Partial derivatives: Worked examples — AI-ML deep dive

Parent: Partial derivatives. Here we don't teach a new idea — we drill the one rule (Freeze Others, Differentiate) against every kind of input it can meet, so no surprise catches you later.

Before we start, one promise: every symbol you see was already earned in the parent note. $\partial f / \partial x$ = "nudge only $x$ , freeze the rest, measure the output's slope". $\nabla f$ = the list of all such slopes. If any line feels unfamiliar, re-read the parent — we build strictly on top of it.

The scenario matrix

Think of every partial-derivative problem as landing in one of these cells. If we work one clean example per cell, you have seen the whole map.

#	Case class	What makes it tricky	Covered by
A	Pure polynomial, both partials	terms with no moving variable must vanish	Ex 1
B	Product of two variables (needs product/constant-multiple care)	is it product rule or constant multiple?	Ex 2
C	Composition $g (inner)$ — chain rule inside	must multiply by inner slope	Ex 3
D	Quotient / negative powers — sign of the slope flips by region	denominator can make slope positive or negative	Ex 4
E	Degenerate / zero input — evaluate at a point, incl. $0$	does a "vanishing" term really vanish numerically?	Ex 5
F	Symmetry & mixed second partials	order of differentiation, does it matter?	Ex 6
G	Real-world word problem (heat / area) with units	keep track of what each slope means physically	Ex 7
H	ML payoff — a full gradient of a loss, both signs of residual	which way does each weight get pushed?	Ex 8
I	Exam twist — variable appears in the exponent AND base	you cannot treat it as a simple power	Ex 9

We now hit every cell.

Worked example Ex 1 — Cell A: pure polynomial, both partials

$f (x, y) = 4 x^{3} y - 7 x + 2 y^{2} + 9$ . Find $f_{x}$ and $f_{y}$ .

Forecast: Which term disappears in $f_{x}$ ? Guess before reading.

$\partial (4 x^{3} y) / \partial x = 12 x^{2} y$ . Why this step? $y$ is frozen (a number), so $4 y$ is just a constant multiplier on $x^{3}$ ; $\frac{d}{d x} x^{3} = 3 x^{2}$ , giving $4 y \cdot 3 x^{2}$ .
$\partial (- 7 x) / \partial x = - 7$ . Why this step? Ordinary slope of a straight line.
$\partial (2 y^{2}) / \partial x = 0$ and $\partial (9) / \partial x = 0$ . Why this step? No $x$ inside → zero slope in the $x$ -direction. This is the "vanishing" rule from the parent.
So $f_{x} = 12 x^{2} y - 7$ .
Now freeze $x$ : $f_{y} = 4 x^{3} (1) + 0 + 4 y + 0 = 4 x^{3} + 4 y$ . Why this step? $4 x^{3}$ multiplies $y^{1}$ → constant times $1$ ; $2 y^{2} \to 4 y$ ; the $- 7 x$ and $9$ have no $y$ .

Verify: At $(x, y) = (1, 1)$ : $f_{x} = 12 - 7 = 5$ , $f_{y} = 4 + 4 = 8$ . Both finite, both correct sign for a surface rising in both directions there. ✅

Worked example Ex 3 — Cell C: chain rule inside

$f (x, y) = sin (x^{2} y)$ . Find $f_{x}$ and $f_{y}$ .

Forecast: Where does the "inner factor" come from?

Let the inside be $u = x^{2} y$ . Then $f = sin u$ . Why this step? Naming the inside makes the chain rule visible: outer function of an inner function.
Why chain rule and not just $cos$ ? Because $u$ itself changes as $x$ changes. The chain rule answers "how fast does the whole composite move" = (how fast $sin$ moves per unit $u$ ) $\times$ (how fast $u$ moves per unit $x$ ).
$f_{x} = cos (u) \cdot \partial u / \partial x = cos (x^{2} y) \cdot (2 x y) = 2 x y cos (x^{2} y)$ . Why this step? Inner slope $\partial (x^{2} y) / \partial x = 2 x y$ (with $y$ frozen).
$f_{y} = cos (u) \cdot \partial u / \partial y = cos (x^{2} y) \cdot x^{2} = x^{2} cos (x^{2} y)$ .

Verify: At $(x, y) = (1, 0)$ : inner $u = 0$ , $cos 0 = 1$ . $f_{x} = 0$ , $f_{y} = 1$ . Since $f = sin (x^{2} y)$ near $y = 0$ looks like $sin (x^{2} y) \approx x^{2} y$ , its $y$ -slope at $x = 1$ is $x^{2} = 1$ . ✅

Worked example Ex 4 — Cell D: quotient, and the slope's sign flips by region

$f (x, y) = \frac{x}{y}$ (with $y \neq = 0$ ). Find both partials and discuss the sign in each quadrant of $y$ .

Forecast: Is $f_{y}$ positive or negative when $y > 0$ ?

Rewrite $f = x \cdot y^{- 1}$ . Why this step? Turning division into a negative power lets us use the plain power rule instead of the quotient rule.
$f_{x} = y^{- 1} = \frac{1}{y}$ . Why this step? $y^{- 1}$ is a frozen constant multiplier on $x$ .
$f_{y} = x \cdot (- 1) y^{- 2} = - \frac{x}{y ^{2}}$ . Why this step? Power rule: $\frac{d}{d y} y^{- 1} = - y^{- 2}$ ; the frozen $x$ stays.
Sign discussion (covering all cases):
- $f_{x} = 1/ y$ is positive when $y > 0$ , negative when $y < 0$ — pushing $x$ up raises $f$ only if the divisor is positive.
- $f_{y} = - x / y^{2}$ . Since $y^{2} > 0$ always, the sign is set by $- x$ : negative when $x > 0$ , positive when $x < 0$ , zero when $x = 0$ .
- Degenerate: at $y = 0$ the function itself is undefined, so no partials exist there — always state the domain first.

Verify: At $(x, y) = (2, - 1)$ : $f_{x} = 1/ (- 1) = - 1$ , $f_{y} = - 2/1 = - 2$ . Sanity: $f = 2/ (- 1) = - 2$ ; nudge $y$ toward $0^{-}$ makes $f$ more negative → $\partial f / \partial y < 0$ locally? Check: moving $y$ from $- 1$ up toward $0$ increases $y$ , and $- 2/ y^{2}$ at $x = 2$ is negative, so $f$ decreases — consistent. ✅

Worked example Ex 5 — Cell E: evaluate at a zero / degenerate point

$f (x, y) = x^{2} e^{y} + x y$ . Find $f_{x}$ , then evaluate at $(0, 0)$ and at $(0, 5)$ .

Forecast: Will $f_{x}$ be $0$ at $(0, 0)$ ? At $(0, 5)$ ?

$f_{x} = 2 x e^{y} + y$ . Why this step? $e^{y}$ is frozen → constant multiplier on $x^{2}$ giving $2 x e^{y}$ ; the term $x y$ has $y$ frozen → slope $y$ .
At $(0, 0)$ : $f_{x} = 2 (0) e^{0} + 0 = 0$ . Why this step? Both terms carry a factor that is $0$ here — a genuine degenerate point where the $x$ -slope flattens.
At $(0, 5)$ : $f_{x} = 2 (0) e^{5} + 5 = 5$ . Why this step? The first term still dies (factor $x = 0$ ) but the $+ y$ term survives as $5$ . Lesson: " $x = 0$ " does not force the whole slope to zero — only terms that contain $x$ vanish.

Verify: $f_{x} (0, 0) = 0$ and $f_{x} (0, 5) = 5$ . Cross-check $f_{y} = x^{2} e^{y} + x$ ; at $(0, 5)$ that is $0$ , so the surface is momentarily flat in $y$ but tilted in $x$ there. ✅

Geometric read of Ex 7: the rectangle's area $A = x y$ . The blue arrow shows growth when you push the width $x$ ; the orange arrow shows growth when you push the height $y$ . The slope in each direction is literally the length of the other side.

Worked example Ex 7 — Cell G: word problem with units

A rectangular metal plate has width $x$ metres and height $y$ metres. Area $A (x, y) = x y$ (m²). At $x = 3 m, y = 4 m$ , how fast does area grow per metre of width? Per metre of height?

Forecast: Which side controls the width-slope — the width or the height?

$\partial A / \partial x = y$ . Why this step? Freeze height $y$ ; area is $y \cdot x$ , a line of slope $y$ . Adding a thin strip of width $d x$ adds area = height × $d x$ .
At the point, $\partial A / \partial x = 4$ m²/m = 4 m. Why this step? Units: (m²)/(m) = m. The height number is the width-slope — pushing width by 1 m adds a strip of area equal to the current height.
$\partial A / \partial y = x = 3$ m²/m at the point. Why this step? Symmetric: freeze width, the added strip's area = current width × $d y$ .

Verify: Grow both by a tiny $0.01$ m: predicted $Δ A \approx 4 (0.01) + 3 (0.01) = 0.07$ m². Exact new area $3.01 \times 4.01 = 12.0701$ , old $= 12$ , so $Δ A = 0.0701$ — matches to first order (the extra $0.0001$ is the tiny corner square $d x d y$ ). ✅

Worked example Ex 8 — Cell H: full ML gradient, both signs of residual

Linear model $\hat{t} = w x + b$ . One data point $x = 2, t = 1$ . Squared loss $L (w, b) = (w x + b - t)^{2}$ . Evaluate the gradient at (a) $w = 1, b = 0$ (under-predict) and (b) $w = 0, b = 3$ (over-predict).

Forecast: In which case does $\partial L / \partial w$ come out negative? (Negative slope → push $w$ up.)

Residual $r = w x + b - t$ . From the parent: $\partial L / \partial w = 2 r x$ , $\partial L / \partial b = 2 r$ . Why this step? Chain rule: $\partial L / \partial w = 2 r \cdot \partial r / \partial w = 2 r \cdot x$ ; likewise $\partial r / \partial b = 1$ .
Case (a) $w = 1, b = 0$ : prediction $= 2$ , so $r = 2 + 0 - 1 = 1$ (over-predict, $r > 0$ ). $\partial L / \partial w = 2 (1) (2) = 4$ , $\partial L / \partial b = 2 (1) = 2$ . Why this step? Positive slopes → descent pushes $w$ and $b$ down, shrinking the too-big prediction. Correct.
Case (b) $w = 0, b = 3$ : prediction $= 3$ , $r = 3 - 1 = 2$ (still over-predict). $\partial L / \partial w = 2 (2) (2) = 8$ , $\partial L / \partial b = 2 (2) = 4$ . Why this step? Bigger residual → bigger slopes → bigger corrective step.
Case (c) to show a negative slope: $w = 0, b = 0$ : prediction $= 0$ , $r = - 1$ (under-predict, $r < 0$ ). $\partial L / \partial w = 2 (- 1) (2) = - 4$ . Why this step? Negative slope → descent adds to $w$ , raising the too-small prediction. Sign of $r$ sets the direction; $x$ 's sign scales/flips $\partial L / \partial w$ .

Verify: In case (a), nudge $w$ down by $0.01$ : new $r = 0.99 \cdot 2 - 1 = 0.98$ , new $L = 0.9604$ vs old $L = 1$ . $Δ L \approx - 0.0396$ ; predicted $\partial L / \partial w \cdot (- 0.01) = 4 \cdot (- 0.01) = - 0.04$ . Match. ✅ See 1.3.01-Gradient-descent.

Worked example Ex 9 — Cell I (exam twist): the variable is in BOTH base and exponent

$f (x, y) = x^{y}$ (with $x > 0$ ). Find $f_{x}$ and $f_{y}$ . Warning: $f_{x} \neq = y x^{y - 1}$ and $f_{y} \neq = x^{y} ln x$ are not both from the same rule — pick the right one for each.

Forecast: Which partial uses the power rule, which uses the exponential rule?

$f_{x}$ : here the base $x$ moves, the exponent $y$ is frozen → it's a power function $x^{(const)}$ . So $f_{x} = y x^{y - 1}$ . Why this step? Power rule applies precisely because the exponent is constant.
$f_{y}$ : here the exponent $y$ moves, the base $x$ is frozen → it's an exponential $x^{y} = e^{y l n x}$ . Differentiate: $f_{y} = e^{y l n x} \cdot ln x = x^{y} ln x$ . Why this step? When the exponent is the mover you must rewrite as $e^{(\cdot)}$ ; that's why $ln x$ appears — it's the inner slope $\partial (y ln x) / \partial y = ln x$ . This is exactly the single-variable $\frac{d}{d y} a^{y} = a^{y} ln a$ with $a = x$ .
The exam trap: writing $f_{y} = y x^{y - 1}$ (power rule on the wrong variable) is wrong — the power rule needs a constant exponent, but here the exponent is the mover.

Verify: At $(x, y) = (2, 3)$ : $f_{x} = 3 \cdot 2^{2} = 12$ , $f_{y} = 2^{3} ln 2 = 8 ln 2 \approx 5.545$ . Numerically nudge $y$ up by $0.001$ : $2^{3.001} = 8 \cdot 2^{0.001} \approx 8.00554$ , so slope $\approx 5.545$ . Match. ✅

Recall Which rule for which mover? (the whole matrix in one card)

Constant exponent, moving base ::: power rule, $y x^{y - 1}$ Constant base, moving exponent ::: exponential rule, $x^{y} ln x$ Term with no copy of the moving variable ::: slope is $0$ , it vanishes Division $x / y$ , take $\partial / \partial y$ ::: rewrite as $x y^{- 1}$ , get $- x / y^{2}$ Composite $sin (inner)$ ::: chain rule, multiply by inner's partial Order of mixed partials $f_{x y}$ vs $f_{y x}$ ::: equal for smooth functions (Clairaut)

Connections

Partial derivatives — the parent this drills.
1.2.02-Derivatives-single-variable — every cell reduces to this once you freeze.
1.2.04-Gradient-and-directional-derivatives — Ex 8's pair of slopes is a gradient.
1.2.05-Chain-rule-multivariable — Ex 3 and Ex 9 lean on it.
1.3.01-Gradient-descent — Ex 8's signs decide the update direction.
Linear-regression — Ex 8 is its exact learning step.

Worked examples — Partial derivatives

The scenario matrix

Connections

Case Map