Single-Variable-Calculus · Unit 4 · Video 2 · Interactive Practice

The Quotient Rule, and Why the Power Rule Works for Negative Exponents

IKey Formulas

Formula	Name	Where it comes from
$\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$	Quotient rule	Proved here; needs $v \neq 0$
$\Delta\!\left(\dfrac{u}{v}\right) = \dfrac{(\Delta u)v - u\,\Delta v}{(v + \Delta v)v}$	Change in a quotient	Common denominator; $uv$ cancels
$\dfrac{d}{dx}\left(\dfrac{1}{v}\right) = -\dfrac{v'}{v^2} = -v^{-2}v'$	Reciprocal rule	Quotient rule with $u = 1$
$\dfrac{d}{dx}x^{-n} = -n\,x^{-n-1}$	Power rule, negative exponent	Reciprocal rule with $v = x^n$

Key Insight: Read $-n$ as a single number: the coefficient in front is that number, and the new exponent is one less than it. The power rule's statement never changed — only the exponents it covers.

IIVisualization 1 — The Proof, Line by Line

Every line of the proof reports the same number; only the last one, the limit, changes it.

Step 1

Step 1 — Write the change

\Delta\!\left(\frac{u}{v}\right) = \frac{u + \Delta u}{v + \Delta v} - \frac{u}{v}

Step 2 — Over one denominator

= \frac{(u + \Delta u)v - u(v + \Delta v)}{(v + \Delta v)v}

Step 3 — $uv$ occurs twice, once with each sign

= \frac{uv + (\Delta u)v - uv - u\,\Delta v}{(v + \Delta v)v}

= \frac{(\Delta u)v - u\,\Delta v}{(v + \Delta v)v}

Step 4 — Divide by $\Delta x$

\frac{\Delta(u/v)}{\Delta x} = \frac{\dfrac{\Delta u}{\Delta x}\,v - u\,\dfrac{\Delta v}{\Delta x}}{(v + \Delta v)v}

Step 5 — Let $\Delta x \to 0$

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

v

is continuous, so

\Delta v \to 0

and the denominator

(v + \Delta v)v

becomes

v^2

IIIVisualization 2 — Predict the Slope of $x^{-n}$

Before computing: how steep is $y = x^{-n}$ at a chosen point?

n in x⁻ⁿ point x 2.00

💡 Challenge: find the point where $y = x^{-2}$ has tangent slope exactly $-\tfrac{1}{4}$ .

IVVisualization 3 — Where the Exponent $-n-1$ Comes From

Differentiating $1/x^n$ multiplies $x^{-2n}$ by $x^{n-1}$ , so the two exponents add.

n 2

VQuiz Questions

Problem 1 · The Rule, Straight Through

Given: $f(x) = \dfrac{x^2}{x + 1}$ — find $f'(x)$ .

✅ Correct!

u'v - uv' = 2x(x+1) - x^2 = x^2 + 2x

, and the old denominator is squared.

❌ That plus belongs to the product rule. The quotient rule subtracts:

2x(x+1) - x^2

, not

2x(x+1) + x^2

❌ The denominator is squared. The rule divides by

v^2 = (x+1)^2

, not by

v

❌ The two terms are in the wrong order. The numerator differentiated comes first,

u'v

, and then you subtract

uv'

❌ Not quite. Name the pieces

u = x^2

v = x + 1

and substitute into

\dfrac{u'v - uv'}{v^2}

Show solution

With $u = x^2$ , $u' = 2x$ , $v = x + 1$ , $v' = 1$ :

f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(x+1) - x^2 \cdot 1}{(x+1)^2}

= \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}

Check all three things before moving on: the minus sign, the order ( $u'v$ first), and the squared denominator. The formula holds where $v \neq 0$ , that is for $x \neq -1$ .

Problem 2 · A Negative Exponent, Sign and Size

Given: $g(x) = \dfrac{1}{x^5}$ — find $g'(x)$ .

✅ Correct!

\dfrac{1}{x^5} = x^{-5}

, and the power rule with exponent

-5

gives

-5x^{-6}

❌ You differentiated the denominator and left it downstairs. The reciprocal rule is

-\dfrac{v'}{v^2}

, not

\dfrac{1}{v'}

❌ The exponent goes down by one, not up. With

n = 5

the new exponent is

-n - 1 = -6

❌ Sign. Differentiating a reciprocal introduces a minus:

\dfrac{d}{dx}\left(\dfrac{1}{v}\right) = -\dfrac{v'}{v^2}

❌ Not quite. Write it as

x^{-5}

and apply

\dfrac{d}{dx}x^{-n} = -n\,x^{-n-1}

Show solution

With the new power rule: $g(x) = x^{-5}$ , so $n = 5$ and

g'(x) = -5x^{-5-1} = -5x^{-6} = -\frac{5}{x^6}

With the reciprocal rule (where it came from), taking $v = x^5$ and $v' = 5x^4$ :

-\frac{v'}{v^2} = -\frac{5x^4}{x^{10}} = -5x^{4-10} = -5x^{-6}

Both routes agree, as they must — the negative-exponent power rule is the reciprocal rule applied to $v = x^n$ .

Problem 3 · Build the Numerator, Then Simplify

Given: $h(x) = \dfrac{x^2 - 1}{x^2 + 1}$ — find the numerator $u'v - uv'$ , then $h'(x)$ .

What is $u'v - uv'$ ?

What is $h'(x)$ ?

✅ Correct! The cubes cancel:

2x^3 + 2x - 2x^3 + 2x = 4x

❌ Reversed.

u'v

leads and

uv'

is subtracted, which flips the sign of this answer.

❌ That is the product rule's pattern. With a plus the

x^3

terms add instead of cancelling.

❌ Derivatives do not simply multiply. The rule pairs

u'

with

v

and

u

with

v'

❌ Not quite. Here

u = x^2 - 1

u' = 2x

v = x^2 + 1

v' = 2x

❌ Check the sign.

2x^3 + 2x - 2x^3 + 2x = +4x

; subtracting

-2x

adds.

❌ The $x^3$ terms cancel.

4x^3

is what a plus sign in the numerator would leave.

❌ The old denominator is squared. It is

(x^2+1)^2

, not

x^2 + 1

❌ Not quite. Expand the numerator you just built, then divide by

v^2

Show solution

With $u = x^2 - 1$ , $u' = 2x$ , $v = x^2 + 1$ , $v' = 2x$ :

h'(x) = \frac{2x(x^2+1) - (x^2-1)(2x)}{(x^2+1)^2}

= \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2+1)^2} = \frac{4x}{(x^2+1)^2}

The $2x^3$ terms cancel only because of the minus sign — with a plus you would be left with $4x^3$ , a different function. The result is positive for $x > 0$ : $h$ climbs from $-1$ toward $1$ .

Problem 4 · Two Routes to One Derivative

Given: $k(x) = \dfrac{x^2 + 3}{x^3}$ — rewrite it with negative exponents, then differentiate.

Which rewrite is correct?

What is $k'(x)$ ?

✅ Correct! The quotient rule gives

\dfrac{2x \cdot x^3 - (x^2+3)3x^2}{x^6} = \dfrac{-x^4 - 9x^2}{x^6}

— the same function.

❌ Off by one. Dividing subtracts exponents once:

\dfrac{x^2}{x^3} = x^{2-3} = x^{-1}

❌ Exponents subtract when you divide.

2 - 3 = -1

, so the first term is

x^{-1}

❌ Moving a power upstairs flips the sign of its exponent.

\dfrac{3}{x^3} = 3x^{-3}

❌ Not quite. Split the fraction:

\dfrac{x^2}{x^3} + \dfrac{3}{x^3}

, then subtract exponents in each piece.

❌ The exponent becomes the coefficient.

\dfrac{d}{dx}3x^{-3} = 3(-3)x^{-4} = -9x^{-4}

❌ Both coefficients are negative — the exponents

-1

and

-3

come out in front as they are.

❌ One less than $-1$ is $-2$ , not $0$ . You subtract 1 from the exponent even when the exponent is negative.

❌ Not quite. Apply

\dfrac{d}{dx}x^{-n} = -n\,x^{-n-1}

to each term of the rewrite.

Show solution

Route 1 — rewrite, then use the negative-exponent power rule:

k(x) = \frac{x^2}{x^3} + \frac{3}{x^3} = x^{-1} + 3x^{-3}

k'(x) = (-1)x^{-2} + 3(-3)x^{-4} = -x^{-2} - 9x^{-4}

Route 2 — quotient rule with $u = x^2 + 3$ , $u' = 2x$ , $v = x^3$ , $v' = 3x^2$ :

k'(x) = \frac{2x \cdot x^3 - (x^2 + 3)\cdot 3x^2}{(x^3)^2} = \frac{2x^4 - 3x^4 - 9x^2}{x^6}

= \frac{-x^4 - 9x^2}{x^6} = -x^{-2} - 9x^{-4}

The routes agree because the negative-exponent power rule was itself derived from the quotient rule. Route 1 is shorter whenever the denominator is a single power of $x$ .

Solved: 0 / 4

The Quotient Rule, and Why the Power Rule Works for Negative Exponents

IKey Formulas

IIVisualization 1 — The Proof, Line by Line

IIIVisualization 2 — Predict the Slope of x−nx^{-n}x−n

IVVisualization 3 — Where the Exponent −n−1-n-1−n−1 Comes From

VQuiz Questions

IIIVisualization 2 — Predict the Slope of $x^{-n}$

IVVisualization 3 — Where the Exponent $-n-1$ Comes From