wahan decrease ho rahi hai.

True.

x = 0

par cosine apni peak par hoti hai aur turant girne lagti hai, isliye uski slope negative hai; chhote

x > 0

ke liye

- sin x < 0

isse confirm karta hai. Minus sign geometrically forced hai.

Do ingredient limits (

\frac{s i n h}{h} \to 1

aur

\frac{c o s h - 1}{h} \to 0

) logically independent facts hain.

False. Doosra pehle se bana hai: conjugate trick

\frac{c o s h - 1}{h}

- \frac{s i n h}{h} \cdot \frac{s i n h}{c o s h + 1}

mein convert karti hai, isliye isse value

- (1) \cdot \frac{0}{2} = 0

milti hai (

cos h \to 1

cosine ki continuity se, isliye

cos h + 1 \to 2

Agar hum sirf

\frac{s i n h}{h} \to 1

jaante hain lekin addition formulas ke baare mein kuch nahi, tab bhi proof complete kar sakte hain.

False. Addition formula hi

sin (x + h)

ko ek "

x

part" aur ek "

h

part" mein split karti hai; iske bina limits plug in karne ki koi jagah hi nahi hai. Dono ingredients load-bearing hain.

\frac{d}{d x} sin (2 x) = cos (2 x)

False. Inner rate matter karta hai:

\frac{d}{d x} sin (2 x) = 2 cos (2 x)

, kyunki

\frac{s i n 2 h}{h} \to 2

2

dekhne ke liye:

u = 2 h

substitute karo, isliye

\frac{s i n 2 h}{h} = \frac{s i n u}{u /2} = 2 \cdot \frac{s i n u}{u}

, aur jab

h \to 0

toh

u \to 0

, giving

2 \cdot 1 = 2

. Ye missing factor of

2

exactly wahi hai jo Chain Rule baad mein formalize karta hai.

Spot the error

\frac{s i n h}{h}

par L'Hôpital apply karo: upar aur neeche differentiate karo to

\frac{c o s h}{1} \to 1

milta hai. Done."

Error circularity hai. L'Hôpital ko

\frac{d}{d h} sin h = cos h

chahiye, jo exactly wahi theorem hai jo hum prove kar rahe hain; ise yahan use karna conclusion assume karna hai. Squeeze argument hi honest foundation hai.

cos h \approx 1 - \frac{h ^{2}}{2}

, isliye

\frac{c o s h - 1}{h} \approx - \frac{h}{2} \to - \frac{1}{2}

- \frac{h}{2}

ka limit jab

h \to 0

hota hai toh

0

hai, $- \frac{1}{2}$ nahi.

\frac{1}{2}

alag limit

\frac{1 - c o s h}{h ^{2}} \to \frac{1}{2}

se belong karta hai — denominator mein

h

ki power dekho.

"Squeeze mein,

\frac{1}{2} sin h \leq \frac{1}{2} h \leq \frac{1}{2} tan h

sabhi

h

ke liye hold karta hai, isliye poora argument kisi bhi

h

ke liye kaam karta hai."

Area inequality chhote positive

h

ke liye derive ki gayi hai (unit circle par ek first-quadrant angle, Figure s02).

h < 0

ke liye hum evenness invoke karte hain —

\frac{s i n h}{h}

ek even function hai — na ki raw area picture, jise positive angle chahiye.

sin (x + h) = sin x + sin h

, isliye derivative

lim_{h \to 0} \frac{s i n h}{h} = 1

hai."

Sine additive nahi hai:

sin (x + h) \neq = sin x + sin h

. Correct expansion hai

sin x cos h + cos x sin h

, aur exactly yahi

cos x

deta hai, constant

1

nahi.

"Kyunki dono limits constants dete hain (

0

aur

1

sin x

ki derivative constant

cos (0) \cdot 1 = 1

hai."

sin x

aur

cos x

factors

h

ke respect mein constants hain, lekin phir bhi

x

par depend karte hain aur rehne chahiye. Answer hai

sin x \cdot 0 + cos x \cdot 1 = cos x

, jo

x

ka function hai, number

1

nahi.

tan h

squeeze mein appear karta hai, isliye humne secretly assume kar liya ki hum

\frac{d}{d x} tan x

jaante hain."

Nahi —

tan h = \frac{s i n h}{c o s h}

purely ek lengths ka ratio ki tarah use hota hai (Figure s02 mein outer triangle ki area), ek static geometric fact. Tangent ki koi derivative invoke nahi hoti; Derivatives of tan, sec, csc, cot baad mein aate hain, is result par build karke.

Why questions

Hum addition formula se expand kyun karte hain, seedha

h = 0

plug in karne ki jagah?

h = 0

plug karne se

\frac{0}{0}

milta hai — zero-width interval par average slope undefined hoti hai. Addition formula quotient ko un pieces mein reshape karti hai jinke limits exist karte hain.

Calculus ke liye angle radians mein kyun hona chahiye, geometrically?

Squeeze ka middle term, sector area

\frac{1}{2} h

, actual circular-sector area ke barabar hota hai sirf unit circle (radius $1$ ) par jab $h$ radians mein ho, jahan arc length angle ke barabar hoti hai. Koi bhi doosri unit har derivative mein ek conversion constant insert kar deti hai.

Limit-flow

cos x cos h - sin x sin h

neatly

cos x \cdot 1 - sin x \cdot 0

mein kyun collapse hoti hai?

Kyunki jab

h \to 0

, cosine continuous hai isliye

cos h \to cos 0 = 1

aur sine continuous hai isliye

sin h \to sin 0 = 0

;

x

-factors bas unchanged ride karte hain. Figure s03 ka companion idea: "

h

-part" clean

1

aur

0

mein fade ho jaata hai, "

x

-part" khada rehta hai.

Differentiating sine/cosine family ko exactly

9 0^{\circ}

har baar kyun rotate karti hai?

Kyunki ek wave ki slope wahan peak karti hai jahan wave zero cross karti hai aur wahan vanish hoti hai jahan wave peak karti hai — ek quarter-cycle offset (Figure s03 mein visible). Iterate karne par cycle

sin \to cos \to - sin \to - cos \to sin

milti hai.

Conjugate multiplier

cos h + 1

kyun choose kiya aur kuch aur kyun nahi?

Kyunki

(cos h - 1) (cos h + 1) = cos^{2} h - 1 = - sin^{2} h

stubborn

cos h - 1

ko ek

sin

mein convert kar deta hai jo hum already control karte hain, jo hume

\frac{s i n h}{h} \to 1

reuse karne deta hai.

Sanity check (cosine ki peak ⇒ negative slope) evidence kyun count hoti hai lekin proof nahi?

Ye confirm karta hai ki sign consistent hai, sign slip rule out karta hai, lekin ye magnitude (

sin x

vs. kuch aur) ke baare mein kuch nahi kehta. Sirf full limit argument exact function pin karta hai.

Hum

cos h - 1

h

se (linear power) prove kyun karte hain,

h^{2}

se kyun nahi?

Kyunki derivative ki definition difference ko interval width

h

ki pehli power se divide karti hai. Wo single power limit ko

0

force karta hai;

h^{2}

version ek alag, higher-order fact hai jo yahan zaruri nahi hai.

Edge cases

x = 0

par, kya

sin

ke liye first-principles limit abhi bhi

cos 0 = 1

hai, aur kya graph isse confirm karta hai?

Haan.

x = 0

par sine graph apni steepest upward slope ke saath zero cross karta hai (Figure s03), aur derivation

cos 0 = 1

deti hai —

sin

ki maximum possible slope, jo picture se bilkul match karti hai.

\frac{s i n h}{h}

left se kya hota hai, yaani jab

h \to 0^{-}

Ye phir bhi

\to 1

hota hai, kyunki

\frac{s i n h}{h}

ek even function hai (

sin (- h) / (- h) = sin h / h

). Two-sided limit exist karta hai aur

1

ke barabar hai, isliye derivative har

x

par well-defined hai.

Kya

sin x

har jagah differentiable hai, including apni peaks

x = \frac{π}{2}

par?

Haan.

x = \frac{π}{2}

par derivation

cos \frac{π}{2} = 0

deti hai — ek smooth horizontal tangent, corner nahi. Sine aur cosine poore

R

par differentiable hain.

Agar hum

h

ko ek sequence jaise

h = \frac{1}{n} \to 0

se replace karein (jahan

n

ek positive integer hai jo bina bound ke badh raha hai,

n \to \infty

), kya hume same limit milegi?

Haan, kyunki limit ek genuine (two-sided) limit ke roop mein exist karti hai, har sequence

h_{k} \to 0

jahan

h_{k} \neq = 0

1

deti hai. Path-independence guaranteed hai jab ek baar limit establish ho jaati hai.

Kya proof break hoti hai agar

x

khud bahut bada ho, maano

x = 1000

Nahi. Limit

h \to 0

par li jaati hai jab

x

ek constant ki tarah fixed hota hai;

x

ki size kabhi do ingredient limits mein enter nahi hoti. Result

cos x

har real

x

ke liye hold karta hai.

Agar koi

\frac{f ( x ) - f ( x - h )}{h}

(backward difference) se derivative define kare to kya hoga?

Tumhe same derivative

cos x

milegi; backward, forward, aur symmetric quotients sab ek differentiable function ke liye identical limit par converge karte hain. Yahan choice cosmetic hai.

Recall Har trap ki ek-line summary

Lagbhag saare traps teen sins par reduce hote hain: wrong units (degrees), wrong power of $h$ ( $h$ ke do cosine limits mix karna), ya circular logic (L'Hôpital / additivity assume karna). Un teeno ko guard karo aur topic airtight hai.

Connections

Squeeze Theorem — wo engine jise "spot the error" items probe karte hain
Trigonometric addition formulas — non-additivity trap
Radian measure — degrees-vs-radians trap
Chain Rule — $sin (2 x)$ factor-of-2 trap
Derivatives of tan, sec, csc, cot — jahan $tan h$ baad mein reappear karta hai
Taylor series of sin and cos — $1 - \frac{h ^{2}}{2}$ approximation trap
Limits — definition and properties — indeterminate-form reasoning