Visual walkthrough — Partial derivatives

Everything here uses one running function: $f(x,y)=x^{2}y+3y$ We will compute $\frac{\partial f}{\partial x}$ — the east–west slope — and get $2xy$, exactly as the parent note claims, but this time picture by picture.

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Step 1 — The output is a HEIGHT above a floor

WHAT. Two inputs $x$ and $y$ live on a flat floor (think east and north). The function $f$ gives one number: how high the surface floats above the point $(x,y)$. We call that height $z=f(x,y)$.

WHY. Before any calculus, we must know what "the answer changes" even means. Here it means the height goes up or down. A partial derivative will end up being a slope, and slope only makes sense once we have a height to be sloped.

PICTURE. The floor is the flat plane; every point on the floor sends a vertical stick up to the curved sheet above. The sheet is our surface $z=x^{2}y+3y$.

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Step 2 — Freeze $y$: slice the hill with a wall

WHAT. Pick one fixed north value, say $y=c$. Slide a vertical wall along that line $y=c$. Where the wall cuts the surface, we get a single curved line — a 1-D curve, no longer a 2-D sheet.

WHY. A partial derivative asks "nudge only $x$." If $y$ is not allowed to move, then $y$ is stuck at $c$ forever. Geometrically, "$y$ stuck at $c$" is the wall. Standing on that wall, we have gone from a hard 2-input problem to an easy 1-input curve — and a 1-input curve has an ordinary slope we already know how to find (see 1.2.02-Derivatives-single-variable).

PICTURE. The magenta wall at $y=c$ slices the sheet. The bright curve on the wall is our sliced height, which now depends on $x$ alone.

On this slice, substitute $y=c$ into $f$: $g(x)=\underset{\text{moves with }x}{\underbrace{x^2}}\underset{\text{frozen number}}{\underbrace{c}}+\underset{\text{frozen, no }x}{\underbrace{3c}}$

• $x^{2}c$ ::: the only piece that changes as we walk east — $c$ is now just a number multiplier.
• $3c$ ::: a flat constant height — it has no $x$, so it lifts the whole curve up but never tilts it.

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Step 3 — Take one tiny step $h$ to the east

WHAT. From position $x$, step a small amount $h$ to the right, landing at $x+h$. Read the new height $f(x+h,y)$. Keep $y$ where it was.

WHY. Slope means "rise over run." To measure rise we need two heights: one here, one a little over. The little run is $h$. We keep $y$ identical in both readings — that is the whole trick of "partial."

PICTURE. Two vertical sticks on the same wall: one at $x$ (height $f(x,y)$), one at $x+h$ (height $f(x+h,y)$). The gap between the stick-tops is the rise; the floor gap is the run $h$.

$\text{average slope}=\frac{\stackrel{\text{height after step}}{\overbrace{f(x+h,y)}}-\stackrel{\text{height before}}{\overbrace{f(x,y)}}}{\underset{\text{how far we stepped}}{\underbrace{h}}}$

• $f(x+h,y)-f(x,y)$ ::: the rise — how much taller (or shorter) we became.
• $h$ ::: the run — the size of the step. Notice the same $y$ sits inside both height terms.

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Step 4 — Plug in the actual function and watch pieces cancel

WHAT. Substitute $f(x,y)=x^{2}y+3y$ into that rise-over-run fraction.

WHY. Now it is pure algebra. Any term that appears identically in both height readings will subtract to zero — because it didn't change when we stepped. Watch which survives.

$\frac{[(x+h{)}^{2}y+3y]-[x^{2}y+3y]}{h}$

Expand $(x+h{)}^2=x^2+2xh+h^2$:

$\frac{(x^2+2xh+h^2)y+3y-x^{2}y-3y}{h}$

Colour-track each survivor and each casualty:

• $x^{2}y$ appears twice with opposite sign → cancels. (It was the same at both sticks.)
• $3y$ appears twice with opposite sign → cancels. (No $x$ inside — flat — never tilted.)
• $2xhy$ and $h^{2}y$ survive — they are the only pieces carrying an $h$.

PICTURE. The two "before/after" bars split into stacked blocks; the identical blocks ($x^{2}y$, $3y$) grey out and disappear, leaving only the $h$-blocks that made the height change.

$=\frac{2xhy+h^{2}y}{h}$

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Step 5 — Cancel the run, then shrink the step to zero

WHAT. Every surviving term has a factor of $h$. Divide it out, then let $h$ shrink toward $0$.

WHY. Dividing by $h$ turns "rise over run" into "slope so far." Letting $h\to 0$ answers the real question: at the exact point (not over a fat step), how steep is it? This shrinking is the limit — the tool we borrow precisely because it converts an average slope over a step into an instant slope at a point. No smaller tool can do that: algebra alone can't reach "infinitely close."

$\frac{2xhy+h^{2}y}{h}=2xy+hy\underset{h\to 0}{\to }2xy$

• $2xy$ ::: survives the shrink — it has no leftover $h$, so it is the true instantaneous slope.
• $hy$ ::: has a lonely $h$, so as $h\to 0$ it vanishes — the fat-step error melts away.

PICTURE. The two sticks slide together. The slanted "secant" line (through both stick-tops) rotates until it just kisses the curve — the tangent. Its steepness settles on $2xy$.

$\boxed{\frac{\partial f}{\partial x}=2xy}$

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Step 6 — Now freeze the OTHER variable (the symmetric case)

WHAT. Repeat everything, but this time freeze $x$ and nudge $y$. The wall now runs the other way, along $x=$const.

WHY. A partial derivative comes in one flavour per input. We must show both so no reader hits a case we skipped. The mechanics are identical — just swap which wall we stand on.

Freeze $x$, differentiate $f=x^{2}y+3y$ in $y$ (here $x^2$ and $3$ are the frozen numbers): $\frac{\partial f}{\partial y}=\underset{\text{const}\times \frac{\partial y}{\partial y}}{\underbrace{x^2}}+\underset{\frac{\partial (3y)}{\partial y}}{\underbrace{3}}=x^2+3$

• $x^2$ ::: the $3y$-term's partner in $x^{2}y$; with $x$ frozen it's a constant slope-per-unit-$y$.
• $3$ ::: now $3y$ does tilt (it has a $y$!), giving slope $3$ — the mirror of Step 4, where it was flat.

PICTURE. Two walls on the same hill: the magenta wall (Step 2) gives $\partial f/\partial x$; the orange wall (this step) gives $\partial f/\partial y$. Different slices, different slopes.

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Step 7 — Degenerate & edge cases (so nothing surprises you)

WHAT & WHY. We check the corners where formulas often break.

• Flat direction, $y=0$. At $y=0$: $\partial f/\partial x=2x\cdot 0=0$ for every $x$. The slice at $y=0$ is $g(x)=0$ — a dead-flat line, zero slope everywhere. Consistent.
• A term with no chosen variable. Any constant-in-$x$ term (like $3y$, or a bare $7$) always dies in $\partial /\partial x$: two equal heights subtract to $0$ before we even divide. This is why the parent says "treat $y$ as ruthlessly as the number $7$."
• The single-input limit. If $f$ secretly depends on only one variable, its partial is the ordinary derivative — the wall slices nothing away. Partial derivatives contain single-variable derivatives as a special case, never contradict them.
• Stacking both slopes. Put the two answers side by side and you have built the gradient: $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)=(2xy,x^2+3)$ This vector is what steepest ascent and gradient descent actually consume; Example 3 in the parent is exactly the linear-regression version.

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The one-picture summary

One hill, two walls, two tangent lines, one gradient arrow — the entire walkthrough in a single frame: freeze one input → slice → measure rise over a step → shrink the step → read the tangent slope → repeat for each input → stack into $\nabla f$.

Recall Feynman: tell it to a 12-year-old

You're standing on a curvy hill. Your height depends on how far east you are ($x$) and how far north you are ($y$). First, promise not to move north — that's like standing against an invisible wall. Now your hill has become a single curvy path running east. Take one baby step east and see how much your head went up; divide that height-gain by how big the step was — that's your steepness "so far." Take smaller and smaller steps and the number settles down to one exact value: that's the east-slope, $\partial f/\partial x=2xy$. Notice a funny thing: the part of your height that had nothing to do with east (the $3y$ part) never changed as you stepped, so it added zero to the slope — it just quietly disappeared. Now turn to face north, put up the other wall, and do it all again to get the north-slope, $\partial f/\partial y=x^2+3$. Two slopes together tell you exactly which way is downhill fastest — and "downhill fastest" is precisely how a computer learns.

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Connections

• Partial derivatives — the parent this walkthrough visualises.
• 1.2.02-Derivatives-single-variable — each slice reduces to exactly this.
• 1.2.04-Gradient-and-directional-derivatives — the two slopes stacked into $\nabla f$.
• 1.2.05-Chain-rule-multivariable — the tool behind $e^{xy}$-style inner factors.
• 1.3.01-Gradient-descent — consumes these partials to update weights.
• Linear-regression — the payoff example.