Derivative from first principles — difference quotient definition: Worked examples

4.1.10 · D3 · Maths › Calculus I — Limits & Derivatives › Derivative from first principles — difference quotient defin

Koi bhi symbol se pehle: yaad karo ki $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$ ka matlab hai "ek buddy point $h$ door lo, unke beech ki line ka slope measure karo, phir buddy ko slide karo jab tak $h$ khatam na ho jaye." Poori difficulty hamesha wahi hai: $h = 0$ par raw fraction $\frac{0}{0}$ padha jaata hai (dekho Indeterminate forms 0 over 0), isliye humein pehle algebra se $h$ cancel karna hoga, tabhi $h \to 0$ karne denge.

Scenario matrix

Har first-principles problem inhi cells mein se kisi ek mein aata hai. Neeche ke examples label kiye gaye hain taaki har cell kam se kam ek baar cover ho.

Cell	Isme kya alag hai	Algebra tool jo chahiye	Example
A. Polynomial (higher power)	$(x + h)^{n}$ ko expand karna padta hai	Binomial expansion, $x^{n}$ cancel karo	Ex 1
B. Negative slope region	derivative negative aata hai	answer ki sign track karo	Ex 2
C. Rational (fraction) function	fractions subtract karna	common denominator	Ex 3
D. Root / irrational	pehle factor karne ke liye koi $h$ nahi	conjugate multiplication	Ex 4
E. Constant function (degenerate)	numerator exactly $0$ hota hai	$0/ h$ ko pehchano	Ex 5
F. Non-differentiable corner	left slope ≠ right slope	one-sided limits	Ex 6
G. Real-world word problem	units + interpretation	recipe apply karo, meaning padho	Ex 7
H. Exam twist (ek point par evaluate karo)	numeric slope + tangent line	differentiate karne ke baad value plug karo	Ex 8

Ex 1 — Cell A: higher power, $f (x) = x^{3}$

Forecast: $x^{2}$ ke liye humne $2 x$ paya tha. $x^{3}$ ke liye kya pattern guess karoge? (Padhne se pehle note karo.) Bahut log $3 x^{2}$ guess karte hain — aao ise earn karte hain.

Step 1 — $f (x + h)$ likho. $f (x + h) = (x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ Yeh step kyun? Humein buddy point par output jaanna hai. Cube ko expand karna $x^{2}$ se compare kiya toh yahi naya kaam hai; har $x$ ki jagah block $(x + h)$ aa jaata hai.

Step 2 — numerator banao. $f (x + h) - f (x) = (x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}) - x^{3} = 3 x^{2} h + 3 x h^{2} + h^{3}$ Yeh step kyun? Akele $x^{3}$ wale terms cancel ho jaate hain. Yahi cancellation reason hai ki ek finite slope survive karti hai — iske bina hum kuch-jo-chota-nahi-hai ko $h$ se divide kar rahe hote.

Step 3 — $h$ factor out karo aur cancel karo. $\frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} = \frac{h ( 3 x ^{2} + 3 x h + h ^{2} )}{h} = 3 x^{2} + 3 x h + h^{2}$ Yeh step kyun? Har surviving term mein pehle se $h$ hai, isliye $h$ cleanly factor ho jaata hai. Use cancel karne se $\frac{0}{0}$ ka trap hata jaata hai — ab expression $h = 0$ par safely evaluate ho sakta hai.

Step 4 — $h \to 0$ shrink karo. $f^{'} (x) = lim_{h \to 0} (3 x^{2} + 3 x h + h^{2}) = 3 x^{2}$ Yeh step kyun? Bache hue dono terms $3 x h$ aur $h^{2}$ mein $h$ hai, isliye woh vanish ho jaate hain. Answer: $f^{'} (x) = 3 x^{2}$ .

Verify: $x = 2$ par, $f^{'} (2) = 3 (2)^{2} = 12$ . Sanity check: $x^{3}$ $x = 2$ par tezi se chad raha hai ( $8$ se $27$ ki taraf agle unit mein), toh steepness $12$ believable hai. Aur $f^{'} (0) = 0$ — curve origin se guzarte hue flat hoti hai. ✓ Yeh Power Rule $\frac{d}{d x} x^{n} = n x^{n - 1}$ se match karta hai.

Ex 2 — Cell B: negative slope ka region, $f (x) = 5 - x^{2}$

Forecast: Parabola neeche ki taraf khulti hai. Apne peak ke daayein woh girती hai. Kya tumhe $x = 3$ par positive ya negative slope expect hai?

Step 1. $f (x + h) = 5 - (x + h)^{2} = 5 - x^{2} - 2 x h - h^{2}$ . Yeh step kyun? Minus sign ka dhyan rakho — poora $(x + h)^{2}$ block subtract ho raha hai.

Step 2. $f (x + h) - f (x) = (5 - x^{2} - 2 x h - h^{2}) - (5 - x^{2}) = - 2 x h - h^{2}$ Yeh step kyun? $5$ aur $- x^{2}$ cancel ho jaate hain, sirf $h$ -wale terms bachte hain — exactly woh sign jo survive karne ke liye chahiye.

Step 3. $\frac{- 2 x h - h ^{2}}{h} = \frac{h ( - 2 x - h )}{h} = - 2 x - h$ Yeh step kyun? $h$ factor karo, cancel karo, $\frac{0}{0}$ se bacho.

Step 4. $f^{'} (x) = lim_{h \to 0} (- 2 x - h) = - 2 x$ Answer: $f^{'} (x) = - 2 x$ .

Verify: $x = 3$ par, $f^{'} (3) = - 6$ — negative, forecast se match karta hai ki curve apne peak ke daayein girती hai. $x = 0$ par slope $0$ hai (hill ka top flat hai). ✓

Ex 3 — Cell C: ek fraction, $f (x) = \frac{1}{x ^{2}}$

Forecast: $1/ x$ ke liye parent ne $- 1/ x^{2}$ paya tha (hamesha negative). Kya tumhe expect hai ki $1/ x^{2}$ bhi $x > 0$ ke liye negatively slope karega?

Step 1. $f (x + h) = \frac{1}{( x + h ) ^{2}}$ .

Step 2 — common denominator. $\frac{1}{( x + h ) ^{2}} - \frac{1}{x ^{2}} = \frac{x ^{2} - ( x + h ) ^{2}}{x ^{2} ( x + h ) ^{2}} = \frac{x ^{2} - ( x ^{2} + 2 x h + h ^{2} )}{x ^{2} ( x + h ) ^{2}} = \frac{- 2 x h - h ^{2}}{x ^{2} ( x + h ) ^{2}}$ Yeh step kyun? Do fractions subtract karne ke liye shared denominator chahiye; aisa karne se numerator mein hidden $h$ surface ho jaata hai taaki hum use eventually cancel kar sakein.

Step 3 — $h$ se divide karo, cancel karo. $\frac{1}{h} \cdot \frac{- 2 x h - h ^{2}}{x ^{2} ( x + h ) ^{2}} = \frac{h ( - 2 x - h )}{h x ^{2} ( x + h ) ^{2}} = \frac{- 2 x - h}{x ^{2} ( x + h ) ^{2}}$ Yeh step kyun? Numerator se $h$ factor karo, denominator ke $h$ se cancel karo — $\frac{0}{0}$ khatam.

Step 4. $f^{'} (x) = lim_{h \to 0} \frac{- 2 x - h}{x ^{2} ( x + h ) ^{2}} = \frac{- 2 x}{x ^{2} \cdot x ^{2}} = \frac{- 2 x}{x ^{4}} = - \frac{2}{x ^{3}}$ Answer: $f^{'} (x) = - \frac{2}{x ^{3}}$ .

Verify: $x = 1$ par: $f^{'} (1) = - 2$ . $x > 0$ ke liye yeh negative hai — haan, $1/ x^{2}$ girता hai jaise $x$ badhta hai, forecast se match karta hai. Power-rule cross-check: $x^{- 2} \Rightarrow - 2 x^{- 3} = - 2/ x^{3}$ . ✓

Ex 4 — Cell D: ek root, $f (x) = 2 x + 1$

Forecast: Parent ne dikhaya tha $x \to \frac{1}{2 x}$ . Is root mein andar $2 x + 1$ hai — kya $2$ ka extra factor aayega?

Step 1. $f (x + h) = 2 (x + h) + 1 = 2 x + 2 h + 1$ .

Step 2. Numerator $= 2 x + 2 h + 1 - 2 x + 1$ — stuck, factor karne ke liye koi $h$ nahi.

Step 3 — conjugate move. Upar aur neeche $2 x + 2 h + 1 + 2 x + 1$ se multiply karo: $\frac{2 x + 2 h + 1 - 2 x + 1}{h} \cdot \frac{2 x + 2 h + 1 + 2 x + 1}{2 x + 2 h + 1 + 2 x + 1} = \frac{( 2 x + 2 h + 1 ) - ( 2 x + 1 )}{h ( 2 x + 2 h + 1 + 2 x + 1 )}$ Yeh step kyun? $(a - b) (a + b) = a - b$ use karke numerator mein ek aisa difference manufacture hota hai jo plain subtraction mein reduce hota hai aur usme $h$ hota hai. $= \frac{2 h}{h ( 2 x + 2 h + 1 + 2 x + 1 )} = \frac{2}{2 x + 2 h + 1 + 2 x + 1}$ $h$ cancel ho gaya — trap se nikal gaye.

Step 4. $f^{'} (x) = lim_{h \to 0} \frac{2}{2 x + 2 h + 1 + 2 x + 1} = \frac{2}{2 2 x + 1} = \frac{1}{2 x + 1}$ Answer: $f^{'} (x) = \frac{1}{2 x + 1}$ . Andar se $2$ ka extra factor aaya, phir half hoke chala gaya.

Verify: $x = 4$ par: $2 x + 1 = 9$ , toh $f^{'} (4) = 1/ 9 = \frac{1}{3}$ . Slope positive hai (root chad rahi hai), aur halka ( $\frac{1}{3}$ ) kyunki roots flat hoti jaati hain. ✓

Ex 5 — Cell E: ek degenerate case, $f (x) = 7$ (constant)

Forecast: Ek horizontal line ki height kabhi nahi badlती. Toh iska slope har jagah kya hona chahiye?

Step 1. $f (x + h) = 7$ — $x + h$ substitute karne se kuch nahi badalta, kyunki formula kabhi $x$ use hi nahi karta. Yeh step kyun? Yahi poori subtlety hai: replace karne ke liye koi $x$ nahi, isliye buddy point ka output same hota hai.

Step 2. $f (x + h) - f (x) = 7 - 7 = 0$ . Yeh step kyun? Rise exactly zero hai — dono points same height par hain.

Step 3. $\frac{0}{h} = 0$ har $h \neq = 0$ ke liye. Yeh step kyun? Ise $\frac{0}{0}$ mat kaho: numerator genuinely $0$ hai jabki $h$ nonzero hai, toh quotient honestly $0$ hai, indeterminate nahi. Yahi Indeterminate forms 0 over 0 se key distinction hai.

Step 4. $f^{'} (x) = h \to 0 lim 0 = 0$ . Answer: $f^{'} (x) = 0$ .

Verify: Flat line ki steepness har point par zero hoti hai. $f^{'} (1) = 0$ , $f^{'} (100) = 0$ . ✓

Ex 6 — Cell F: ek corner jahan derivative FAIL ho jaata hai, $f (x) = ∣ x ∣$

Forecast: $∣ x ∣$ ka graph origin par ek sharp "V" hai. Agar tum us spike par ek single tangent line rakhne ki koshish karo, kya ek line dono sides fit kar sakti hai?

Figure — Derivative from first principles — difference quotient definition

Humein difference quotient $x = 0$ par compute karna hai: $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{∣ h ∣ - ∣0∣}{h} = \frac{∣ h ∣}{h}$

Step 1 — daayein se approach karo ( $h \to 0^{+}$ , matlab $h > 0$ ). Jab $h > 0$ , $∣ h ∣ = h$ , toh $\frac{∣ h ∣}{h} = \frac{h}{h} = 1$ . Yeh step kyun? Daayein se V slope $+ 1$ ke saath utha hai; ek one-sided limit (dekho Limits — formal definition and one-sided limits) exactly wahi capture karta hai. Toh right-hand slope $+ 1$ hai.

Step 2 — baayein se approach karo ( $h \to 0^{-}$ , matlab $h < 0$ ). Jab $h < 0$ , $∣ h ∣ = - h$ , toh $\frac{∣ h ∣}{h} = \frac{- h}{h} = - 1$ . Yeh step kyun? Baayein se V slope $- 1$ ke saath girता hai. Toh left-hand slope $- 1$ hai.

Step 3 — compare karo. $lim_{h \to 0^{+}} \frac{∣ h ∣}{h} = + 1 \neq = lim_{h \to 0^{-}} \frac{∣ h ∣}{h} = - 1$ Yeh step kyun? Two-sided limit tab exist karta hai tabhi jab dono one-sided limits same hon. Yahan woh alag hain.

Step 4 — conclusion. Limit exist nahi karti, isliye $f$ $0$ par differentiable nahi hai. Answer: $f^{'} (0)$ exist nahi karta (ek corner hai).

Verify: Right slope $= 1$ , left slope $= - 1$ , aur $1 \neq = - 1$ . ✓

Ex 7 — Cell G: ek real-world word problem (instantaneous speed)

Forecast: $t = 0$ aur $t = 3$ ke beech ball $45$ m gira, jo ki average speed $15$ m/s hai. Kyunki woh accelerate karta rehta hai, kya tumhe expect hai ki $t = 3$ par speed $15$ se zyada hogi ya kam?

Step 1 — buddy point. $s (t + h) = 5 (t + h)^{2} = 5 (t^{2} + 2 t h + h^{2}) = 5 t^{2} + 10 t h + 5 h^{2}$ . Yeh step kyun? Yahi average-vs-instantaneous idea hai: hum position abhi se position thodi $h$ seconds baad compare karte hain.

Step 2 — position mein change (the rise). $s (t + h) - s (t) = (5 t^{2} + 10 t h + 5 h^{2}) - 5 t^{2} = 10 t h + 5 h^{2}$ Yeh step kyun? Yeh chhote interval $h$ mein giri hui distance hai.

Step 3 — $h$ (the run = elapsed time) se divide karo aur cancel karo. $\frac{10 t h + 5 h ^{2}}{h} = \frac{h ( 10 t + 5 h )}{h} = 10 t + 5 h$ Yeh step kyun? $\frac{distance}{time}$ ek speed hai; units m/s hain. $h$ factor karna us $\frac{0}{0}$ se bachata hai jo ek single instant deta.

Step 4 — interval shrink karo. $s^{'} (t) = lim_{h \to 0} (10 t + 5 h) = 10 t m/s$ $t = 3$ par: $s^{'} (3) = 10 (3) = 30$ m/s. Answer: $t = 3$ s par instantaneous speed $30$ m/s hai.

Verify (units + sanity): $s^{'} (t) = 10 t$ ki units (m)/(s) hain, jo speed hai. $t = 3$ par instantaneous $30$ m/s, $[0, 3]$ par average $15$ m/s se zyada hai — sahi hai, kyunki ball speed up karta rehta hai. ✓

Ex 8 — Cell H: exam twist, $f (x) = x^{2} - 4 x$ ke liye $x = 1$ par tangent line nikalo

Forecast: Tangent $(1, f (1))$ par curve ko touch karta hai. Pehle guess karo: kya curve $x = 1$ par chad raha hai ya gir raha hai? (Iska vertex $x = 2$ par hai, toh $x = 1$ bottom ke baayein hai.)

Step 1. $f (x + h) = (x + h)^{2} - 4 (x + h) = x^{2} + 2 x h + h^{2} - 4 x - 4 h$ . Yeh step kyun? Har $x$ ko block $(x + h)$ se replace karo — $x^{2}$ mein bhi aur $- 4 x$ mein bhi.

Step 2. $f (x + h) - f (x) = (x^{2} + 2 x h + h^{2} - 4 x - 4 h) - (x^{2} - 4 x) = 2 x h + h^{2} - 4 h$ Yeh step kyun? $x^{2}$ aur $- 4 x$ terms cancel ho jaate hain, sirf $h$ -wale terms bachte hain.

Step 3. $\frac{2 x h + h ^{2} - 4 h}{h} = \frac{h ( 2 x + h - 4 )}{h} = 2 x + h - 4$ Yeh step kyun? $h$ factor karo aur cancel karo $\frac{0}{0}$ se bachne ke liye.

Step 4. $f^{'} (x) = lim_{h \to 0} (2 x + h - 4) = 2 x - 4$ Toh $f^{'} (1) = 2 (1) - 4 = - 2$ — $x = 1$ par tangent slope.

Step 5 — tangent line banao. Touch point hai $(1, f (1)) = (1, 1 - 4) = (1, - 3)$ . Point-slope $y - y_{0} = m (x - x_{0})$ use karte hue: $y - (- 3) = - 2 (x - 1) \Rightarrow y = - 2 x - 1$ Yeh step kyun? Ek line ek point aur ek slope se fix hoti hai; derivative slope deta hai, function value point deta hai. Answer: tangent line $y = - 2 x - 1$ .

Verify: Slope $f^{'} (1) = - 2$ negative hai — curve $x = 1$ par gir raha hai ( $x = 2$ par apne vertex ke baayein), forecast se match karta hai. Line $(1, - 3)$ se guzarti hai: $- 2 (1) - 1 = - 3$ ✓. Aur $f^{'} (2) = 0$ vertex par, expected hai. ✓

Recall Kaun sa tool kaun si cell ke liye?

Higher power ::: binomial expand karo, $x^{n}$ cancel karo Fraction ::: hidden $h$ surface karne ke liye common denominator Root ::: $h$ manufacture karne ke liye conjugate se multiply karo Constant ::: numerator genuinely $0$ hai, quotient $0$ hai ( $0/0$ nahi) Corner (jaise $∣ x ∣$ ) ::: left aur right one-sided limits check karo — agar differ karein, koi derivative nahi Word problem ::: same recipe; phir units padho aur interpret karo

Connections

Derivative from first principles — difference quotient definition — woh recipe jo yeh examples drill karte hain.
Limits — formal definition and one-sided limits — corner case (Ex 6) aur har "shrink" step ko power karta hai.
Indeterminate forms 0 over 0 — kyun constants (Ex 5) indeterminate nahi hote jabki true derivatives hote hain.
Secant and Tangent lines — Ex 8 mein tangent-line construction.
Power Rule — Ex 1 aur Ex 3 ko cross-check karta hai.
Continuity and Differentiability — Ex 6 ka moral.
Average vs Instantaneous Rate of Change — word problem Ex 7 ka core.