Multivariable-Calculus · Unit 13 · Video 1 · Interactive Practice

Implicit Partial Derivatives: Differentiate the Constraint

IKey Formulas

Formula	Name	For $x^2 + yz + z^3 = 8$ at $(2, 3, 1)$
$dg = g_x\,dx + g_y\,dy + g_z\,dz = 0$	Differentiate the constraint $g(x, y, z) = c$	$4\,dx + dy + 6\,dz = 0$
$\dfrac{\partial z}{\partial x} = -\dfrac{g_x}{g_z}$ , $\ \dfrac{\partial z}{\partial y} = -\dfrac{g_y}{g_z}$ $\ (g_z \neq 0)$	Rates of $z = z(x, y)$ , read off as the coefficients of $dz$	$\dfrac{\partial z}{\partial x} = -\dfrac{2}{3}$ , $\ \dfrac{\partial z}{\partial y} = -\dfrac{1}{6}$
$\left(\dfrac{\partial y}{\partial x}\right)_{z} = -\dfrac{g_x}{g_y}$	Any other rate: set the held variable's differential to $0$	$dz = 0 \Rightarrow dy = -4\,dx$ , so $-4$

Key Insight: Partial derivatives do not cancel like fractions. $\dfrac{g_x}{g_z} = \dfrac{\partial g/\partial x}{\partial g/\partial z}$ looks like $+\dfrac{\partial z}{\partial x}$ , but the true rate carries a minus sign: when $x$ pushes $g$ up, $z$ must move to push it back down, because $g$ is held constant.

IIVisualization 1 — Rates Without Solving the Cubic

Near $(2, 3, 1)$ , the linear relation $4\,dx + dy + 6\,dz = 0$ predicts $z$ without solving $z^3 + yz + x^2 = 8$ .

IIIVisualization 2 — One Equation, Three Held Variables

Holding one variable fixed strikes its term from $4\,dx + dy + 6\,dz = 0$ ; the rest is the slope of a slice curve.

Held constant

IVVisualization 3 — Why the Minus Sign

With $y$ fixed, $x^2$ grows by about $4\,dx$ ; since $g_z = 6 > 0$ , $z$ must fall to keep $g = 8$ .

dx 0.30 z responds with

VQuiz Questions

Problem 1 · Read Off ∂z/∂x

Given: $xy + yz + zx = 11$ , which passes through $(1, 2, 3)$ . Treating $z$ as a function of $x$ and $y$ , find $\dfrac{\partial z}{\partial x}$ at that point.

✅ Correct!

dg = 5\,dx + 4\,dy + 3\,dz = 0

; with

dy = 0

dz = -\tfrac{5}{3}\,dx

❌ Check the sign. That is

g_x/g_z

with the partials "cancelled". Moving

5\,dx

across the equation

5\,dx + 3\,dz = 0

makes it negative.

❌ Upside down.

-\tfrac{3}{5}

\partial x/\partial z

with

y

fixed. You divide by

g_z

, the coefficient of the differential you isolate.

❌ Wrong numerator.

g_y = x + z = 4

belongs to

\partial z/\partial y

. For

\partial z/\partial x

use

g_x = y + z = 5

❌ Not quite. Compute

g_x

and

g_z

(1, 2, 3)

, then use

\partial z/\partial x = -g_x/g_z

Show solution

Check the point: $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 1 = 2 + 6 + 3 = 11$ . With $g = xy + yz + zx$ :

g_x = y + z = 5, \qquad g_y = x + z = 4, \qquad g_z = y + x = 3.

So $dg = 5\,dx + 4\,dy + 3\,dz = 0$ . Holding $y$ constant sets $dy = 0$ :

3\,dz = -5\,dx \quad\Rightarrow\quad \frac{\partial z}{\partial x} = -\frac{g_x}{g_z} = -\frac{5}{3}.

Problem 2 · Which Variable Is Held?

Given: near $(2, 3, 1)$ the constraint $x^2 + yz + z^3 = 8$ gives $4\,dx + dy + 6\,dz = 0$ . Find the rate of change of $y$ with respect to $z$ when $x$ is held constant, $\left(\dfrac{\partial y}{\partial z}\right)_{x}$ .

✅ Correct!

dx = 0

leaves

dy + 6\,dz = 0

, so

dy = -6\,dz

❌ That is the reciprocal rate.

-\tfrac{1}{6}

\partial z/\partial y

(z depends on y). Here

y

is the dependent variable, so isolate

dy

❌ Check the sign. Moving

6\,dz

to the other side of

dy + 6\,dz = 0

makes it

-6\,dz

❌ Wrong variable held.

-\tfrac{6}{4} = -\tfrac{3}{2}

\partial x/\partial z

with

y

fixed. With

x

fixed, the

4\,dx

term is the one that drops.

❌ Not quite. Set

dx = 0

and solve the remaining equation for

dy

Show solution

Holding $x$ constant means $dx = 0$ :

4 \cdot 0 + dy + 6\,dz = 0 \quad\Rightarrow\quad dy = -6\,dz.

So $\left(\dfrac{\partial y}{\partial z}\right)_{x} = -\dfrac{g_z}{g_y} = -\dfrac{6}{1} = -6$ . It is the reciprocal of $\partial z/\partial y = -\tfrac{1}{6}$ , since both describe the same slice $x = 2$ .

Problem 3 · Rate, Then Estimate

Given: $xz^2 + y^2 z + xy = 7$ , which passes through $(1, 2, 1)$ , with $z = z(x, y)$ . Find $\dfrac{\partial z}{\partial y}$ there, then estimate $z$ at $(x, y) = (1.1, 1.9)$ .

What is ∂z/∂y at (1, 2, 1)?

Estimate of z at (1.1, 1.9)

✅ Correct!

dz = -\tfrac{1}{2}(0.1) - \tfrac{5}{6}(-0.1) = \tfrac{1}{30}

, so

z \approx 1.033

(solving the quadratic in

z

exactly gives

1.034

❌ Check the sign.

g_y/g_z

is not

\partial z/\partial y

: moving

5\,dy

across

5\,dy + 6\,dz = 0

gives

dz = -\tfrac{5}{6}\,dy

❌ Upside down. Divide by

g_z = 6

, the coefficient of

dz

, the differential you isolate.

❌ That is ∂z/∂x.

-g_x/g_z = -\tfrac{3}{6}

. For

\partial z/\partial y

use

g_y = 2yz + x = 5

❌ Check the rate. Find

g_y

and

g_z

(1, 2, 1)

, then use

-g_y/g_z

❌ Sign slip.

3\,dx + 5\,dy = 0.3 - 0.5 = -0.2

, and

dz = -(-0.2)/6

is positive:

z

goes up.

❌ Check $dy$ .

y

moves from

2

1.9

, so

dy = -0.1

, not

+0.1

❌ Missing the division.

6\,dz = 0.2

gives

dz = 0.2/6 \approx 0.033

, not

0.2

❌ Check the estimate. Use

dz = -(3\,dx + 5\,dy)/6

with

dx = 0.1

dy = -0.1

Show solution

Check the point: $1 \cdot 1^2 + 2^2 \cdot 1 + 1 \cdot 2 = 1 + 4 + 2 = 7$ . With $g = xz^2 + y^2 z + xy$ :

g_x = z^2 + y = 3, \qquad g_y = 2yz + x = 5, \qquad g_z = 2xz + y^2 = 6.

So $3\,dx + 5\,dy + 6\,dz = 0$ , and $\dfrac{\partial z}{\partial y} = -\dfrac{g_y}{g_z} = -\dfrac{5}{6}$ (also $\dfrac{\partial z}{\partial x} = -\dfrac{1}{2}$ ).

For the estimate, $dx = 0.1$ and $dy = -0.1$ :

dz = -\frac{3(0.1) + 5(-0.1)}{6} = -\frac{-0.2}{6} = \frac{1}{30} \approx 0.033, \qquad z \approx 1 + 0.033 = 1.033.

Problem 4 · Three Rates of the Gas Law

Given: the ideal gas law $PV = nRT$ with $n$ and $R$ constant. Find the product $\left(\dfrac{\partial P}{\partial V}\right)_{T} \left(\dfrac{\partial V}{\partial T}\right)_{P} \left(\dfrac{\partial T}{\partial P}\right)_{V}$ .

✅ Correct!

\left(-\tfrac{P}{V}\right)\left(\tfrac{nR}{P}\right)\left(\tfrac{V}{nR}\right) = -1

, every time.

❌ That is the cancellation trap. Each factor is taken with a different variable held fixed, so the

\partial P

\partial V

\partial T

do not cancel. The first factor alone is negative.

❌ No factor vanishes. On

PV = nRT

with

P, V, T > 0

, the three rates are

-P/V

nR/P

and

V/(nR)

, all nonzero.

❌ Only the first factor.

-P/V

\left(\partial P/\partial V\right)_T

; the other two factors are

nR/P

and

V/(nR)

❌ Not quite. Write

g = PV - nRT

, compute each rate as

-(\text{coefficient})/(\text{coefficient})

, and multiply.

Show solution

Let $g = PV - nRT = 0$ . Then $dg = V\,dP + P\,dV - nR\,dT = 0$ , so $g_P = V$ , $g_V = P$ , $g_T = -nR$ .

$T$ held ( $dT = 0$ ): $V\,dP + P\,dV = 0$ , so $\left(\dfrac{\partial P}{\partial V}\right)_{T} = -\dfrac{P}{V}$ .

$P$ held ( $dP = 0$ ): $P\,dV = nR\,dT$ , so $\left(\dfrac{\partial V}{\partial T}\right)_{P} = \dfrac{nR}{P}$ .

$V$ held ( $dV = 0$ ): $V\,dP = nR\,dT$ , so $\left(\dfrac{\partial T}{\partial P}\right)_{V} = \dfrac{V}{nR}$ .

\left(-\frac{P}{V}\right)\left(\frac{nR}{P}\right)\left(\frac{V}{nR}\right) = -1.

Treating the partials as fractions would predict $+1$ ; the minus sign from $dg = 0$ survives.

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