Multivariable-Calculus · Unit 7 · Video 4 · Interactive Practice

Partial Derivatives: Change One Variable, Hold the Other Fixed

IKey Formulas

Formula	Name	What stays fixed
$f_x(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x,\, y_0) - f(x_0, y_0)}{\Delta x}$	Partial derivative in $x$	$y = y_0$ in both terms
$f_y(x_0, y_0) = \lim_{\Delta y \to 0} \frac{f(x_0,\, y_0 + \Delta y) - f(x_0, y_0)}{\Delta y}$	Partial derivative in $y$	$x = x_0$ in both terms
$\frac{\partial f}{\partial x} = f_x, \quad \frac{\partial f}{\partial y} = f_y$	Two notations, same meaning	—
$f = x^3 y + y^2 \;\Rightarrow\; f_x = 3x^2 y,\ \ f_y = x^3 + 2y$	Video example	the other variable, treated as a constant

Key Insight: $f_x(x_0, y_0)$ is the slope of the slice $y = y_0$ of the graph, and $f_y(x_0, y_0)$ is the slope of the slice $x = x_0$ : freeze one variable, let the other vary.

IIVisualization 1 — Two Slices, Two Slopes

On the dome $z = 4 - x^2 - y^2$ , the partials at $(x_0, y_0)$ are the slopes of two vertical slices.

x_0

0.5

y_0

1.0

IIIVisualization 2 — The Limit with One Input Frozen

In the difference quotient only one input moves; as the step shrinks, the secant on the slice becomes its tangent.

Variable that moves Step size 1.000

IVVisualization 3 — Freeze, Then Differentiate

With the other variable boxed as a constant, each term follows the ordinary one-variable rules.

Function Differentiate in Step 1

VQuiz Questions

Problem 1 · Compute $f_x$

Given: $f(x, y) = x^2 y + 3y^2$ — find $\frac{\partial f}{\partial x}$ .

✅ Correct! With

y

held constant,

x^2 y

differentiates to

2xy

and

3y^2

is a constant, so it contributes

0

❌ That is $f_y$ . You differentiated in

y

with

x

held fixed; here

x

is the variable and

y

is the constant.

❌ Not quite. The term

3y^2

contains no

x

, so with respect to

x

it is a constant and its derivative is

0

❌ Not quite. The constant

y

multiplies

x^2

, so it stays as a factor:

\frac{\partial}{\partial x}(x^2 y) = 2x \cdot y

Show solution

Treat $y$ as a constant and differentiate each term in $x$ :

\frac{\partial}{\partial x}(x^2 y) = 2x \cdot y = 2xy, \qquad \frac{\partial}{\partial x}(3y^2) = 0.

So $f_x = 2xy$ .

Problem 2 · The Coefficient That Stays

Given: $f(x, y) = x^4 y^2 - 5x + y^3$ — find $\frac{\partial f}{\partial y}$ .

✅ Correct! The constant

x^4

stays as the coefficient of

\frac{d}{dy}(y^2) = 2y

; only

-5x

, which has no

y

, differentiates to

0

❌ The $x^4$ term did not vanish.

x^4 y^2

contains

y

; the constant

x^4

multiplies

y^2

, so it stays:

\frac{\partial}{\partial y}(x^4 y^2) = x^4 \cdot 2y

. Only a term with no

y

at all gives

0

❌ Not quite.

-5x

contains no

y

, so with respect to

y

it is a constant: its derivative is

0

, not

-5

❌ Not quite.

8x^3 y

differentiates

x^4

as well as

y^2

. In

\frac{\partial}{\partial y}

only

y

varies;

x^4

is a constant factor.

Show solution

Hold $x$ constant and take each term's ordinary derivative in $y$ :

\frac{\partial}{\partial y}(x^4 y^2) = x^4 \cdot 2y = 2x^4 y, \qquad \frac{\partial}{\partial y}(-5x) = 0, \qquad \frac{\partial}{\partial y}(y^3) = 3y^2.

f_y = 2x^4 y + 3y^2

Problem 3 · Slope of a Slice

Given: the graph $z = 6 - x^2 - 2y^2$ and the point $(2, -1, 0)$ on it. The vertical plane $x = 2$ cuts the graph in a curve — find the slope of that curve's tangent line at the point.

Which partial derivative is this slope?

What is its value?

✅ Correct! Along

x = 2

the slice is

z = 2 - 2y^2

, whose slope at

y = -1

-4y = 4

❌ Check which variable moves. On the plane

x = 2

the value of

x

is frozen; the slice is traced out by changing

y

, so its slope is a rate of change with respect to

y

❌ That is the height.

f(2, -1) = 0

is where the point sits, not how steeply the slice climbs through it.

❌ That is the other slice.

f_x(2, -1) = -2x = -4

is the slope of the slice

y = -1

. The plane

x = 2

holds

x

fixed, so you need

f_y

❌ Check the coefficient.

\frac{\partial}{\partial y}(-2y^2) = -4y

, not

-2y

❌ Not quite. Differentiate

6 - x^2 - 2y^2

y

with

x

fixed, then substitute

y = -1

Show solution

The plane $x = 2$ freezes $x$ , so the slice is the one-variable curve

z = 6 - 2^2 - 2y^2 = 2 - 2y^2.

Its slope is the partial derivative in $y$ :

f_y = \frac{\partial}{\partial y}(6 - x^2 - 2y^2) = -4y, \qquad f_y(2, -1) = -4(-1) = 4.

For comparison, $f_x(2, -1) = -2(2) = -4$ is the slope of the other slice, $y = -1$ .

Problem 4 · From the Definition to a Number

Given: $f(x, y) = x y^3 - 2x^2$ — identify the limit that defines $f_y(3, 1)$ , then compute it.

Which limit is $f_y(3, 1)$ ?

What is $f_y(3, 1)$ ?

✅ Correct! Only the second input moves, and

f_y = 3xy^2

gives

3 \cdot 3 \cdot 1 = 9

❌ Both inputs move here. That quotient changes

x

and

y

together, which measures change along the diagonal, not a partial derivative. For

f_y

the first input must stay

3

in both terms.

❌ That limit is $f_x(3, 1)$ . Renaming the step does not change which input it is added to; here the step is added to

x

while

y

stays at

1

❌ Not quite. In the definition of

f_y

, the first input stays at

x_0 = 3

in both terms.

❌ That is $f_x(3, 1)$ .

f_x = y^3 - 4x

gives

1 - 12 = -11

; you need the rate of change in

y

❌ Not quite.

-2x^2

contains no

y

, so its

y

-derivative is

0

; it should not contribute

-4x = -12

❌ Check the power rule.

\frac{\partial}{\partial y}(x y^3) = x \cdot 3y^2

; the factor

3

comes down from the exponent.

❌ Not quite. Hold

x

constant, differentiate in

y

, then substitute

(3, 1)

Show solution

The limit. $f_y(x_0, y_0)$ holds the first input at $x_0$ and moves only the second:

f_y(3, 1) = \lim_{\Delta y \to 0} \frac{f(3,\, 1 + \Delta y) - f(3, 1)}{\Delta y}.

The value. With $x$ held constant,

\frac{\partial}{\partial y}(x y^3) = 3xy^2, \qquad \frac{\partial}{\partial y}(-2x^2) = 0, \qquad f_y = 3xy^2.

f_y(3, 1) = 3 \cdot 3 \cdot 1^2 = 9.

Check from the limit: $f(3, 1 + \Delta y) - f(3, 1) = 3(1 + \Delta y)^3 - 3 = 9\Delta y + 9\Delta y^2 + 3\Delta y^3$ , so the quotient is $9 + 9\Delta y + 3\Delta y^2 \to 9$ .

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