Interpretation — instantaneous rate of change, slope of tangent: Worked examples

4.1.11 · D3 · Maths › Calculus I — Limits & Derivatives › Interpretation — instantaneous rate of change, slope of tang

Yeh page ek drill hall hai. Parent note ne machinery banayi thi (secant → limit → tangent). Yahan hum us machinery par har tarah ka input daalte hain taaki koi bhi exam question tumhe surprise na kar sake. Hum sab kuch difference quotient se compute karte hain

$f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$

aur kabhi "why" skip nahi karte. Pehle scenario matrix padho, phir har cell dhundo.

The scenario matrix

Derivative-at-a-point ka har question inhi case classes mein se ek hota hai. Last column batata hai ki neeche kaun sa worked example isko cover karta hai.

#	Case class	Tricky kyun hai	Example
A	Positive slope (curve upar jaati hai)	baseline sanity check	Ex 1
B	Negative slope (curve neeche jaati hai)	answer ka sign $< 0$ aana chahiye	Ex 2
C	Zero slope (flat tangent, koi peak/valley)	answer exactly $0$ hota hai	Ex 3
D	Denominator function $1/ x$	$h$ ek fraction ke andar chupta hai	Ex 4
E	Root function $x$	rationalise karna padta hai, sirf expand nahi	Ex 5
F	Word / rate problem with units	sign interpret karo + units lagao	Ex 6
G	Degenerate: corner $∣ x ∣$	limit fail hoti hai, derivative nahi	Ex 7
H	Degenerate: vertical tangent $x^{1/3}$	limit $+ \infty$ hai, finite slope nahi	Ex 8
I	Exam twist: tangent parallel to a given line	$f^{'} (a) = m$ solve karo $a$ ke liye	Ex 9

Prerequisites jo tum khule rakhna chahte ho: Average rate of change (secant slope), Limits — formal definition (engine), aur Velocity and acceleration (word problem ke liye).

Ex 1 — Case A: positive slope

Worked example

$f (x) = x^{2} - 2 x$ ka slope $x = 3$ par Forecast: parabola $y = x^{2} - 2 x$ upar ki taraf khurti hai aur $x = 3$ uske lowest point ke daayein hai, isliye curve upar ja rahi hai. Guess: ek positive number.

Step 1. Difference quotient banao. $\frac{f ( 3 + h ) - f ( 3 )}{h} = \frac{[ ( 3 + h ) ^{2} - 2 ( 3 + h ) ] - [ 9 - 6 ]}{h}$ Yeh step kyun? Yeh do real points ke beech secant ka honest slope hai — abhi division by zero nahi.

Step 2. Top expand karo: $(3 + h)^{2} = 9 + 6 h + h^{2}$ aur $- 2 (3 + h) = - 6 - 2 h$ , toh bracket hai $9 + 6 h + h^{2} - 6 - 2 h = 3 + 4 h + h^{2}$ . $f (3) = 3$ subtract karo: $numerator = 4 h + h^{2} .$ Yeh step kyun? Humein $\frac{0}{0}$ wali problem ( $h$ neeche) ko top par ek $h$ se match karna hai taaki cancel kar sakein.

Step 3. Cancel karo aur limit lo. $\frac{4 h + h ^{2}}{h} = 4 + h (h \neq = 0), lim_{h \to 0} (4 + h) = 4.$ Yeh step kyun? Cancel karna illegal $0/0$ ko pehle khatam karta hai, tab $h \to 0$ jaane dete hain — "Simplify Before Limit".

Toh $f^{'} (3) = 4$ .

Verify: positive, jaise forecast tha. Tiny secant $h = 0.001$ se sanity check: $4 + 0.001 = 4.001 \approx 4$ . ✓ Tangent line: $y - 3 = 4 (x - 3)$ .

Figure — Interpretation — instantaneous rate of change, slope of tangent

Ex 2 — Case B: negative slope

Ex 3 — Case C: zero slope (vertex par flat tangent)

Ex 4 — Case D: ek reciprocal function

Worked example

$f (x) = \frac{1}{x}$ ka slope $x = 2$ par Forecast: $1/ x$ $x > 0$ ke liye decreasing curve hai (bada $x$ → chhota output), isliye hum negative slope expect karte hain, size mein chhoti.

Step 1. Difference quotient: $\frac{\frac{1}{2 + h} - \frac{1}{2}}{h} .$ Yeh step kyun? Standard secant, lekin $h$ ab ek fraction ke andar daba hua hai, isliye abhi cancel nahi ho sakta — pehle combine karna hoga.

Step 2. Top fractions ko common denominator $2 (2 + h)$ par combine karo: $\frac{1}{2 + h} - \frac{1}{2} = \frac{2 - ( 2 + h )}{2 ( 2 + h )} = \frac{- h}{2 ( 2 + h )} .$ Yeh step kyun? Yeh top par ek $h$ manufacture karta hai neeche ke $h$ se match karne ke liye. Fractions combine karna hi woh ek tool hai jo buried $h$ tak pahunchta hai.

Step 3. $h$ se divide karo (yaani $1/ h$ se multiply karo): $\frac{- h}{2 ( 2 + h )} \cdot \frac{1}{h} = \frac{- 1}{2 ( 2 + h )}, lim_{h \to 0} \frac{- 1}{2 ( 2 + h )} = \frac{- 1}{4} .$ Yeh step kyun? Ab $h$ cleanly cancel hota hai aur $h = 0$ plug karna legal hai. Toh $f^{'} (2) = - \frac{1}{4}$ .

Verify: negative aur chhoti, jaise forecast tha. Tiny secant $h = 0.001$ : $\frac{1/2.001 - 1/2}{0.001} \approx - 0.2499 \approx - 0.25$ . ✓

Ex 5 — Case E: ek square root (rationalising trick)

Worked example

$f (x) = x$ ka slope $x = 9$ par Forecast: $x$ badhta hai lekin $x$ badhne par flat hota jaata hai. $x = 9$ par yeh ek chhota positive number hona chahiye.

Step 1. Difference quotient: $\frac{9 + h - 9}{h} = \frac{9 + h - 3}{h} .$ Yeh step kyun? Expand karna root ke liye kaam nahi karega; $h$ radical ke andar fansa hua hai.

Step 2. Top aur bottom ko conjugate $9 + h + 3$ se multiply karo: $\frac{9 + h - 3}{h} \cdot \frac{9 + h + 3}{9 + h + 3} = \frac{( 9 + h ) - 9}{h ( 9 + h + 3 )} = \frac{h}{h ( 9 + h + 3 )} .$ Yeh step kyun? $(A - B) (A + B) = A - B$ roots ke difference ko ek plain difference mein badal deta hai, jo top par ek $h$ free karta hai.

Step 3. $h$ cancel karo aur limit lo: $\frac{1}{9 + h + 3} h \to 0 \frac{1}{9 + 3} = \frac{1}{6} .$ Toh $f^{'} (9) = \frac{1}{6}$ .

Verify: chhota positive, jaise forecast tha. Tiny secant $h = 0.001$ : $\frac{9.001 - 3}{0.001} \approx 0.16666 \approx \frac{1}{6}$ . ✓ (Yeh Power Rule result $\frac{1}{2} x^{- 1/2}$ at $x = 9$ disguise mein hai.)

Ex 6 — Case F: signs aur units ke saath ek rate word problem

Worked example Ek ball upar phenki jaati hai; height

$s (t) = 20 t - 5 t^{2}$ metres, $t$ seconds mein. $t = 1$ s aur $t = 3$ s par velocity nikalo. Forecast: ball pehle upar jaati hai phir neeche aati hai. $t = 1$ s par (abhi chadh rahi hai) velocity positive honi chahiye; $t = 3$ s tak (top ke baad) yeh negative honi chahiye. Units: metres per second.

Step 1 (ek baar banao). $\frac{s ( a + h ) - s ( a )}{h}$ with $s (t) = 20 t - 5 t^{2}$ . Kyun? Velocity hai hi position ka instantaneous rate of change — dekho Velocity and acceleration.

Step 2 (general slope). Top $= 20 (a + h) - 5 (a + h)^{2} - [20 a - 5 a^{2}]$ . $- 5 (a + h)^{2} = - 5 a^{2} - 10 ah - 5 h^{2}$ expand karo: $top = 20 h - 10 ah - 5 h^{2} .$ $h$ se divide karo: $20 - 10 a - 5 h$ . Kyun? General $a$ ke saath karna dono times ke liye kaam reuse karne deta hai.

Step 3 (limit, phir substitute). $h \to 0 lim (20 - 10 a - 5 h) = 20 - 10 a$ . Toh $v (a) = 20 - 10 a$ .

$a = 1$ par: $v = 20 - 10 = + 10$ m/s.
$a = 3$ par: $v = 20 - 30 = - 10$ m/s.

Verify: signs forecast se match karte hain — $+ 10$ chadh raha, $- 10$ utar raha. Units: $[m] / [s]$ = m/s. ✓ Peak (jahan $v = 0$ ) $a = 2$ s par hai, exactly hamare do times ke beech, jo sign flip ke saath consistent hai.

Ex 7 — Case G: degenerate corner (derivative nahi)

Worked example Kya

$f (x) = ∣ x - 1∣$ ka $x = 1$ par slope hai? Forecast: $∣ x - 1∣$ $x = 1$ par apni point ke saath ek sharp V banata hai. Corner ka koi single tangent nahi hota — guess: derivative exist nahi karta.

Step 1. $a = 1$ par difference quotient: kyunki $f (1) = 0$ , $\frac{∣1 + h - 1∣ - 0}{h} = \frac{∣ h ∣}{h} .$ Yeh step kyun? Hume pehle test karna hoga ki limit exist bhi karti hai ya nahi, phir koi number claim karein.

Step 2. $h$ ke sign ke according split karo — yeh "cover all cases" move hai:

$h > 0$ : $∣ h ∣ = h \Rightarrow \frac{h}{h} = + 1.$
$h < 0$ : $∣ h ∣ = - h \Rightarrow \frac{- h}{h} = - 1.$

Yeh step kyun? Absolute value dono sides par alag behave karta hai, isliye one-sided limits disagree kar sakte hain.

Step 3. Right limit $= + 1$ , left limit $= - 1$ . Yeh disagree karte hain, isliye $h \to 0 lim \frac{∣ h ∣}{h}$ exist nahi karta. Isliye $f^{'} (1)$ exist nahi karta.

Verify: forecast se match — ek corner. Yeh Differentiability and continuity ka lesson hai: har jagah continuous, lekin us point par differentiable nahi.

Ex 8 — Case H: degenerate vertical tangent (limit infinite hai)

Worked example Kya

$f (x) = x^{1/3}$ (cube root) ka $x = 0$ par slope hai? Forecast: cube-root curve origin se almost seedha upar guzarti hai. Vertical tangent ka koi finite slope nahi hota — guess: limit $+ \infty$ tak blow up karti hai, isliye koi finite derivative nahi.

Step 1. $a = 0$ par, $f (0) = 0$ , toh difference quotient hai $\frac{( 0 + h ) ^{1/3} - 0}{h} = \frac{h ^{1/3}}{h} = h^{1/3 - 1} = h^{- 2/3} = \frac{1}{h ^{2/3}} .$ Yeh step kyun? Ise single power ke roop mein likhne se $h \to 0$ hone par behaviour expose hota hai.

Step 2. Jaise $h \to 0$ (kisi bhi side se), $h^{2/3} = (h^{1/3})^{2} \geq 0$ $0^{+}$ ki taraf shrink karta hai, isliye $\frac{1}{h ^{2/3}}$ without bound badhta hai: $lim_{h \to 0} \frac{1}{h ^{2/3}} = + \infty.$ Yeh step kyun? Ek negative number ka cube root negative hota hai, lekin use square karne se denominator dono sides par positive ho jaata hai — isliye dono one-sided limits saath $+ \infty$ ki taraf jaate hain.

Step 3. Limit ek finite number nahi hai, isliye $f^{'} (0)$ exist nahi karta (tangent vertical hai, slope " $\infty$ ").

Verify: forecast se match. $h = 0.001$ par magnitude check: $h^{- 2/3} = 0.00 1^{- 2/3} = 100$ , aur $h = 1 0^{- 6}$ par yeh $1 0^{4}$ hai — clearly badhta ja raha hai. ✓

Ex 9 — Case I: exam twist (tangent parallel to a given line)

Worked example

$f (x) = x^{2}$ ke liye, kis point par tangent parallel hai line $y = 8 x - 3$ ke? Forecast: parallel lines ka slope same hota hai. Line ka slope $8$ hai, isliye humein $f^{'} (a) = 8$ chahiye. Kyunki $y = x^{2}$ $x$ badhne par steep hota jaata hai, ek single positive $a$ expect karo.

Step 1. $x^{2}$ ka general slope nikalo. Difference quotient $\frac{( a + h ) ^{2} - a ^{2}}{h} = \frac{2 ah + h ^{2}}{h} = 2 a + h \to 2 a$ . Toh $f^{'} (a) = 2 a$ . Yeh step kyun? Humein slope position ke function ke roop mein chahiye taaki ise ek target ke barabar set kar sakein.

Step 2. $f^{'} (a) = 8$ set karo: $2 a = 8 \Rightarrow a = 4.$ Yeh step kyun? "Parallel" directly "equal slope" mein translate hota hai; $a$ ke liye solve karna point locate karta hai.

Step 3. Point hai $(4, f (4)) = (4, 16)$ . Tangent line: $y - 16 = 8 (x - 4) \Rightarrow y = 8 x - 16$ . Yeh step kyun? Ek baar point aur slope mil jaaye toh point–slope form kaam khatam karta hai.

Verify: slope $8$ diye gaye line ke slope $8$ se match karta hai ✓, aur dono lines alag hain ( $- 16 \neq = - 3$ ), isliye yeh parallel hain, same line nahi. ✓

Recall Matrix par rapid self-test

Kaun sa case negative answer deta hai, aur kyun? ::: Case B — girती curve; minus sign squared term par distribute hota hai. Kin do cases mein "derivative does not exist" hai, aur yeh kaise alag hain? ::: Corner (Ex 7, one-sided limits $+ 1$ vs $- 1$ ) aur vertical tangent (Ex 8, limit $\to + \infty$ ). Reciprocal ya root ke liye, kaun sa algebra buried $h$ ko unblock karta hai? ::: Fractions combine karo (reciprocal) ya conjugate se multiply karo (root). "Tangent parallel to line $y = m x + c$ " kaun si equation ban jaati hai? ::: $f^{'} (a) = m$ , phir $a$ ke liye solve karo.

Connections

Average rate of change — har example limit se pehle secant slope ke roop mein shuru hua.
Limits — formal definition — Ex 7 aur Ex 8 actually limit-existence ke questions hain.
Derivative as a function — Ex 6 aur Ex 9 ne general $a$ ke liye $f^{'} (a)$ nikala, yaani $f^{'} (x)$ .
Power Rule — Ex 5 aur Ex 9 secretly ise confirm karte hain.
Differentiability and continuity — Ex 7, 8 dikhate hain ki continuity derivative ke liye kaafi nahi hai.
Velocity and acceleration — Ex 6 usi limit ka physics wala roop hai.