Chain rule — proof, composite function derivatives: Exercises

4.1.16 · D4 · Maths › Calculus I — Limits & Derivatives › Chain rule — proof, composite function derivatives

Ek-line ka tool jis par hum poore time rely karte hain:

Level 1 — Recognition

Kya tum outer function $f$ aur inner function $g$ identify kar sakte ho, aur rule ek baar apply kar sakte ho?

Problem 1.1

$h (x) = (2 x + 7)^{4}$ ko differentiate karo.

Recall Solution 1.1

Do machines identify karo. Outer $f (u) = u^{4}$ , inner $g (x) = 2 x + 7$ . Dono ko differentiate karo. $f^{'} (u) = 4 u^{3}$ (shell par power rule); $g^{'} (x) = 2$ (ek line ki derivative). Multiply karo, inside ko wapas plug in karo. $h^{'} (x) = 4 (2 x + 7)^{3} \cdot 2 = 8 (2 x + 7)^{3} .$

Problem 1.2

$h (x) = cos (5 x)$ ko differentiate karo.

Recall Solution 1.2

Outer $f (u) = cos u \Rightarrow f^{'} (u) = - sin u$ . Inner $g (x) = 5 x \Rightarrow g^{'} (x) = 5$ . $h^{'} (x) = - sin (5 x) \cdot 5 = - 5 sin (5 x) .$ Notice karo ki inside $5 x$ hi rehta hai $sin$ ke andar — sirf outer rate change hoti hai.

Problem 1.3

$h (x) = e^{3 x}$ ko differentiate karo.

Recall Solution 1.3

Outer $f (u) = e^{u} \Rightarrow f^{'} (u) = e^{u}$ (exponential apni khud ki derivative hai). Inner $g (x) = 3 x \Rightarrow g^{'} (x) = 3$ . $h^{'} (x) = e^{3 x} \cdot 3 = 3 e^{3 x} .$

Level 2 — Application

Chain rule ko power, product-of-constants, aur known derivatives ke saath combine karo.

Problem 2.1

$h (x) = 9 - x^{2}$ ko differentiate karo.

Recall Solution 2.1

$(9 - x^{2})^{1/2}$ ki tarah rewrite karo. Outer $f (u) = u^{1/2} \Rightarrow f^{'} (u) = \frac{1}{2} u^{- 1/2}$ . Inner $g (x) = 9 - x^{2} \Rightarrow g^{'} (x) = - 2 x$ . $h^{'} (x) = \frac{1}{2} (9 - x^{2})^{- 1/2} \cdot (- 2 x) = \frac{- x}{9 - x ^{2}} .$ $\frac{1}{2}$ aur $- 2$ mil kar $- 1$ banaate hain — clean cancellation.

Problem 2.2

$h (x) = tan (x^{3})$ ko differentiate karo.

Recall Solution 2.2

Outer $f (u) = tan u \Rightarrow f^{'} (u) = sec^{2} u$ . Inner $g (x) = x^{3} \Rightarrow g^{'} (x) = 3 x^{2}$ . $h^{'} (x) = sec^{2} (x^{3}) \cdot 3 x^{2} = 3 x^{2} sec^{2} (x^{3}) .$

Problem 2.3

$h (x) = ln (x^{2} + 1)$ ko differentiate karo.

Recall Solution 2.3

Outer $f (u) = ln u \Rightarrow f^{'} (u) = \frac{1}{u}$ . Inner $g (x) = x^{2} + 1 \Rightarrow g^{'} (x) = 2 x$ . $h^{'} (x) = \frac{1}{x ^{2} + 1} \cdot 2 x = \frac{2 x}{x ^{2} + 1} .$ Note karo ki $x^{2} + 1 > 0$ hamesha hota hai, isliye $ln$ saare $x$ ke liye defined hai aur koi domain ki chinta nahi hoti.

Level 3 — Analysis

Multiple layers, aur chain rule ko Product rule / Quotient rule ke saath combine karna.

Problem 3.1

$h (x) = sin^{3} (2 x)$ ko differentiate karo (matlab $[sin (2 x)]^{3}$ ).

Recall Solution 3.1

Teen layers hain: sabse baahri cube, phir $sin$ , phir $2 x$ . Bahar se andar peel karo.

Cube: $\frac{d}{d ( \cdot )} (\cdot)^{3} = 3 (\cdot)^{2}$ , $sin (2 x)$ par evaluate kiya ⇒ $3 sin^{2} (2 x)$ .
$sin$ : derivative $cos (2 x)$ .
$2 x$ : derivative $2$ . Teeno ko multiply karo: $h^{'} (x) = 3 sin^{2} (2 x) \cdot cos (2 x) \cdot 2 = 6 sin^{2} (2 x) cos (2 x) .$

Problem 3.2

$h (x) = x^{2} e^{- x^{2}}$ ko differentiate karo.

Recall Solution 3.2

Yeh ek product hai $x^{2} \cdot e^{- x^{2}}$ , aur doosre factor ko chain rule ki zaroorat hai. Product rule: $(uv)^{'} = u^{'} v + u v^{'}$ jahan $u = x^{2}$ , $v = e^{- x^{2}}$ .

$u^{'} = 2 x$ .
$v^{'} = e^{- x^{2}} \cdot (- 2 x)$ (chain rule: outer $e^{(\cdot)}$ , inner $- x^{2} \Rightarrow - 2 x$ ). Assemble karo: $h^{'} (x) = 2 x e^{- x^{2}} + x^{2} (- 2 x e^{- x^{2}}) = 2 x e^{- x^{2}} - 2 x^{3} e^{- x^{2}} = 2 x (1 - x^{2}) e^{- x^{2}} .$

Problem 3.3

$h (x) = \frac{cos x}{1 + x ^{2}}$ ko differentiate karo.

Recall Solution 3.3

Quotient rule: $(\frac{p}{q})^{'} = \frac{p ^{'} q - p q ^{'}}{q ^{2}}$ jahan $p = cos x$ , $q = (1 + x^{2})^{1/2}$ .

$p^{'} = - sin x$ .
$q^{'} = \frac{1}{2} (1 + x^{2})^{- 1/2} \cdot 2 x = \frac{x}{1 + x ^{2}}$ (chain rule).
$q^{2} = 1 + x^{2}$ . $h^{'} (x) = \frac{- s i n x 1 + x ^{2} - c o s x \cdot \frac{x}{1 + x ^{2}}}{1 + x ^{2}} .$ Andar wali fraction clear karne ke liye upar aur neeche $1 + x^{2}$ se multiply karo: $h^{'} (x) = \frac{- s i n x ( 1 + x ^{2} ) - x c o s x}{( 1 + x ^{2} ) ^{3/2}} .$

Level 4 — Synthesis

Composite khud banao: implicit differentiation aur inverse-function derivatives chain rule hi hain chhupe hue.

Problem 4.1

Curve $x^{2} + y^{2} = 25$ , $(3, 4)$ se guzarti hai. Implicit differentiation use karke, wahan $\frac{d y}{d x}$ nikalo.

Recall Solution 4.1

$y$ ko $x$ ka function maano: tab $y^{2}$ ek composite $[y (x)]^{2}$ hai, isliye iske derivative ko chain rule chahiye: $\frac{d}{d x} (y^{2}) = 2 y \cdot \frac{d y}{d x} .$ $x^{2} + y^{2} = 25$ ke dono sides differentiate karo: $2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} .$ $(3, 4)$ par: $\frac{d y}{d x} = - \frac{3}{4}$ . Geometry check: $(3, 4)$ tak radius ki slope $\frac{4}{3}$ hai; tangent perpendicular hai, slope $- \frac{3}{4}$ . ✓ (figure dekho).

Figure — Chain rule — proof, composite function derivatives

Problem 4.2

Maano $f (x) = x^{3} + x$ (strictly increasing hai, isliye invertible hai). Inverse function derivative use karke $(f^{- 1})^{'} (2)$ nikalo.

Recall Solution 4.2

Identity $f (f^{- 1} (x)) = x$ ek composite hai. Dono sides ko chain-rule karo: $f^{'} (f^{- 1} (x)) \cdot (f^{- 1})^{'} (x) = 1 \Rightarrow (f^{- 1})^{'} (x) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))} .$ $f^{- 1} (2)$ nikalo: $x^{3} + x = 2$ solve karo. $x = 1$ try karo: $1 + 1 = 2$ . ✓ Toh $f^{- 1} (2) = 1$ . $f^{'} (x) = 3 x^{2} + 1$ , isliye $f^{'} (1) = 4$ . $(f^{- 1})^{'} (2) = \frac{1}{f ^{'} ( 1 )} = \frac{1}{4} .$

Problem 4.3

Ek balloon ek sphere hai jiska volume $V = \frac{4}{3} π r^{3}$ hai. Hawa pump ki jaati hai taaki $\frac{d r}{d t} = 0.2 cm/s$ ho. Related rates use karke, $r = 5 cm$ par $\frac{d V}{d t}$ nikalo.

Recall Solution 4.3

$V$ , $r$ par depend karta hai, aur $r$ , $t$ par depend karta hai — ek composite $V (r (t))$ . Leibniz form mein chain rule: $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t} .$ $\frac{d V}{d r} = 4 π r^{2}$ . $r = 5$ par: $4 π (25) = 100 π$ . $\frac{d V}{d t} = 100 π \cdot 0.2 = 20 π \approx 62.83 cm^{3} / s .$

Level 5 — Mastery

Subtle cases: repeated nesting, proof ki apni machinery, aur jahan naïve cancellation break ho jaati.

Problem 5.1

$h (x) = e^{s i n (c o s x)}$ ko differentiate karo (yeh genuinely triple nest hai).

Recall Solution 5.1

Layers, bahar se andar: $e^{(\cdot)}$ , phir $sin (\cdot)$ , phir $cos x$ .

$e^{(\cdot)}$ ki derivative $sin (cos x)$ par: $e^{s i n (c o s x)}$ .
$sin (\cdot)$ ki derivative $cos x$ par: $cos (cos x)$ .
$cos x$ ki derivative: $- sin x$ . Teeno ko multiply karo: $h^{'} (x) = e^{s i n (c o s x)} \cdot cos (cos x) \cdot (- sin x) = - sin x cos (cos x) e^{s i n (c o s x)} .$

Problem 5.2

Maano $g (x) = x^{2} sin (1/ x)$ jab $x \neq = 0$ aur $g (0) = 0$ . Yeh jaana jaata hai ki $g^{'} (0) = 0$ . Yeh $g$ , $x \to 0$ jaate waqt infinitely baar $g (x) = 0$ hit karta hai, isliye $Δ u = g (x + Δ x) - g (0)$ infinitely baar zero hota hai nonzero $Δ x$ ke liye. Ek paragraph mein explain karo ki $h (x) = sin (g (x))$ ke liye chain rule phir bhi $h^{'} (0) = cos (g (0)) \cdot g^{'} (0) = 0$ kyun deta hai, bhale hi naïve "divide by $Δ u$ " derivation yahan illegal hai.

Recall Solution 5.2

Error-function proof kabhi $Δ u$ se divide nahi karta. Parent note se yaad karo: $u = g (0) = 0$ aur $f = sin$ ke saath, likho $f (u + k) - f (u) = (f^{'} (u) + E (k)) k, E (k) \to 0 as k \to 0, E (0) = 0.$ $k = Δ u$ set karo. Yeh equation ek multiplication hai, valid tab bhi jab $Δ u = 0$ ho (dono sides tab $0 = 0$ hain). $Δ x \neq = 0$ se divide karo: $\frac{h ( Δ x ) - h ( 0 )}{Δ x} = (f^{'} (0) + E (Δ u)) \frac{Δ u}{Δ x} .$ Jaise $Δ x \to 0$ : $g$ continuous ⇒ $Δ u \to 0$ ⇒ $E (Δ u) \to 0$ , aur $\frac{Δ u}{Δ x} \to g^{'} (0) = 0$ . Toh poora product $\to (cos 0 + 0) \cdot 0 = 0$ ho jaata hai. Infinitely baar zero hone wale $Δ u$ koi nuksaan nahi pahulichate kyunki woh kabhi denominator mein nahi hote. Yahi woh pathology hai jiske baare mein parent note mein [!mistake] "steel-man" ne warning di thi.

Problem 5.3

Rigorous proof mein term $E (Δ u) \to 0$ iss baat par rely karta hai ki $Δ u \to 0$ ho. $g$ ki kaun si ek property guarantee karti hai ki $Δ x \to 0$ hone par $Δ u \to 0$ ho, aur kaun sa theorem yeh supply karta hai?

Recall Solution 5.3

$g$ ki continuity $x$ par: $lim_{Δ x \to 0} (g (x + Δ x) - g (x)) = 0$ , yaani $Δ u \to 0$ . Yeh guaranteed hai kyunki $g$ differentiable hai, aur differentiable ⇒ continuous — dekho Continuity implies differentiability fails converse. (Converse fail hota hai: continuity akele humein $g^{'} (x)$ nahi dega, jo hume $\frac{Δ u}{Δ x}$ factor ke liye bhi chahiye.)

Connections

Chain rule — proof, composite function derivatives — parent note; har problem iske rule ka ek application hai.
Derivative — limit definition — proof ke peeche difference quotient ka source (5.2, 5.3).
Product rule · Quotient rule — L3 mein chain rule ke saath combine kiye gaye.
Implicit differentiation — $y (x)$ par chain rule (4.1).
Inverse function derivative — $f (f^{- 1} (x)) = x$ par chain rule (4.2).
Related rates — $\frac{d V}{d t} = \frac{d V}{d r} \frac{d r}{d t}$ mein chain rule (4.3).
Continuity implies differentiability fails converse — $Δ u \to 0$ supply karta hai (5.3).