Partial derivatives — notation, calculation, geometric meaning: Exercises

4.4.3 · D4 · Maths › Multivariable Calculus › Partial derivatives — notation, calculation, geometric meani

Har solution parent parent topic ke ek hi rule se use karta hai: doosre variables ko freeze karo, normally differentiate karo.

Level 1 — Recognition

Goal: yeh identify karna ki kaun se terms differentiation variable par depend karte hain aur kaun se frozen constants hain.

Problem 1.1

$f (x, y) = 5 x^{2} + 3 y - 8$ ke liye, $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 1.1

Hum kya karte hain: $x$ se pehle differentiate karo, $y$ ko ek number maan ke treat karo.

$f_{x}$ : $5 x^{2} \to 10 x$ ; term $3 y$ mein ==no $x$ == hai isliye yeh ek constant hai $\Rightarrow 0$ ; $- 8 \to 0$ . $f_{x} = 10 x$
$f_{y}$ : ab $5 x^{2}$ ek frozen constant hai $\Rightarrow 0$ ; $3 y \to 3$ ; $- 8 \to 0$ . $f_{y} = 3$

Problem 1.2

$f (x, y) = x^{3} y$ ke liye, $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 1.2

$f_{x}$ : $y$ ek constant multiplier hai. $\frac{d}{d x} x^{3} = 3 x^{2}$ , $y$ ko rakho: $f_{x} = 3 x^{2} y$
$f_{y}$ : $x^{3}$ constant multiplier hai. $\frac{d}{d y} y = 1$ : $f_{y} = x^{3}$

Problem 1.3

$f (x, y, z) = 7$ (ek constant function) ke liye, $f_{x}, f_{y}, f_{z}$ nikalo.

Recall Solution 1.3

Yeh kaisa dikhta hai: bilkul flat surface — kisi bhi direction mein koi slope nahi. $f_{x} = 0, f_{y} = 0, f_{z} = 0$ Kyun: constant ka derivative $0$ hota hai chahe tum koi bhi variable wiggle karo.

Level 2 — Application

Goal: product, quotient, aur chain rules use karo jabki ek variable frozen ho.

Problem 2.1

$f (x, y) = sin (x) e^{y}$ . $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 2.1

$f_{x}$ : $e^{y}$ ek constant coefficient hai. $\frac{d}{d x} sin x = cos x$ : $f_{x} = cos (x) e^{y}$
$f_{y}$ : $sin x$ constant coefficient hai. $\frac{d}{d y} e^{y} = e^{y}$ : $f_{y} = sin (x) e^{y}$

Problem 2.2

$f (x, y) = e^{x^{2} y}$ . $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 2.2

Kaun sa tool aur kyun? Exponent $x^{2} y$ khud ek function hai, isliye hume chain rule chahiye: $e^{(inner)}$ ka derivative $e^{(inner)} \cdot (inner)^{'}$ hota hai.

$f_{x}$ : inner $= x^{2} y$ , iska $x$ -derivative hai $2 x y$ ( $y$ frozen): $f_{x} = e^{x^{2} y} \cdot 2 x y$
$f_{y}$ : inner ka $y$ -derivative hai $x^{2}$ ( $x$ frozen): $f_{y} = e^{x^{2} y} \cdot x^{2}$

Problem 2.3

$f (x, y) = \frac{x}{x + y}$ . $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 2.3

Kaun sa tool aur kyun? Ek ratio jisme dono parts variable contain karte hain → quotient rule $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

$f_{x}$ : $u = x, u_{x} = 1$ ; $v = x + y, v_{x} = 1$ : $f_{x} = \frac{1 \cdot ( x + y ) - x \cdot 1}{( x + y ) ^{2}} = \frac{y}{( x + y ) ^{2}}$
$f_{y}$ : ab $u = x$ ek constant hai isliye $u_{y} = 0$ ; $v = x + y, v_{y} = 1$ : $f_{y} = \frac{0 \cdot ( x + y ) - x \cdot 1}{( x + y ) ^{2}} = \frac{- x}{( x + y ) ^{2}}$

Level 3 — Analysis

Goal: points par evaluate karo, slopes geometrically padho, degenerate/zero inputs handle karo.

Problem 3.1

$f (x, y) = x^{2} + y^{2}$ . $f_{x} (3, - 1)$ aur $f_{y} (3, - 1)$ compute karo, aur batao kaun sa direction zyada steep hai.

Recall Solution 3.1

Pehle differentiate karo, baad mein evaluate karo (kabhi ulta mat karo). $f_{x} = 2 x \Rightarrow f_{x} (3, - 1) = 6, f_{y} = 2 y \Rightarrow f_{y} (3, - 1) = - 2$ Steepness $∣ slope ∣$ se measure hoti hai: $∣6∣ > ∣ - 2∣$ , isliye surface yahan $x$ -direction mein zyada steep hai. Negative $f_{y}$ ka matlab hai ki $+ y$ direction mein move karne se height decrease hoti hai (hum $y = - 1$ par hain, bowl ki down-slope side par).

Problem 3.2

$f (x, y) = ln (x^{2} + y^{2})$ . $f_{x}$ aur $f_{y}$ nikalo, phir $(1, 0)$ par evaluate karo. Origin $(0, 0)$ par kya hota hai?

Recall Solution 3.2

Chain rule: $\frac{d}{d x} ln (u) = \frac{u _{x}}{u}$ jahan $u = x^{2} + y^{2}$ . $f_{x} = \frac{2 x}{x ^{2} + y ^{2}}, f_{y} = \frac{2 y}{x ^{2} + y ^{2}}$ $(1, 0)$ par: $f_{x} = \frac{2}{1} = 2$ , $f_{y} = \frac{0}{1} = 0$ . Degenerate case $(0, 0)$ : wahan $x^{2} + y^{2} = 0$ hai, isliye dono partials mein $\frac{0}{0}$ hai — undefined. Saath hi $ln 0$ hai $- \infty$ , isliye $f$ khud bhi origin par defined nahi hai. Partials har jagah exist karte hain sirf us ek hole ko chhod ke.

Problem 3.3

$f (x, y) = x^{2} y^{3}$ . $(1, 2)$ par $z = f (x, y)$ ka tangent plane nikalo.

Recall Solution 3.3

Formula (parent se): $z = f (a, b) + f_{x} (a, b) (x - a) + f_{y} (a, b) (y - b)$ .

$f (1, 2) = 1^{2} \cdot 2^{3} = 8$ .
$f_{x} = 2 x y^{3} \Rightarrow f_{x} (1, 2) = 2 \cdot 1 \cdot 8 = 16$ .
$f_{y} = 3 x^{2} y^{2} \Rightarrow f_{y} (1, 2) = 3 \cdot 1 \cdot 4 = 12$ . $z = 8 + 16 (x - 1) + 12 (y - 2)$ Do slice-slopes $16$ aur $12$ exactly plane ka $x$ aur $y$ mein tilt hain — neeche figure dekho.

Figure — Partial derivatives — notation, calculation, geometric meaning

Level 4 — Synthesis

Goal: partials combine karo — higher-order, mixed, aur identities/theorems verify karo.

Problem 4.1

$f (x, y) = x^{3} y^{2} + sin (x y)$ . Mixed partials $f_{x y}$ aur $f_{y x}$ compute karo aur Clairaut's theorem verify karo.

Recall Solution 4.1

Step 1 — $f_{x}$ : ( $y$ frozen) $f_{x} = 3 x^{2} y^{2} + cos (x y) \cdot y$ . Step 2 — $f_{x y} = \partial_{y} (f_{x})$ : isko $y$ mein differentiate karo.

$3 x^{2} y^{2} \to 6 x^{2} y$ .
$cos (x y) y$ : $cos (x y)$ aur $y$ ka product, dono $y$ par depend karte hain. Product rule: $\partial_{y} [cos (x y)] \cdot y + cos (x y) \cdot 1 = - sin (x y) \cdot x \cdot y + cos (x y)$ $f_{x y} = 6 x^{2} y - x y sin (x y) + cos (x y)$ Step 3 — doosra order. $f_{y} = 2 x^{3} y + cos (x y) \cdot x$ , phir $f_{y x} = \partial_{x} (f_{y})$ :
$2 x^{3} y \to 6 x^{2} y$ .
$cos (x y) x$ : $x$ mein product rule: $- sin (x y) \cdot y \cdot x + cos (x y)$ . $f_{y x} = 6 x^{2} y - x y sin (x y) + cos (x y)$ Dono match karte hain ✔ — jaise Clairaut's theorem guarantee karta hai, kyunki yeh second partials continuous hain.

Problem 4.2

Dikhao ki $f (x, y) = ln (x^{2} + y^{2})$ Laplace's equation $f_{xx} + f_{y y} = 0$ satisfy karta hai (origin se door).

Recall Solution 4.2

Problem 3.2 se, $f_{x} = \frac{2 x}{x ^{2} + y ^{2}}$ . Dobara $x$ mein differentiate karo (quotient rule, $u = 2 x$ , $v = x^{2} + y^{2}$ ): $f_{xx} = \frac{2 ( x ^{2} + y ^{2} ) - 2 x \cdot 2 x}{( x ^{2} + y ^{2} ) ^{2}} = \frac{2 y ^{2} - 2 x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ Symmetry $x \leftrightarrow y$ se (roles swap karo), $f_{y y} = \frac{2 x ^{2} - 2 y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Jodo: $f_{xx} + f_{y y} = \frac{( 2 y ^{2} - 2 x ^{2} ) + ( 2 x ^{2} - 2 y ^{2} )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{0}{( x ^{2} + y ^{2} ) ^{2}} = 0 ✓$

Problem 4.3

$f (x, y) = x^{y}$ ( $x > 0$ ke saath). $f_{x}$ aur $f_{y}$ nikalo.

Recall Solution 4.3

Do alag tools kyun? $f_{x}$ ke liye variable base mein hai (power rule) lekin $f_{y}$ ke liye exponent mein hai (exponential rule) — same expression ko alag rules ki zaroorat padti hai depending on kaun sa slot wiggle karo.

$f_{x}$ : $y$ ko fixed exponent maano → power rule $\frac{d}{d x} x^{y} = y x^{y - 1}$ . $f_{x} = y x^{y - 1}$
$f_{y}$ : $x$ ko fixed base maano. $x^{y} = e^{y l n x}$ rewrite karo; phir $\partial_{y} = e^{y l n x} \cdot ln x = x^{y} ln x$ . $f_{y} = x^{y} ln x$

Level 5 — Mastery

Goal: limit definition se derive karo aur pathological points ke baare mein reason karo.

Problem 5.1

Sirf limit definition use karke, $f (x, y) = x^{2} y$ ke liye $f_{x} (2, 3)$ compute karo.

Recall Solution 5.1

Kaun sa tool aur kyun? Humse kaha ja raha hai ki rules bypass karo aur definition use karo — yahi hai jo rules shorthand hain. $f_{x} (2, 3) = lim_{h \to 0} \frac{f ( 2 + h , 3 ) - f ( 2 , 3 )}{h}$ Pieces compute karo ( $y = 3$ frozen):

$f (2 + h, 3) = (2 + h)^{2} \cdot 3 = 3 (4 + 4 h + h^{2}) = 12 + 12 h + 3 h^{2}$ .
$f (2, 3) = 4 \cdot 3 = 12$ . Numerator $= 12 h + 3 h^{2}$ . $h$ se divide karo: $\frac{12 h + 3 h ^{2}}{h} = 12 + 3 h h \to 0 12$ $f_{x} (2, 3) = 12$ Rule se check karo: $f_{x} = 2 x y \Rightarrow 2 \cdot 2 \cdot 3 = 12$ ✔.

Problem 5.2

Piecewise function ke liye $f (x, y) = ⎩ ⎨ ⎧ \frac{x y}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0)$ limit definition se $f_{x} (0, 0)$ aur $f_{y} (0, 0)$ nikalo.

Recall Solution 5.2

Definition kyun, rules kyun nahi? Origin par formula badal jaata hai; quotient rule stitched point par apply nahi hota. Sirf limit dono pieces dekhta hai. $f_{x} (0, 0) = lim_{h \to 0} \frac{f ( h , 0 ) - f ( 0 , 0 )}{h}$ Ab $f (h, 0) = \frac{h \cdot 0}{h ^{2} + 0} = 0$ aur $f (0, 0) = 0$ , isliye quotient hai $\frac{0 - 0}{h} = 0$ har $h \neq = 0$ ke liye: $f_{x} (0, 0) = 0$ $y$ ke saath identical computation se (set $x = 0$ ): $f (0, k) = \frac{0 \cdot k}{0 + k ^{2}} = 0$ , isliye $f_{y} (0, 0) = 0$ Punchline: dono partials exist karte hain aur $0$ ke barabar hain, phir bhi yeh $f$ origin par famously not continuous hai ( $y = x$ ke along approach karne par $\frac{1}{2}$ milta hai, axes ke along $0$ milta hai). Isliye partials ka exist karna smoothness guarantee nahi karta — yeh Tangent plane and linear approximation ke liye ek gehri warning hai.

Problem 5.3

$f (x, y, z) = \frac{1}{x ^{2} + y ^{2} + z ^{2}}$ (3D "inverse distance", $r = x^{2} + y^{2} + z^{2}$ ). $f_{x}$ nikalo aur $(1, 2, 2)$ par evaluate karo.

Recall Solution 5.3

$f = (x^{2} + y^{2} + z^{2})^{- 1/2}$ likhо. Chain rule inner $u = x^{2} + y^{2} + z^{2}$ , $u_{x} = 2 x$ ke saath: $f_{x} = - \frac{1}{2} (x^{2} + y^{2} + z^{2})^{- 3/2} \cdot 2 x = \frac{- x}{( x ^{2} + y ^{2} + z ^{2} ) ^{3/2}}$ $(1, 2, 2)$ par: denominator base $= 1 + 4 + 4 = 9$ , aur $9^{3/2} = (3^{2})^{3/2} = 3^{3} = 27$ . Isliye $f_{x} (1, 2, 2) = \frac{- 1}{27}$

Recall summary

Recall Kaun sa rule kaun se slot ke liye?

Variable base mein hai ( $x^{y}$ , $f_{x}$ nikalo) ::: power rule → $y x^{y - 1}$ . Variable exponent mein hai ( $x^{y}$ , $f_{y}$ nikalo) ::: $e^{y l n x}$ rewrite karo → $x^{y} ln x$ . Dono factors variable par depend karte hain ::: product rule (do terms). Ratio, dono parts variable par depend karte hain ::: quotient rule. Piecewise / singular $f$ ka origin par point ::: limit definition par wapas jao.

Connections

Single-variable derivative — har partial ek baar freeze karne ke baad inhi mein se ek ban jaata hai.
Tangent plane and linear approximation — Problems 3.3 & 5.2 directly isme feed karte hain.
Clairaut's theorem — Problem 4.1 mein numerically verify kiya gaya.
Gradient vector — yahan compute kiye gaye $f_{x}, f_{y}, f_{z}$ ko bundle karta hai.
Directional derivative — axis directions se aage ki agli generalization.
Chain rule (multivariable) — Problems 2.2, 3.2, 5.3 ke peeche ka engine.
Total differential — partials ko $df$ mein combine karta hai.