Derivative from first principles — difference quotient definition: Exercises

4.1.10 · D4 · Maths › Calculus I — Limits & Derivatives › Derivative from first principles — difference quotient defin

Parent note se recipe yaad karo: Plug, Subtract, Cancel, Shrink.

Figure — Derivative from first principles — difference quotient definition

Har problem isi picture ki story hai: do kareeb points se guzarti ek secant line dheere dheere rotate hoti hai aur tangent ban jaati hai jab doosra point slide karta hai. Agar yeh picture fuzzy lagti hai toh Secant and Tangent lines padho.

Level 1 — Recognition

(Kya tum pieces spot kar sakte ho aur machine ko simplest inputs par chala sakte ho?)

Exercise 1.1

$f (x) = 5 x - 2$ ke liye difference quotient likho, phir first principles se $f^{'} (x)$ nikalo.

Recall Solution 1.1

Step 1 — Plug. $f (x + h) = 5 (x + h) - 2 = 5 x + 5 h - 2$ . Step 2 — Subtract. $f (x + h) - f (x) = (5 x + 5 h - 2) - (5 x - 2) = 5 h$ Step 3 — Cancel. $\frac{5 h}{h} = 5$ . Step 4 — Shrink. $f^{'} (x) = lim_{h \to 0} 5 = 5$ . Answer: $f^{'} (x) = 5$ . Ek straight line ka ek constant slope hota hai har jagah — exactly uska coefficient. ✓

Exercise 1.2

Constant $f (x) = 7$ ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 1.2

Plug. $f (x + h) = 7$ (replace karne ke liye koi $x$ nahi). Subtract. $7 - 7 = 0$ . Cancel. $\frac{0}{h} = 0$ har $h \neq = 0$ ke liye. Shrink. $f^{'} (x) = lim_{h \to 0} 0 = 0$ . Answer: $f^{'} (x) = 0$ . Ek flat horizontal line kabhi upar nahi jaati — har jagah zero steepness. ✓

Exercise 1.3

$f (x) = x^{2}$ ke liye (jo parent note mein $f^{'} (x) = 2 x$ derive hua tha), $x = - 3$ par tangent line ki slope kya hai?

Recall Solution 1.3

Hamhare paas already $f^{'} (x) = 2 x$ hai. Sirf ek point substitute karo: $f^{'} (- 3) = 2 (- 3) = - 6$ Answer: $- 6$ . Parabola ke left arm par curve downhill slope karta hai, isliye negative slope bilkul expected hai. ✓

Level 2 — Application

(Genuinely curved functions par poori chaar-step machine chalao.)

Exercise 2.1

$f (x) = x^{2} + 3 x$ ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 2.1

Plug. $f (x + h) = (x + h)^{2} + 3 (x + h) = x^{2} + 2 x h + h^{2} + 3 x + 3 h$ . Subtract. $f (x + h) - f (x) = (x^{2} + 2 x h + h^{2} + 3 x + 3 h) - (x^{2} + 3 x) = 2 x h + h^{2} + 3 h$ $x^{2}$ aur $3 x$ terms cancel ho jaate hain — yahi cancellation hai jis wajah se ek finite slope bachta hai. Cancel. $h$ factor karo: $\frac{h ( 2 x + h + 3 )}{h} = 2 x + h + 3$ Shrink. $f^{'} (x) = lim_{h \to 0} (2 x + h + 3) = 2 x + 3$ . Answer: $f^{'} (x) = 2 x + 3$ . ✓

Exercise 2.2

$f (x) = x^{3}$ ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 2.2

Plug. Use karo $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ . Subtract. $f (x + h) - f (x) = 3 x^{2} h + 3 x h^{2} + h^{3}$ Cancel. Har term mein $h$ hai: $\frac{h ( 3 x ^{2} + 3 x h + h ^{2} )}{h} = 3 x^{2} + 3 x h + h^{2}$ Shrink. Jab $h \to 0$ toh $3 x h$ aur $h^{2}$ dono gayab ho jaate hain: $f^{'} (x) = 3 x^{2}$ Answer: $f^{'} (x) = 3 x^{2}$ . Yeh Power Rule pattern $\frac{d}{d x} x^{n} = n x^{n - 1}$ se match karta hai jab $n = 3$ . ✓

Exercise 2.3

$f (x) = \frac{1}{x ^{2}}$ ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 2.3

Plug. $f (x + h) = \frac{1}{( x + h ) ^{2}}$ . Subtract (common denominator). $\frac{1}{( x + h ) ^{2}} - \frac{1}{x ^{2}} = \frac{x ^{2} - ( x + h ) ^{2}}{x ^{2} ( x + h ) ^{2}} = \frac{x ^{2} - ( x ^{2} + 2 x h + h ^{2} )}{x ^{2} ( x + h ) ^{2}} = \frac{- 2 x h - h ^{2}}{x ^{2} ( x + h ) ^{2}}$ Cancel. $h$ se divide karo; upar se $h$ factor karo: $\frac{1}{h} \cdot \frac{- h ( 2 x + h )}{x ^{2} ( x + h ) ^{2}} = \frac{- ( 2 x + h )}{x ^{2} ( x + h ) ^{2}}$ Shrink. $h \to 0$ karo: $f^{'} (x) = \frac{- 2 x}{x ^{2} \cdot x ^{2}} = \frac{- 2 x}{x ^{4}} = - \frac{2}{x ^{3}}$ Answer: $f^{'} (x) = - \frac{2}{x ^{3}}$ . ✓

Level 3 — Analysis

(Algebra $h$ ko chhupaati hai — tumhe use engineer karke saamne laana hoga.)

Exercise 3.1

$f (x) = x + 1$ (ek shifted square root) ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 3.1

Plug. $f (x + h) = x + h + 1$ . Subtract. $x + h + 1 - x + 1$ — atak gaye, koi $h$ factor nahi kar sakte. Conjugate trick use karo (upar aur neeche dono ko $x + h + 1 + x + 1$ se multiply karo, jo $a - b$ ko $a - b$ mein badal deta hai aur $h$ manufacture karta hai): $\frac{x + h + 1 - x + 1}{h} \cdot \frac{x + h + 1 + x + 1}{x + h + 1 + x + 1} = \frac{( x + h + 1 ) - ( x + 1 )}{h ( x + h + 1 + x + 1 )} = \frac{h}{h ( x + h + 1 + x + 1 )}$ Cancel. $= \frac{1}{x + h + 1 + x + 1}$ . Shrink. $h \to 0$ dono roots ko equal bana deta hai: $f^{'} (x) = \frac{1}{2 x + 1}$ Answer: $f^{'} (x) = \frac{1}{2 x + 1}$ . ✓

Exercise 3.2

$f (x) = \frac{x}{x + 1}$ ke liye first principles se $f^{'} (x)$ nikalo.

Recall Solution 3.2

Plug. $f (x + h) = \frac{x + h}{x + h + 1}$ . Subtract (common denominator). $\frac{x + h}{x + h + 1} - \frac{x}{x + 1} = \frac{( x + h ) ( x + 1 ) - x ( x + h + 1 )}{( x + h + 1 ) ( x + 1 )}$ Upar expand karo: $(x + h) (x + 1) = x^{2} + x + h x + h$ aur $x (x + h + 1) = x^{2} + h x + x$ . Subtract karo: $(x^{2} + x + h x + h) - (x^{2} + h x + x) = h$ Toh numerator sirf $h$ hai: $= \frac{h}{( x + h + 1 ) ( x + 1 )}$ Cancel. $h$ se divide karo: $\frac{1}{( x + h + 1 ) ( x + 1 )}$ . Shrink. $h \to 0$ : $f^{'} (x) = \frac{1}{( x + 1 ) ^{2}}$ Answer: $f^{'} (x) = \frac{1}{( x + 1 ) ^{2}}$ . Hamesha positive — yeh function jahan defined hai wahan sirf chadhaai hi karta hai. ✓

Exercise 3.3

First principles se dikhao ki $f (x) = ∣ x ∣$ $x = 0$ par differentiable nahi hai, aur identify karo kya galat ho raha hai. (Limits — formal definition and one-sided limits use karta hai.)

Recall Solution 3.3

$x = 0$ par difference quotient hai $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{∣ h ∣ - 0}{h} = \frac{∣ h ∣}{h}$ Ab $h$ ki sign ke hisaab se split karo (yeh ek one-sided-limit question hai — dono sides ka agree karna zaruri nahi):

Right se ( $h \to 0^{+}$ ): $∣ h ∣ = h$ , toh $\frac{h}{h} = + 1$ . Limit $= + 1$ .
Left se ( $h \to 0^{-}$ ): $∣ h ∣ = - h$ , toh $\frac{- h}{h} = - 1$ . Limit $= - 1$ .

Right-hand slope $(+ 1)$ aur left-hand slope $(- 1)$ agree nahi karte, isliye two-sided limit exist nahi karta. $f$ , $0$ par differentiable nahi hai. Geometrically graph mein ek sharp corner hai — secant har side se ek alag tangent approach karta hai, isliye koi single tangent slope exist nahi karta. (Note karo $∣ x ∣$ $0$ par continuous hai; dekho Continuity and Differentiability — continuous hone ka matlab differentiable hona zaruri nahi.) ✓

Level 4 — Synthesis

(Machine ko unknown constants aur general structure ke saath combine karo.)

Exercise 4.1

General quadratic $f (x) = a x^{2} + b x + c$ ke liye, first principles se prove karo ki $f^{'} (x) = 2 a x + b$ .

Recall Solution 4.1

Plug. $f (x + h) = a (x + h)^{2} + b (x + h) + c = a x^{2} + 2 a x h + a h^{2} + b x + bh + c$ . Subtract. $f (x + h) - f (x) = (a x^{2} + 2 a x h + a h^{2} + b x + bh + c) - (a x^{2} + b x + c) = 2 a x h + a h^{2} + bh$ Har "position" term ( $a x^{2}, b x, c$ ) cancel ho jaati hai; sirf "change" terms bachte hain — jinmein har ek $h$ carry karta hai. Cancel. $\frac{h ( 2 a x + ah + b )}{h} = 2 a x + ah + b$ Shrink. $h \to 0$ se $ah$ khatam: $f^{'} (x) = 2 a x + b$ Answer: $f^{'} (x) = 2 a x + b$ . Exercise 2.1 ( $a = 1, b = 3, c = 0$ ) se check karo: $2 x + 3$ . ✓

Exercise 4.2

First principles se sum rule prove karo: agar $f (x) = u (x) + v (x)$ aur dono $u, v$ differentiable hain, toh $f^{'} (x) = u^{'} (x) + v^{'} (x)$ .

Recall Solution 4.2

Plug & Subtract. Difference quotient likho aur $u$ -parts aur $v$ -parts group karo: $\frac{f ( x + h ) - f ( x )}{h} = \frac{[ u ( x + h ) + v ( x + h ) ] - [ u ( x ) + v ( x ) ]}{h} = \frac{u ( x + h ) - u ( x )}{h} + \frac{v ( x + h ) - v ( x )}{h}$ Shrink. Sum ka limit, limits ka sum hota hai (ek limit law from Limits — formal definition and one-sided limits), aur har piece definition ke hisaab se ek derivative hai: $f^{'} (x) = lim_{h \to 0} \frac{u ( x + h ) - u ( x )}{h} + lim_{h \to 0} \frac{v ( x + h ) - v ( x )}{h} = u^{'} (x) + v^{'} (x)$ Answer: $f^{'} (x) = u^{'} (x) + v^{'} (x)$ . ✓ Koi algebra tricks nahi — poora proof sirf regrouping plus ek limit law hai.

Exercise 4.3

First-principles derivative use karke $f (x) = x^{2}$ par $x = 2$ par tangent line ki equation nikalo. (Secant and Tangent lines se connected.)

Recall Solution 4.3

Parent note se $f^{'} (x) = 2 x$ , isliye $x = 2$ par tangent slope hai $m = f^{'} (2) = 2 (2) = 4$ Tangency ka point hai $(2, f (2)) = (2, 4)$ . Point–slope form $y - y_{0} = m (x - x_{0})$ : $y - 4 = 4 (x - 2) ⟹ y = 4 x - 4$ Answer: $y = 4 x - 4$ . Check ke taur par, yeh line $(2, 4)$ se guzarti hai aur slope $4$ hai; $x = 2.5$ tak ek nearby secant ka slope $\frac{f ( 2.5 ) - f ( 2 )}{0.5} = \frac{6.25 - 4}{0.5} = 4.5$ hai, jo already $4$ ke kareeb hai aur $h \to 0$ ke saath usiki taraf shrink karta hai. ✓

Level 5 — Mastery

(Subtle limits, degenerate points, aur forecasting.)

Exercise 5.1

Piecewise function ke liye first principles se $f^{'} (0)$ nikalo $f (x) = {x^{2} sin (\frac{1}{x}) 0 x \neq = 0 x = 0$ aur note karo ki ordinary "formula mein plug karo" instinct yahan kyun fail hoti hai.

Recall Solution 5.1

Hame $x = 0$ par definition zaroor use karni padegi kyunki $f$ wahan ek special rule se define hai. $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{h ^{2} s i n ( \frac{1}{h} ) - 0}{h} = h sin (\frac{1}{h})$ Ab isse bound karo. Kyunki $- 1 \leq sin (\cdot) \leq 1$ har input ke liye, $- ∣ h ∣ \leq h sin (\frac{1}{h}) \leq ∣ h ∣$ Jab $h \to 0$ toh dono outer bounds $0$ ho jaate hain, isliye beech wala $0$ par squeeze ho jaata hai: $f^{'} (0) = lim_{h \to 0} h sin (\frac{1}{h}) = 0$ Answer: $f^{'} (0) = 0$ . Derivative exist karta hai chahe $sin (1/ h)$ wildly oscillate kare, kyunki upar ka $h$ oscillation ko crush kar deta hai. Tum yeh "formula differentiate karke $0$ plug karne" se nahi nikaal sakte — formula $x^{2} sin (1/ x)$ $0$ par defined hi nahi hai. Sirf difference quotient answer tak pahuncha sakta hai. ✓

Exercise 5.2

Pehle forecast karo, phir verify karo. $f (x) = x^{4}$ ke liye, us pattern se $f^{'} (x)$ forecast karo jo tumne dekha hai ( $x^{2} \to 2 x$ , $x^{3} \to 3 x^{2}$ ). Phir first principles se verify karo.

Recall Solution 5.2

Forecast. Pattern $x^{n} \to n x^{n - 1}$ predict karta hai $f^{'} (x) = 4 x^{3}$ . Verify. Use karo $(x + h)^{4} = x^{4} + 4 x^{3} h + 6 x^{2} h^{2} + 4 x h^{3} + h^{4}$ . Subtract. $f (x + h) - f (x) = 4 x^{3} h + 6 x^{2} h^{2} + 4 x h^{3} + h^{4}$ . Cancel. $h$ factor karo: $\frac{h ( 4 x ^{3} + 6 x ^{2} h + 4 x h ^{2} + h ^{3} )}{h} = 4 x^{3} + 6 x^{2} h + 4 x h^{2} + h^{3}$ Shrink. $h$ wali har term gayab ho jaati hai: $f^{'} (x) = 4 x^{3}$ Answer confirmed: $f^{'} (x) = 4 x^{3}$ . ✓ Forecast sahi nikla — yeh exactly Power Rule hai jo first principles se build ho raha hai.

Exercise 5.3

$h = 0$ par difference quotient $\frac{f ( x + h ) - f ( x )}{h}$ indeterminate form $0/0$ hai. $f (x) = x^{2}$ use karke $x = 3$ par, quotient ko $h = 1, 0.1, 0.01$ par evaluate karo aur batao yeh kis limit ki taraf approach karta hai. Explain karo ki value $0/0$ kyun nahi hai chahe fraction dikhne mein $0/0$ banta ho.

Recall Solution 5.3

$f (x) = x^{2}$ ke saath, $x = 3$ par quotient simplify ho jaata hai (Step 3) $2 x + h = 6 + h$ mein. Tabulate karo:

h & \frac{f(3+h)-f(3)}{h}=6+h\\ \hline 1 & 7\\ 0.1 & 6.1\\ 0.01 & 6.01 \end{array}$$ Values $\mathbf{6}$ ki taraf badh rahi hain, jo $f'(3)=2(3)=6$ ke equal hai. ✓ **$0/0$ kyun nahi?** $h=0$ *exactly* par, rise aur run dono $0$ hain — genuinely undefined. Lekin limit kabhi bhi $h=0$ *par* value nahi poochhta; yeh poochhta hai ki fraction $h=0$ *ke kareeb* kya *approach* karta hai. $h$ cancel karna (Step 3) usi expression ko aisi form mein rewrite karta hai jo wahan *defined* hai, aur woh form $6$ par settle karti hai. $0/0$ sirf ek disguise tha jo algebra ne hata diya. ✓

Self-test recap

Constant

f (x) = c

ka derivative hota hai

0

har jagah.

f (x) = x^{2}

ki slope

x = - 3

par

- 6

First principles se

f (x) = x^{3}

ke liye

f^{'} (x)

3 x^{2}

f (x) = 1/ x^{2}

ke liye

f^{'} (x)

- 2/ x^{3}

∣ x ∣

0

par differentiable kyun nahi?

Right slope

+ 1

hai aur left slope

- 1

hai; yeh agree nahi karte, isliye limit fail ho jaata hai — ek corner.

y = x^{2}

par

x = 2

par tangent

y = 4 x - 4

General quadratic

a x^{2} + b x + c

ka derivative hota hai

2 a x + b

x^{2} sin (1/ x)

ke liye

f^{'} (0)

(jab

f (0) = 0

)

0

, squeeze

- ∣ h ∣ \leq h sin (1/ h) \leq ∣ h ∣

se.

Connections

Limits — formal definition and one-sided limits — Exercises 3.3, 4.2, 5.1 ko power karta hai.
Indeterminate forms 0 over 0 — Exercise 5.3 ka dil.
Secant and Tangent lines — 4.3 ke peeche ki picture.
Power Rule — 2.2, 4.1, 5.2 se build hota hai.
Continuity and Differentiability — Exercise 3.3 (continuous but not differentiable).
Average vs Instantaneous Rate of Change — 5.3 ka shrinking-secant view.
Parent: Derivative from first principles — difference quotient definition.