Multivariable-Calculus · Unit 10 · Video 1 · Interactive Practice

The Total Differential: Why df Is Not Δf, and Its Three Uses

IKey Formulas

Formula	Name	What it does
$dy = f'(x)\,dx$	Single-variable differential	Implicit differentiation: $x = \sin y$ gives $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$
$df = f_x\,dx + f_y\,dy + f_z\,dz$	Total differential	Exact relation between differentials: how variations of $x$ , $y$ , $z$ change $f$
$\Delta f \approx f_x\,\Delta x + f_y\,\Delta y + f_z\,\Delta z$	Linear (tangent plane) approximation	Replace each $d$ by a small number $\Delta$
$\frac{df}{dt} = f_x\,\frac{dx}{dt} + f_y\,\frac{dy}{dt} + f_z\,\frac{dz}{dt}$	Rate of change	Divide by $dt$ when $x$ , $y$ , $z$ all depend on $t$

Key Insight: $df \neq \Delta f$ . The differential $df$ has no numerical value and is expressed only through other differentials; $\Delta f$ is an actual number. A $d$ and a $\Delta$ never appear in the same formula.

IIVisualization 1 — Differentiating sin⁻¹ with Differentials

From $x = \sin y$ , the differentials obey $dx = \cos y\,dy$ , so the slope of $y = \sin^{-1}x$ is $1/\cos y$ .

1 $\;y = \sin^{-1}x \;\Rightarrow\; x = \sin y$

2 $\;dx = \cos y\,dy \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{1}{\cos y}$

3 $\;\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ , the positive root because $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ makes $\cos y \ge 0$

4 $\;\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}, \qquad \dfrac{dx}{dy} = \cos y = \sqrt{1 - x^2}$

IIIVisualization 2 — Replacing dx, dy by Δx, Δy

For the area $f = xy$ , the exact relation $df = y\,dx + x\,dy$ becomes a number only after $dx \to \Delta x$ , $dy \to \Delta y$ .

IVVisualization 3 — Dividing by dt

Along the path $x = 2 + \cos t$ , $y = 2 + \sin t$ , dividing $df = y\,dx + x\,dy$ by $dt$ gives the rate of change of $f = xy$ .

t 0.40

VQuiz Questions

Problem 1 · Write the Total Differential

Given: $f(x, y, z) = x^2 y + y z^3$ — find the total differential $df$ .

✅ Correct! Each term is one variable's contribution: its partial derivative times its differential.

❌ Close, but…

y

appears in both terms of

f

, so

f_y = x^2 + z^3

, not just

x^2

❌ Not quite. That is the sum of the partial derivatives; each one must multiply its own differential

dx

dy

dz

❌ Not quite. This sets the differential

df

equal to a number built from

\Delta

's. Never mix a

d

with a

\Delta

in the same formula.

❌ Not quite. Compute

f_x

f_y

f_z

and attach

dx

dy

dz

Show solution

The partial derivatives of $f = x^2 y + y z^3$ are

f_x = 2xy, \qquad f_y = x^2 + z^3, \qquad f_z = 3yz^2.

The total differential multiplies each by its own differential and adds:

df = 2xy\,dx + (x^2 + z^3)\,dy + 3yz^2\,dz.

It is an exact equality between differentials; no $\Delta$ appears.

Problem 2 · Which Formula Is Correct?

Given: a function $f(x, y)$ and small numbers $\Delta x$ , $\Delta y$ — which formula is correctly written?

✅ Correct! Both sides are numbers, and the relation is an approximation: the tangent plane approximation.

❌ Not quite. The left side is a differential and the right side is a number. A

d

and a

\Delta

never appear in the same formula.

❌ Not quite. The left side is a number and the right side is a differential, which has no numerical value. Never mix a

d

with a

\Delta

❌ Close, but… Every symbol is a number now, but the equality is not exact. For

f = xy

the true change is

y\,\Delta x + x\,\Delta y + \Delta x\,\Delta y

, so the formula misses

\Delta x\,\Delta y

❌ Not quite. Check whether both sides are the same kind of object, and whether the relation is exact.

Show solution

Two rules decide it. First, differentials and numbers never share a formula, which rules out (a) and (b). Second, the exact statement is the one between differentials, $df = f_x\,dx + f_y\,dy$ ; replacing $dx$ , $dy$ by numbers $\Delta x$ , $\Delta y$ produces a number that only approximates the actual change $\Delta f$ . That rules out the equals sign in (c).

For example, with $f = xy$ :

\Delta f = (x + \Delta x)(y + \Delta y) - xy = y\,\Delta x + x\,\Delta y + \Delta x\,\Delta y,

which differs from $f_x\,\Delta x + f_y\,\Delta y = y\,\Delta x + x\,\Delta y$ by $\Delta x\,\Delta y$ . So the correct formula is $\Delta f \approx f_x\,\Delta x + f_y\,\Delta y$ .

Problem 3 · Estimate a Change in Three Variables

Given: $f(x, y, z) = x y^2 z$ at $(1, 2, 3)$ , where $f = 12$ . The variables change by $\Delta x = 0.1$ , $\Delta y = -0.05$ , $\Delta z = 0.2$ — estimate $\Delta f$ with the linear approximation.

✅ Correct!

12(0.1) + 12(-0.05) + 4(0.2) = 1.4

; the exact change is

1.3848

❌ Close, but…

\Delta y

is negative, so the

y

-term is

12(-0.05) = -0.6

, not

+0.6

❌ Not quite.

28 = f_x + f_y + f_z

: each partial derivative must be multiplied by its own

\Delta

❌ Not quite.

13.4

is the estimated new value

f + \Delta f

, not the change

\Delta f

❌ Not quite. Evaluate

f_x

f_y

f_z

(1, 2, 3)

first.

Show solution

Step 1: partial derivatives at $(1, 2, 3)$ .

f_x = y^2 z = 12, \qquad f_y = 2xyz = 12, \qquad f_z = x y^2 = 4.

Step 2: replace $dx$ , $dy$ , $dz$ by the increments.

\Delta f \approx 12(0.1) + 12(-0.05) + 4(0.2) = 1.2 - 0.6 + 0.8 = 1.4.

Check: $f(1.1, 1.95, 3.2) = 1.1 \cdot 1.95^2 \cdot 3.2 = 13.3848$ , so the actual change is $\Delta f = 1.3848$ . The estimate is close but not exact, as an approximation should be.

Problem 4 · Inverse Tangent by Differentials

Given: $y = \tan^{-1} x$ , so $x = \tan y$ with $-\frac{\pi}{2} \lt y \lt \frac{\pi}{2}$ — find $\frac{dy}{dx}$ by taking differentials.

✅ Correct!

dx = \sec^2 y\,dy

and

\sec^2 y = 1 + \tan^2 y = 1 + x^2

❌ Close, but…

1 + x^2 = \sec^2 y

is the reciprocal rate

\frac{dx}{dy}

. Divide

dx = \sec^2 y\,dy

dx

and by

\sec^2 y

❌ Not quite. That is the derivative of

\sin^{-1}x

. Here

x = \tan y

, so the differential of the right side is

\sec^2 y\,dy

❌ Not quite. The identity is

\sec^2 y = 1 + \tan^2 y

with no square root: it is

\sec^2 y

, not

\sec y

, that appears in

dx

❌ Not quite. Take differentials of

x = \tan y

first.

Show solution

Take differentials of both sides of $x = \tan y$ :

dx = \sec^2 y\,dy \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{\sec^2 y}.

To write this in terms of $x$ , use $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ (no sign choice is needed, because it is a square):

\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}.

The reciprocal rate is $\frac{dx}{dy} = \sec^2 y = 1 + x^2$ .

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