Single-Variable-Calculus · Unit 5 · Video 1 · Interactive Practice

The Power Rule for Rational Exponents, by Implicit Differentiation

IKey Formulas

Formula	Step	Why it is legal
$y = x^{m/n} \;\Longrightarrow\; y^n = x^m$	Raise both sides to the $n$ -th power	$(x^{m/n})^n = x^{(m/n)\cdot n} = x^m$ : integer exponents only
$n\,y^{\,n-1}\dfrac{dy}{dx} = m\,x^{m-1}$	Differentiate both sides	Chain rule on the left, power rule on the right
$\dfrac{dy}{dx} = \dfrac{m}{n}\,x^{(m-1)-\frac{m(n-1)}{n}}$	Solve, then substitute $y = x^{m/n}$	Needs $y \neq 0$ ; dividing powers subtracts exponents
$\dfrac{d}{dx}x^{a} = a\,x^{a-1},\quad a = \dfrac{m}{n}$	Combine the exponents	$(m-1)-\dfrac{m(n-1)}{n} = \dfrac{m}{n} - 1$

Key Insight: Differentiate the equation you can handle. $y^n = x^m$ carries only integer powers, and the chain rule makes the unknown $\frac{dy}{dx}$ appear as a factor you can solve for — everything after that is algebra.

IIVisualization 1 — Two Steps to the Derivative of $\sqrt{x}$

Squaring turns $y = \sqrt{x}$ into $y^2 = x$ , an equation whose every exponent is an integer.

Step 1

Step 1 — The function no rule covers yet

y = x^{m/n} \qquad\text{here}\quad y = x^{1/2} = \sqrt{x}

Step 2 — Raise both sides to the $n$ -th power

y^n = \left(x^{m/n}\right)^{n} = x^{m} \qquad y^2 = x

Step 3 — Differentiate both sides

n\,y^{\,n-1}\frac{dy}{dx} = m\,x^{m-1} \qquad 2y\,\frac{dy}{dx} = 1

The chain rule contributes the factor

\frac{dy}{dx}

— the unknown being solved for. Dropping it differentiates with respect to

y

Step 4 — Solve for $\frac{dy}{dx}$ (where $y \neq 0$ )

\frac{dy}{dx} = \frac{m}{n}\cdot\frac{x^{m-1}}{y^{\,n-1}} \qquad \frac{dy}{dx} = \frac{1}{2y} = \frac{1}{4}\ \text{ at } (4,2)

Step 5 — Substitute $y = x^{m/n}$

\frac{dy}{dx} = \frac{m}{n}\cdot\frac{x^{m-1}}{x^{m(n-1)/n}} = \frac{m}{n}\,x^{(m-1)-\frac{m(n-1)}{n}}

\frac{1}{2x^{1/2}} = \tfrac{1}{2}x^{-1/2}

Step 6 — Combine the exponents

(m-1) - \frac{m(n-1)}{n} = m - 1 - m + \frac{m}{n} = \frac{m}{n} - 1

\frac{d}{dx}x^{a} = a\,x^{a-1}, \qquad a = \frac{m}{n} = \tfrac{1}{2}

IIIVisualization 2 — Where the Exponent Lands

The two exponent moves always finish exactly one unit to the left of $a = m/n$ .

m 2 n 3

IVVisualization 3 — One Formula for Every Rational Power

The tangent slope at every point of $y = x^{a}$ is $a\,x^{a-1}$ , for each rational $a = m/n$ .

Exponent

💡 The argument assumes $x^{m/n}$ has a derivative in the first place; given that, implicit differentiation forces that derivative to be $a\,x^{a-1}$ .

VQuiz Questions

Problem 1 · The Rule, Straight Through

Given: $f(x) = x^{5/3}$ — find $f'(x)$ .

✅ Correct! The old exponent comes down in front and the new one is

\tfrac{5}{3} - 1 = \tfrac{2}{3}

❌ The exponent drops by one, it does not rise.

\tfrac{5}{3} - 1 = \tfrac{5}{3} - \tfrac{3}{3} = \tfrac{2}{3}

❌ The coefficient is $a$ itself, not $1/a$ . The rule reads

a\,x^{a-1}

with

a = \tfrac{5}{3}

❌ That uses the new exponent twice. The coefficient is the original

a = \tfrac{5}{3}

; only the exponent becomes

a - 1

❌ Not quite. Apply

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

with

a = \tfrac{5}{3}

Show solution

The rational power rule applies directly with $a = \tfrac{5}{3}$ :

f'(x) = \tfrac{5}{3}x^{\frac{5}{3} - 1} = \tfrac{5}{3}x^{2/3}

The same answer from scratch, the way the video derives it ( $m = 5$ , $n = 3$ ): from $y = x^{5/3}$ , cube both sides to get $y^3 = x^5$ . Differentiating,

3y^2\frac{dy}{dx} = 5x^4 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{5x^4}{3y^2} = \frac{5x^4}{3x^{10/3}} = \tfrac{5}{3}x^{4 - \frac{10}{3}} = \tfrac{5}{3}x^{2/3}

Both routes give $\tfrac{5}{3}x^{2/3}$ , since $4 - \tfrac{10}{3} = \tfrac{12}{3} - \tfrac{10}{3} = \tfrac{2}{3}$ .

Problem 2 · The Factor That Gets Dropped

Given: $y = x^{2/3}$ , rewritten as $y^3 = x^2$ — differentiate both sides with respect to $x$ , then solve for $\dfrac{dy}{dx}$ in terms of $x$ .

Differentiating both sides gives

So $\dfrac{dy}{dx}$ equals

✅ Correct!

\dfrac{dy}{dx} = \dfrac{2x}{3y^2} = \dfrac{2x}{3x^{4/3}} = \tfrac{2}{3}x^{-1/3}

, which is

a\,x^{a-1}

for

a = \tfrac{2}{3}

❌ That is the classic mistake. It differentiates

y^3

with respect to

y

, not

x

; the chain rule supplies the extra factor

\dfrac{dy}{dx}

❌ Only the left side depends on $x$ through $y$ . The right side is

x^2

, whose derivative is

2x

with no extra factor.

❌ The extra factor belongs to the side containing $y$ . The chain rule applies to

\dfrac{d}{dx}(y^3)

, giving

3y^2\dfrac{dy}{dx}

❌ Not quite. Differentiate

y^3

as an outside power of

y

times the inside derivative

\dfrac{dy}{dx}

❌ Check the exponent's sign.

x^{1}/x^{4/3} = x^{1 - 4/3} = x^{-1/3}

; the denominator wins.

❌ The coefficient is $\tfrac{m}{n} = \tfrac{2}{3}$ , the ratio that multiplies, not its reciprocal.

❌ The $x$ in the numerator still counts. Exponents subtract:

1 - \tfrac{4}{3} = -\tfrac{1}{3}

, not

-\tfrac{4}{3}

❌ Not quite. Substitute

y = x^{2/3}

, so

y^2 = x^{4/3}

, and then subtract exponents.

Show solution

Here $m = 2$ and $n = 3$ , so $y^n = x^m$ reads $y^3 = x^2$ . Differentiating both sides with respect to $x$ , the left side needs the chain rule because $y$ depends on $x$ :

3y^2\frac{dy}{dx} = 2x

Solve (legal where $y \neq 0$ ) and substitute $y = x^{2/3}$ , so $y^2 = x^{4/3}$ :

\frac{dy}{dx} = \frac{2x}{3y^2} = \frac{2x}{3x^{4/3}} = \tfrac{2}{3}x^{1 - \frac{4}{3}} = \tfrac{2}{3}x^{-1/3}

And $-\tfrac{1}{3} = \tfrac{2}{3} - 1 = a - 1$ , exactly what the power rule predicts for $a = \tfrac{2}{3}$ .

Problem 3 · The Exponent, Before It Is Simplified

Given: $m = 5$ and $n = 2$ , so $y = x^{5/2}$ — evaluate the exponent $(m-1) - \dfrac{m(n-1)}{n}$ that the substitution produces.

✅ Correct!

4 - \tfrac{5}{2} = \tfrac{8}{2} - \tfrac{5}{2} = \tfrac{3}{2}

, which is

a - 1

for

a = \tfrac{5}{2}

❌ The whole $m(n-1)/n$ is subtracted. With

m = 5

n = 2

that is

\tfrac{5 \cdot 1}{2} = \tfrac{5}{2}

, not

\tfrac{1}{2}

❌ That is $a$ , not $a - 1$ . The exponent of the derivative is one less than the exponent of the function.

❌ Check the sign. The substituted power sits in the denominator, so its exponent is subtracted:

4 - \tfrac{5}{2}

, not

4 + \tfrac{5}{2}

❌ Not quite. Evaluate

m - 1

first, then subtract

\dfrac{m(n-1)}{n}

over a common denominator.

Show solution

Substituting the numbers directly:

(m-1) - \frac{m(n-1)}{n} = (5-1) - \frac{5(2-1)}{2} = 4 - \frac{5}{2} = \frac{8}{2} - \frac{5}{2} = \frac{3}{2}

The general simplification, which the video does once and for all: distribute inside the fraction, $\dfrac{m(n-1)}{n} = m - \dfrac{m}{n}$ , so

(m-1) - \left(m - \frac{m}{n}\right) = m - 1 - m + \frac{m}{n} = \frac{m}{n} - 1 = a - 1

With $a = \tfrac{5}{2}$ this reads $\tfrac{5}{2} - 1 = \tfrac{3}{2}$ , matching the arithmetic above. The derivative is therefore $\tfrac{5}{2}x^{3/2}$ .

Problem 4 · A Root in the Denominator

Given: $f(x) = \dfrac{1}{\sqrt[4]{x}}$ — write it as a single power of $x$ , differentiate, and evaluate $f'(16)$ .

What is $f'(x)$ ?

What is $f'(16)$ ?

✅ Correct!

f(x) = x^{-1/4}

, so

f'(x) = -\tfrac{1}{4}x^{-5/4}

and

16^{5/4} = 2^5 = 32

, giving

-\tfrac{1}{4}\cdot\tfrac{1}{32} = -\tfrac{1}{128}

❌ Subtract one from the exponent.

-\tfrac{1}{4} - 1 = -\tfrac{5}{4}

, so the power stays in the denominator.

❌ The coefficient is $a = -\tfrac{1}{4}$ itself, so the derivative is negative:

f

decreases as

x

grows.

❌ Check that subtraction.

-\tfrac{1}{4} - 1 = -\tfrac{1}{4} - \tfrac{4}{4} = -\tfrac{5}{4}

, not

-\tfrac{3}{4}

❌ Not quite. A reciprocal fourth root is

x^{-1/4}

; now apply

a\,x^{a-1}

with

a = -\tfrac{1}{4}

❌ A fractional exponent is not multiplication.

16^{5/4}

means

\left(16^{1/4}\right)^5 = 2^5 = 32

, not

16 \cdot \tfrac{5}{4} = 20

❌ That used $16^{-1/4} = \tfrac{1}{2}$ . The exponent in the derivative is

-\tfrac{5}{4}

, giving

\tfrac{1}{32}

❌ A negative exponent puts the power downstairs.

16^{-5/4} = \tfrac{1}{32}

, so the answer is a small negative number.

❌ Not quite. Evaluate

16^{5/4}

by taking the fourth root first, then the fifth power.

Show solution

Step 1 — Write it as one power. A fourth root is the exponent $\tfrac{1}{4}$ , and a reciprocal flips the sign:

f(x) = \frac{1}{\sqrt[4]{x}} = \frac{1}{x^{1/4}} = x^{-1/4}

Step 2 — Differentiate with $a = -\tfrac{1}{4}$ (the video's argument allows $m$ or $n$ to be negative, here $m = -1$ , $n = 4$ ):

f'(x) = -\tfrac{1}{4}x^{-\frac{1}{4} - 1} = -\tfrac{1}{4}x^{-5/4}

Step 3 — Evaluate at $x = 16$ . Take the fourth root first, then the fifth power:

16^{5/4} = \left(16^{1/4}\right)^{5} = 2^{5} = 32 \quad\Longrightarrow\quad f'(16) = -\tfrac{1}{4}\cdot\frac{1}{32} = -\frac{1}{128}

The sign is a useful check: $f(x) = x^{-1/4}$ falls as $x$ grows, so $f'(16)$ must be negative, and it is tiny because the curve has almost flattened out by $x = 16$ .

Solved: 0 / 4

The Power Rule for Rational Exponents, by Implicit Differentiation

IKey Formulas

IIVisualization 1 — Two Steps to the Derivative of x\sqrt{x}x​

IIIVisualization 2 — Where the Exponent Lands

IVVisualization 3 — One Formula for Every Rational Power

VQuiz Questions

IIVisualization 1 — Two Steps to the Derivative of $\sqrt{x}$