Worked examples — Partial derivatives — notation, calculation, geometric meaning

Before we compute anything, we need a map of the territory.

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The scenario matrix

Here is every class of problem partial derivatives throw at you. Each later example is tagged with the cell it covers.

Cell	Situation	What could bite you	Covered by
A	Plain polynomial, plug in a point	dropping the frozen variable	Ex 1
B	Product / chain mixed together	forgetting product rule and chain rule at once	Ex 2
C	Quotient, 3 variables, negative sign	sign of $-y^{-2}$; domain $y\ne 0$	Ex 3
D	Exponential & log — every sign of input	$\ln$ needs positive input; sign of result	Ex 4
E	Evaluate at a point where a partial is zero (flat direction)	thinking "zero slope = wrong answer"	Ex 5
F	Degenerate / limiting input (denominator $\to 0$)	undefined vs infinite; where the formula breaks	Ex 6
G	Real-world word problem with units	keeping units, reading the question	Ex 7
H	Exam twist: mixed & pure higher-order, check Clairaut	order of differentiation, sign bookkeeping	Ex 8

Every cell A–H gets a fully worked example below.

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Setup for figures

Several examples are geometric, and each of those carries its own figure. Every figure shows the same core idea in the theme's chalk colours: a surface or curve $z=f(x,y)$ being probed in one direction, so that the partial derivative appears as the slope of what you see. When a figure is present, the text points you to specific coloured strokes ("look at the pink curve", "the blue curve") — the picture carries the argument, the words only narrate it.

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Example 1 — Cell A: polynomial, plug in a point

Steps.

1. $f_x=6x^{2}y-5y^2+0$. Why this step? Freeze $y$. Then $2x^{3}y\to 6x^{2}y$ (the $y$ rides along as a constant coefficient), $-5x{y}^2\to -5y^2$ (again $y^2$ frozen), and $6y$ has no $x$, so its $x$-derivative is $0$ — that is the term that vanishes.

2. $f_y=2x^3-10xy+6$. Why this step? Now freeze $x$. Here $2x^{3}y\to 2x^3$, $-5x{y}^2\to -10xy$, and $6y\to 6$. Every term contained a $y$, so none vanishes this time.

3. Evaluate at $(1,-2)$: $f_x(1,-2)=6(1{)}^2(-2)-5(-2{)}^2=-12-20=-32.$ $f_y(1,-2)=2(1{)}^3-10(1)(-2)+6=2+20+6=28.$ Why this step? Differentiate first, substitute last — the parent note's warning about plugging in too early.

Verify: Direction check — $f_x(1,-2)=-32<0$ means as you step in the $+x$ direction from $(1,-2)$ the surface goes down; $f_y=28>0$ means stepping in $+y$ it climbs steeply. Two different signs is perfectly normal.

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Example 2 — Cell B: product AND chain together

Steps.

1. $f_x$: treat $y$ as constant. Both $x^2$ and $e^{xy}$ depend on $x$, so use the product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$: $f_x=\underset{(x^2{)}^{\prime }}{\underbrace{2x}}{e}^{xy}+x^2\underset{(e^{xy}{)}^{\prime }}{\underbrace{(e^{xy}\cdot y)}}.$ Why this step? $\frac{d}{dx}{e}^{xy}=e^{xy}\cdot \frac{d}{dx}(xy)=e^{xy}\cdot y$ — the chain rule, with inner derivative $y$ (constant times $x$). $\boxed{f_x=2x{e}^{xy}+x^{2}y{e}^{xy}=x{e}^{xy}(2+xy).}$

2. $f_y$: treat $x$ as constant. Now $x^2$ is a frozen coefficient — only $e^{xy}$ has a $y$, so no product rule, just chain: $f_y=x^2\cdot e^{xy}\cdot x=x^3{e}^{xy}.$ Why this step? $\frac{d}{dy}(xy)=x$, so $\frac{d}{dy}{e}^{xy}=x{e}^{xy}$; multiply by the constant $x^2$.

Verify: At $(1,0)$: $e^{xy}=e^0=1$. $f_x(1,0)=1\cdot 1\cdot (2+0)=2$, $f_y(1,0)=1\cdot 1=1$. Sanity: near $(1,0)$, $f\approx x^2(1+xy)$, so wiggling $x$ from the $x^2$ alone gives slope $2x=2$ ✓, wiggling $y$ gives slope $x^3=1$ ✓.

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Example 3 — Cell C: quotient, three variables, negative sign

Steps.

0. Domain first. Because $y$ sits in the denominator, $f$ is defined only for $y\ne 0$. Along the whole plane $y=0$ the value $x^{2}z/y$ blows up (or is $0/0$ if $x^{2}z=0$) — undefined. So every formula below is valid only where $y\ne 0$, and none of the partials exist on $y=0$. Why this step? A quotient can only be differentiated where its denominator is nonzero; skipping this lets a later "answer at $(x,0,z)$" slip through as though it were legal. State the exclusion before touching the rule.

1. Rewrite once and for all: $f=x^{2}z{y}^{-1}$ (for $y\ne 0$). Why this step? A negative power turns the quotient into a product, so the power rule handles $y$ cleanly.

2. $f_x$: freeze $y,z$. Coefficient is $z{y}^{-1}$; $\frac{d}{dx}{x}^2=2x$: $f_x=\frac{2xz}{y}.$ Why this step? With $y$ and $z$ frozen, only the $x^2$ factor changes; the whole $z{y}^{-1}$ block sits out front as a constant multiplier while we apply the power rule to $x^2$.

3. $f_z$: freeze $x,y$. Coefficient $x^2{y}^{-1}$; $\frac{d}{dz}z=1$: $f_z=\frac{x^2}{y}.$ Why this step? Now $x^2{y}^{-1}$ is the frozen block and $z$ appears only to the first power, so $\frac{d}{dz}z=1$ leaves the block untouched.

4. $f_y$: freeze $x,z$. Coefficient $x^{2}z$; $\frac{d}{dy}{y}^{-1}=-y^{-2}$: $f_y=-\frac{x^{2}z}{y^2}.$ Why the minus? $\frac{d}{dy}{y}^{-1}=-1\cdot y^{-2}$ — the exponent drops by one to $-2$ and the old exponent $-1$ comes down as a factor, giving the minus. (Note $y^{-2}$ still requires $y\ne 0$, consistent with Step 0.)

Verify: At $(2,1,3)$ (allowed, since $y=1\ne 0$): $f_x=\frac{2\cdot 2\cdot 3}{1}=12$, $f_z=\frac{4}{1}=4$, $f_y=-\frac{4\cdot 3}{1}=-12$. Sanity for $f_y$: increasing $y$ (the denominator) must shrink $f$, so the slope in $y$ should be negative — and $-12<0$ ✓.

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Example 4 — Cell D: exponential and log, watch every sign

Steps.

1. $f_x=\frac{1}{x^2+y^2}\cdot 2x=\frac{2x}{x^2+y^2}$. Why this step? Chain rule: $\frac{d}{dx}\ln (u)=\frac{1}{u}{u}^{\prime }$ with $u=x^2+y^2$, $u_x=2x$.

2. $f_y=\frac{2y}{x^2+y^2}$. Why this step? Same chain rule with the other slot: $u=x^2+y^2$ but now the inner derivative is $u_y=2y$ ($x^2$ is frozen). The structure is identical to $f_x$ with $x$ swapped for $y$ — that is the "by symmetry" you often hear, spelled out.

3. Domain / degenerate point. $\ln$ needs $x^2+y^2>0$, which holds everywhere except the origin $(0,0)$, where $x^2+y^2=0$ and $\ln 0$ is undefined. So $f$ and its partials do not exist at $(0,0)$. Why this step? Cell D demands we check the sign of the input to a log — this is where it degenerates.

4. Sign of $f_x$ across quadrants (in the figure, follow the pink arrows: the horizontal component of each arrow is $f_x$). The denominator $x^2+y^2$ is always $>0$, so $f_x$ has the sign of $x$:

   • Quadrant I ($x>0,y>0$): $f_x>0$.
   • Quadrant II ($x<0,y>0$): $f_x<0$.
   • Quadrant III ($x<0,y<0$): $f_x<0$.
   • Quadrant IV ($x>0,y<0$): $f_x>0$. And $f_x=0$ exactly on the $y$-axis ($x=0$). Why this step? A fraction's sign is decided entirely by its numerator when the denominator is a fixed positive number; here the numerator is $2x$, so we just read off the sign of $x$. This is why in the figure every pink arrow on the right half leans rightward and every one on the left half leans leftward.

5. Sign of $f_y$ across quadrants (in the same figure, the vertical component of each pink arrow is $f_y$). Same denominator, so $f_y$ has the sign of $y$:

   • Quadrant I ($x>0,y>0$): $f_y>0$.
   • Quadrant II ($x<0,y>0$): $f_y>0$.
   • Quadrant III ($x<0,y<0$): $f_y<0$.
   • Quadrant IV ($x>0,y<0$): $f_y<0$. And $f_y=0$ exactly on the $x$-axis ($y=0$). Why this step? By the identical fraction argument, the numerator $2y$ now controls the sign, so $f_y$ tracks the sign of $y$. Combining Steps 4 and 5: horizontal part follows $x$, vertical part follows $y$, so the arrow always points radially outward from the origin — exactly what the pink arrows do in every quadrant of the figure.

Verify: At $(1,1)$: $f_x=\frac{2}{2}=1$, $f_y=\frac{2}{2}=1$ (both $+$, Quadrant I ✓). At $(-3,4)$: denominator $=25$, $f_x=\frac{-6}{25}=-0.24$ ($x<0$ so $f_x<0$ ✓), $f_y=\frac{8}{25}=0.32$ ($y>0$ so $f_y>0$ ✓).

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Example 5 — Cell E: a point where a partial is zero (flat direction)

Steps.

1. $f_x=2x$, $f_y=2y$. Why this step? Each variable's square differentiates independently; the other is a constant that vanishes.

2. At $(0,3)$: $f_x(0,3)=2\cdot 0=0$, $f_y(0,3)=2\cdot 3=6$. Why this step? Substitute last.

3. Geometry. In the figure, look at the pink slice-curve cut by the plane $y=3$: at $x=0$ it sits at the bottom of a parabola, so the short yellow tangent segment drawn there is horizontal — slope $0$. That is exactly $f_x(0,3)=0$. Meanwhile the blue slice cut by the plane $x=0$ is still climbing at that point with slope $6$ — the surface is rising, just not in the $x$-direction. Why this step? A zero partial does not mean the surface is flat; it means flat along that one axis. Reading the pink curve's horizontal tangent versus the blue curve's steep rise makes this concrete — Cell E's whole point.

Verify: $f_x(0,3)=0$ and $f_y(0,3)=6$. The point $(0,3,9)$ lies at the low point of the $y=3$ slice ($z=x^2+9$, minimized at $x=0$), confirming a horizontal tangent there ✓.

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Example 6 — Cell F: degenerate / limiting input

Steps.

1. $f_x=\frac{1}{y}$ (freeze $y$, so $\frac{d}{dx}(x\cdot y^{-1})=y^{-1}$), valid for $y\ne 0$.

2. $f_y=x\cdot (-y^{-2})=-\frac{x}{y^2}$ (freeze $x$, power rule on $y^{-1}$), also valid for $y\ne 0$.

3. Limiting behaviour of $f_x$ at $x=1$. In the figure the blue curve is $f_x=\frac{1}{y}$: as $y\to 0^{+}$ (approach from the right of the dashed yellow line $y=0$) the blue curve shoots up to $+\infty$; as $y\to 0^{-}$ (from the left) it plunges to $-\infty$. So $f_x$ flips sign depending on the side you come from. Why this step? $\frac{1}{y}$ has the same sign as $y$ itself, and its magnitude $\left|\frac{1}{y}\right|$ grows without bound as $|y|\to 0$. Sign of $y$ therefore decides which infinity — that is exactly why the blue curve lives above the axis on the right and below it on the left.

4. Limiting behaviour of $f_y$ at $x=1$. In the figure the pink curve is $f_y=-\frac{1}{y^2}$: as $y\to 0$ from either side, the pink curve dives to $-\infty$ — both branches point downward. Why this step? $y^2>0$ for every $y\ne 0$, so $\frac{1}{y^2}$ is always positive and blows up as $|y|\to 0$; the leading minus sign then forces the whole thing to $-\infty$ regardless of the approach direction. Unlike $f_x$, there is no sign-flip because squaring erased the sign of $y$.

Verify: At $(1,0.1)$: $f_x=10$, $f_y=-\frac{1}{0.01}=-100$. At $(1,-0.1)$: $f_x=-10$ (sign flipped ✓), $f_y=-100$ (still negative ✓, matches "both sides go to $-\infty$").

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Example 7 — Cell G: word problem with units

Steps.

1. $T_x=-4x$, $T_y=-2y$. Why this step? The constant $100$ vanishes; $-2x^2\to -4x$; $-y^2\to -2y$; each other variable frozen.

2. At $(3,4)$: $T_x(3,4)=-12$, $T_y(3,4)=-8$.

3. Units. $T$ is in °C, $x$ in m, so $T_x$ is in $\frac{\text{°C}}{\text{m}}$. Thus:

   • $-12\text{°C/m}$: moving $+x$, temperature drops $12$°C per metre.
   • $-8\text{°C/m}$: moving $+y$, it drops $8$°C per metre. Why this step? A word problem is only answered when the number has the right unit and a plain-English meaning.

4. Answer the actual question: the $x$-direction cools faster ($12>8$).

Verify: $T_x(3,4)=-4(3)=-12$, $T_y(3,4)=-2(4)=-8$. Both negative (cooling) ✓, magnitude in $x$ larger ✓. Cross-check with a tiny step: $T(3,4)=100-18-16=66$; $T(3.01,4)=100-2(9.0601)-16=100-18.1202-16=65.8798$, change $-0.1202$ over $0.01$ m $\approx -12.02$ °C/m ✓.

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Example 8 — Cell H: exam twist, mixed AND pure higher-order + Clairaut

Steps.

1. First $f_x$ (freeze $y$): $f_x=3x^2{y}^2+\cos (xy)\cdot y=3x^2{y}^2+y\cos (xy)$. Why? Power rule on $x^3$; chain on $\sin (xy)$ with inner $x$-derivative $y$.

2. Pure $f_{xx}={\partial }_x(f_x)$: differentiate $3x^2{y}^2+y\cos (xy)$ in $x$ again.

   • $3x^2{y}^2\to 6x{y}^2$.
   • $y\cos (xy)$: $y$ is a constant coefficient here, chain on $\cos (xy)$ gives $-\sin (xy)\cdot y$, so $y\cdot (-y\sin (xy))=-y^2\sin (xy)$. $f_{xx}=6x{y}^2-y^2\sin (xy).$ Why this step? Cell H asks for pure higher-order too — differentiating $f_x$ once more in the same variable $x$.

3. $f_{xy}={\partial }_y(f_x)$: differentiate $3x^2{y}^2+y\cos (xy)$ in $y$.

   • $3x^2{y}^2\to 6x^{2}y$.
   • $y\cos (xy)$: product rule in $y$ — $(y{)}^{\prime }\cos (xy)+y(\cos (xy){)}^{\prime }=\cos (xy)+y(-\sin (xy)\cdot x)$. $f_{xy}=6x^{2}y+\cos (xy)-xy\sin (xy).$

4. Now the other route — first $f_y$ (freeze $x$): $f_y=2x^{3}y+\cos (xy)\cdot x=2x^{3}y+x\cos (xy)$.

5. Pure $f_{yy}={\partial }_y(f_y)$: differentiate $2x^{3}y+x\cos (xy)$ in $y$ again.

   • $2x^{3}y\to 2x^3$.
   • $x\cos (xy)$: $x$ constant coefficient, chain on $\cos (xy)$ gives $-\sin (xy)\cdot x$, so $x\cdot (-x\sin (xy))=-x^2\sin (xy)$. $f_{yy}=2x^3-x^2\sin (xy).$ Why this step? The $y$-side pure second partial, completing the "higher-order" promise of the heading.

6. $f_{yx}={\partial }_x(f_y)$:

   • $2x^{3}y\to 6x^{2}y$.
   • $x\cos (xy)$: product rule in $x$ — $\cos (xy)+x(-\sin (xy)\cdot y)=\cos (xy)-xy\sin (xy)$. $f_{yx}=6x^{2}y+\cos (xy)-xy\sin (xy).$ Why this step? Doing the mixed partial both ways is the exam's Clairaut check — you see the symmetry, not just quote it.

7. Compare mixed: $f_{xy}$ (Step 3) and $f_{yx}$ (Step 6) are identical. ✓ Clairaut holds. The pure partials $f_{xx}\ne f_{yy}$ in general — and they are not supposed to match; only the mixed pair must.

Verify (numeric): At $(1,\pi )$: $\cos (xy)=\cos \pi =-1$, $\sin \pi =0$. Mixed: $f_{xy}=6(1)(\pi )+(-1)-0=6\pi -1$, $f_{yx}=6\pi -1$ (equal ✓, $\approx 17.85$). Pure: $f_{xx}=6{\pi }^2-0=6{\pi }^2\approx 59.22$, $f_{yy}=2-0=2$.

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Active Recall

Recall Which cell did each example cover?

Ex1 → A (polynomial+point) · Ex2 → B (product+chain) · Ex3 → C (quotient, sign, domain $y\ne 0$) · Ex4 → D (log/exp signs) · Ex5 → E (zero partial) · Ex6 → F (degenerate limit) · Ex7 → G (units) · Ex8 → H (mixed + pure higher-order, Clairaut).

Recall Zero slope in one direction — flat surface?

No. A zero partial means flat along that one axis only; the surface can still rise steeply in the other direction (Ex 5: $f_x(0,3)=0$ but $f_y(0,3)=6$).

Recall When can a partial "blow up", and does the direction of approach matter?

Near a point the formula isn't defined — a zero denominator (Ex 6: $y\to 0$ in $x/y$). Whether the sign of approach matters depends on the power of the offending variable: $\frac{1}{y}$ flips sign ($\pm \infty$), but $-\frac{1}{y^2}$ does not ($-\infty$ from both sides) because squaring erases the sign.

Recall Do all second partials have to be equal?

No — only the mixed pair $f_{xy}=f_{yx}$ (when continuous, by Clairaut). The pure ones $f_{xx}$ and $f_{yy}$ measure different things and generally differ (Ex 8: $f_{xx}=6x{y}^2-y^2\sin (xy)$ vs $f_{yy}=2x^3-x^2\sin (xy)$).

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Connections

• Parent (Hinglish): Partial derivatives — the rule these examples exercise.
• Gradient vector — bundles the partials from Ex 5 & 7 into $\nabla f$.
• Directional derivative — combine $f_x,f_y$ to move in any direction.
• Tangent plane and linear approximation — Ex 5's two slopes span it.
• Chain rule (multivariable) — the engine inside Ex 2, 4, 8.
• Total differential — $df=f_{x}dx+f_{y}dy$, used in the units of Ex 7.
• Single-variable derivative — every "freeze" step reduces to this.
• Clairaut's theorem — proven by hand in Ex 8.

Concept Map