Single-Variable-Calculus · Unit 12 · Video 1 · Interactive Practice

Differentials: dy = f'(x) dx and linear approximation

IKey Formulas

Formula	Name	Meaning
$dy = f'(x)\,dx$	Differential of $y = f(x)$	Rise along the tangent line for a step $dx$
$\dfrac{dy}{dx} = f'(x)$	Leibniz's ratio	The definition divided through by $dx$
$\Delta y = f(x + dx) - f(x)$	True change	Rise along the curve; $\Delta y \approx dy$ for a small step
$f(x) \approx f(a) + f'(a)(x - a)$	Linear approximation	The same estimate, with $y = f(a)$ and $dx = x - a$

Key Insight: $y + dy$ only approximates the exact value $y + \Delta y$ : $dy$ climbs the tangent line, $\Delta y$ climbs the curve.

IIVisualization 1 — dy Along the Tangent, Δy Along the Curve

For $y = x^{1/3}$ , the tangent-line rise $dy$ and the true rise $\Delta y$ agree more closely as $dx$ shrinks.

Base point a 1.00 Step dx 3.00

IIIVisualization 2 — Same Method, New Notation

Estimating a cube root by differentials and by the linear-approximation formula produces the same numbers, row for row.

Base point (perfect cube) Step dx 0.1

Differentials · $dy = f'(x)\,dx$

Linear approximation · $f(x) \approx f(a) + f'(a)(x - a)$

IVQuiz Questions

Problem 1 · Write the Differential

Given: $y = \sqrt{x}$ — find the differential $dy$ .

✅ Correct!

dy = f'(x)\,dx

with

f'(x) = \tfrac{1}{2}x^{-1/2}

; dividing by

dx

gives back

\tfrac{dy}{dx} = \tfrac{1}{2\sqrt{x}}

❌ Close, but the $dx$ is missing. That is

f'(x)

, the derivative. The differential is the derivative multiplied by

dx

dy = f'(x)\,dx

❌ Not quite.

\tfrac{2}{3}x^{3/2}

is an antiderivative of

\sqrt{x}

. The differential uses the derivative of

\sqrt{x}

❌ Not quite. Power rule: the exponent drops by one,

\tfrac{1}{2} - 1 = -\tfrac{1}{2}

Show solution

Write $y = x^{1/2}$ . By the power rule, $f'(x) = \tfrac{1}{2}x^{1/2 - 1} = \tfrac{1}{2}x^{-1/2}$ . The differential is the derivative times $dx$ :

dy = f'(x)\,dx = \tfrac{1}{2}x^{-1/2}\,dx = \frac{dx}{2\sqrt{x}}.

Dividing both sides by $dx$ returns the familiar $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$ .

Problem 2 · Estimate a Square Root

Given: $y = \sqrt{x}$ — use differentials at the base point $x = 36$ to estimate $\sqrt{36.6}$ .

✅ Correct!

dy = \tfrac{1}{12}(0.6) = 0.05

, so

\sqrt{36.6} \approx 6.05

(the true value is

6.04979\ldots

❌ Not quite. You used

dy = \tfrac{1}{2}\,dx

, dropping the factor

x^{-1/2}

. At

x = 36

the derivative is

\tfrac{1}{2}\cdot 36^{-1/2} = \tfrac{1}{12}

❌ Not quite. That adds the step

dx = 0.6

itself to

y

, as if

dy = dx

. The rise is

dy = f'(36)\,dx

, and

f'(36) = \tfrac{1}{12}

, not

1

❌ Close, but check the derivative.

f'(36) = \dfrac{1}{2\sqrt{36}} = \dfrac{1}{12}

, not

\dfrac{1}{2\cdot 36} = \dfrac{1}{72}

Show solution

Take the differential: $dy = \tfrac{1}{2}x^{-1/2}\,dx$ . Choose the perfect square $x = 36$ , where $y = \sqrt{36} = 6$ .

Since $x + dx = 36.6$ , the step is $dx = 0.6$ . At $x = 36$ , $36^{-1/2} = \tfrac{1}{6}$ , so

dy = \tfrac{1}{2}\cdot\tfrac{1}{6}\cdot 0.6 = \tfrac{1}{12}\cdot 0.6 = 0.05.

\sqrt{36.6} \approx y + dy = 6 + 0.05 = 6.05.

The true value is $\sqrt{36.6} = 6.04979\ldots$ ; the estimate is slightly too large because $\sqrt{x}$ bends below its tangent line.

Problem 3 · A Step to the Left

Given: $y = x^{1/3}$ with base point $x = 64$ — estimate $(63.9)^{1/3}$ using differentials, and decide how the estimate compares with the true value.

What is the estimate $y + dy$ ?

Compared with the true value of $(63.9)^{1/3}$ , the estimate is…

✅ Correct!

4 - \tfrac{1}{480} = 3.9979167\ldots

versus the true

3.9979156\ldots

: too large by about

10^{-6}

, because the cube-root curve lies below its tangent line on both sides of

64

❌ Check the sign of $dx$ . The target

63.9

is to the left of

64

, so

dx = -\tfrac{1}{10}

and

dy

is negative.

❌ Close, but you dropped $dx$ .

dy = \tfrac{1}{48}\,dx

, and

dx = -\tfrac{1}{10}

, so

dy = -\tfrac{1}{480}

❌ Not quite. At

x = 64

dy = \tfrac{1}{3}\cdot 64^{-2/3}\,dx = \tfrac{1}{48}\,dx

with

dx = 63.9 - 64

❌ Not quite. A negative

dx

does not flip the comparison. The estimate lies on the tangent line, and for

x > 0

the tangent line of

x^{1/3}

lies above the curve on both sides of the base point.

❌ Not quite.

y + dy

is a point on the tangent line; the exact value

y + \Delta y

is on the curve. They differ unless the curve is itself a line.

❌ Not quite. Ask whether the tangent line lies above or below the curve.

Show solution

As in the video, $dy = \tfrac{1}{3}x^{-2/3}\,dx$ , and at $x = 64$ : $64^{-2/3} = \tfrac{1}{16}$ , so $dy = \tfrac{1}{48}\,dx$ and $y = 4$ .

Now $x + dx = 63.9$ , so $dx = -\tfrac{1}{10}$ and

(63.9)^{1/3} \approx y + dy = 4 + \tfrac{1}{48}\cdot\left(-\tfrac{1}{10}\right) = 4 - \tfrac{1}{480} \approx 3.9979167.

The true value is $(63.9)^{1/3} = 3.9979156\ldots$ , so the estimate is too large by about $1.1 \times 10^{-6}$ . The graph of $x^{1/3}$ is concave down for $x > 0$ , so it bends below its tangent line to the left of $64$ as well as to the right; every tangent-line estimate overshoots.

Problem 4 · Area of a Growing Circle

Given: $A = \pi r^2$ . The radius grows from $r = 10$ to $r = 10.05$ — find the differential $dA$ that estimates the change in area.

✅ Correct!

dA = 2\pi r\,dr = 2\pi(10)(0.05) = \pi

; the exact change is

\Delta A = 1.0025\pi

, only

0.0025\pi

more.

❌ Not quite.

\pi(dr)^2 = 0.0025\pi

is the small leftover term, the gap between

\Delta A

and

dA

. The differential is

dA = A'(r)\,dr

❌ Close, but the $dr$ is missing.

2\pi r = 20\pi

is the derivative

A'(10)

; multiply by

dr = 0.05

❌ Close, but that is $\Delta A$ .

\pi(10.05)^2 - \pi(10)^2 = 1.0025\pi

is the exact change along the curve. The differential is the tangent-line rise

A'(10)\,dr

❌ Not quite. Use

dA = A'(r)\,dr

with

A'(r) = 2\pi r

Show solution

The differential of $A = \pi r^2$ is $dA = A'(r)\,dr = 2\pi r\,dr$ . With $r = 10$ and $dr = 10.05 - 10 = 0.05$ :

dA = 2\pi(10)(0.05) = \pi \approx 3.14.

For comparison, the exact change is

\Delta A = \pi(10.05)^2 - \pi(10)^2 = \pi\left(2\cdot 10\cdot 0.05 + 0.05^2\right) = 1.0025\pi.

The difference $\Delta A - dA = \pi(dr)^2 = 0.0025\pi$ is tiny: a thin ring of width $dr$ has area close to circumference $\times$ width, $2\pi r\,dr$ .

Solved: 0 / 4