A. Rate of Change · Derivatives and Differentiation Rules

The Power Rule for Negative and Rational Exponents

A3.4About 65 minutes

Learning goals

I can rewrite roots and reciprocals of x as powers with rational exponents.
I can verify that the power rule works for negative and rational exponents, numerically and with tangent slopes.
I can differentiate functions with radicals and powers of x in the denominator.
I can identify points where such functions are not differentiable.

Words to know

rational exponent
An exponent that is a fraction, such as x^(2/3) = ∛(x²).

negative exponent
x⁻ⁿ = 1/xⁿ, for x ≠ 0.

radical function
A function involving a root of x, such as √x or ∛x.

cusp
A sharp point on a graph where the tangent slopes on the two sides approach +∞ and −∞; the function is not differentiable there.

More than whole-number powers

In the lesson The Derivative from First Principles, two challenge questions gave surprising answers:

d/dx (1/x) = −1/x²
d/dx (√x) = 1/(2√x)

Now rewrite them with exponents, using the exponent laws from Grade 11 Functions (MCR3U, unit Exponential Functions): 1/x = x⁻¹ and √x = x^(1/2).

The power rule would predict d/dx (x⁻¹) = −1x⁻² = −1/x². ✓
The power rule would predict d/dx (x^(1/2)) = (1/2)x^(−1/2) = 1/(2√x). ✓

The power rule works for these exponents too!

The power rule, extended

For any rational number n, d/dx (xⁿ) = nxⁿ⁻¹ (wherever xⁿ⁻¹ is defined).

(In fact it is true for every real exponent, which you can prove in first-year university calculus.)

Rewriting is the key step

Before you differentiate, rewrite every root and every power in a denominator as a power of x:

√x = x^(1/2), ∛x = x^(1/3), ∛(x²) = x^(2/3), √(x³) = x^(3/2)
1/x = x⁻¹, 1/x³ = x⁻³, 1/√x = x^(−1/2)
5/x² = 5x⁻² (the constant stays in the numerator)

After differentiating, you may rewrite the answer with positive exponents or roots so it is easy to read. A useful fact: subtracting 1 from a fraction like 1/2 gives −1/2, and from −1/3 gives −4/3.

y = √x with tangents at x = 1, 4 and 9. The slopes 1/2, 1/4 and 1/6 match 1/(2√x). Near the origin the tangent becomes vertical, so the derivative does not exist at x = 0.

Worked example

Example 1: Verifying the rule numerically

Verify the extended power rule for f(x) = √x at x = 4, and for g(x) = x^(2/3) at x = 8, using centred estimates with h = 0.001.

Power rule: f′(x) = (1/2)x^(−1/2), so f′(4) = 1/(2 × 2) = 0.25.
Estimate: [√4.001 − √3.999] ÷ 0.002 = (2.000249984 − 1.999749984) ÷ 0.002 ≈ 0.2500000.
Power rule: g′(x) = (2/3)x^(−1/3), so g′(8) = (2/3)(1/2) = 1/3 ≈ 0.333333.
Estimate: [8.001^(2/3) − 7.999^(2/3)] ÷ 0.002 ≈ 0.3333333.

Both agree to at least six decimal places: f′(4) = 0.25 and g′(8) = 1/3. ✓

Worked example

Example 2: Differentiating roots and reciprocals

Differentiate: (a) y = ∛x (b) y = 1/x⁴ (c) y = x√x (d) y = 6/√x

(a) y = x^(1/3), so dy/dx = (1/3)x^(−2/3) = 1/(3∛(x²)).
(b) y = x⁻⁴, so dy/dx = −4x⁻⁵ = −4/x⁵.
(c) y = x¹ · x^(1/2) = x^(3/2), so dy/dx = (3/2)x^(1/2) = (3/2)√x.
(d) y = 6x^(−1/2), so dy/dx = 6(−1/2)x^(−3/2) = −3x^(−3/2) = −3/(x√x).

(a) 1/(3∛(x²)) (b) −4/x⁵ (c) (3/2)√x (d) −3/(x√x)

Worked example

Example 3: Several terms

Differentiate f(x) = 4√x − 3/x² + 2x.

Rewrite: f(x) = 4x^(1/2) − 3x⁻² + 2x.
Differentiate term by term: 4(1/2)x^(−1/2) − 3(−2)x⁻³ + 2.
Simplify: f′(x) = 2x^(−1/2) + 6x⁻³ + 2 = 2/√x + 6/x³ + 2.

f′(x) = 2/√x + 6/x³ + 2 (for x > 0).

y = 1/x with tangents at x = 1, 2 and −1. Every tangent slopes downward, which matches dy/dx = −1/x² < 0 for all x ≠ 0. Points the same distance from 0 have equal slopes.

Worked example

Example 4: Divide first, then differentiate

Differentiate y = (x² + 3)/√x.

Split the fraction: y = x²/x^(1/2) + 3/x^(1/2) = x^(3/2) + 3x^(−1/2).
dy/dx = (3/2)x^(1/2) + 3(−1/2)x^(−3/2) = (3/2)x^(1/2) − (3/2)x^(−3/2).
Optional tidy form: (3/2)[√x − 1/(x√x)] = 3(x² − 1)/(2x√x).

dy/dx = (3/2)√x − 3/(2x√x). Splitting a fraction over a single-term denominator turns it into powers; for longer denominators you will need the rules in The Quotient Rule and Rational Functions.

Worked example

Example 5: A tangent line and a non-differentiable point

(a) Find the tangent to y = √x at x = 9. (b) Where is y = x^(2/3) not differentiable?

(a) dy/dx = 1/(2√x) = 1/6 at x = 9. Point (9, 3). Tangent: y − 3 = (1/6)(x − 9), so y = x/6 + 3/2.
(b) dy/dx = (2/3)x^(−1/3) = 2/(3∛x). This is undefined at x = 0, where the graph has a sharp point (a cusp): slopes are very negative just left of 0 and very positive just right.

(a) y = x/6 + 3/2 (b) at x = 0.

Just for fun

Square roots are shy: they always hide their exponent. Writing √x as x^(1/2) is like asking them to take off their hat so the power rule can recognize them.

Hands-on activity

Check the rule with a measuring tape

You need: a measuring tape or ruler, some square tiles or square sheets of paper of different sizes (sticky notes work), and a calculator.

For a square of area A, the side is s = √A. Measure three squares and record A and s.
Imagine the area grows by a little bit. The power rule says ds/dA = 1/(2√A) = 1/(2s). Calculate this for each square.
Check with numbers: for a 7.6 cm sticky note, A = 57.76 cm². Find √(57.76 + 1) − √57.76. This is the extra side length for 1 cm² of extra area.
Compare with 1/(2 × 7.6).

What to look for: √58.76 − 7.6 ≈ 0.0655 cm and 1/15.2 ≈ 0.0658 cm, almost equal (1 cm² is a fairly big step, so the match is close but not perfect). Bigger squares give smaller values of ds/dA: a big square needs only a tiny change in side length to gain 1 cm² of area, because that area is spread along a long edge.

Your turn

Rewrite with exponents: ⁴√x, 1/x², 7/∛x.

Differentiate y = x^(5/2).

Differentiate f(x) = 1/x³.

Differentiate g(x) = 8∜x.

Differentiate y = 3x² − 2/x + √x.

Find the slope of the tangent to y = 1/x at x = −2.

Find the equation of the tangent to y = √x at x = 4.

More practice: warm-up

Write √x as a power.

Write 1/x as a power.

d/dx (x⁻²) = ?

d/dx (x^(1/3)) = ?

What is 1/2 − 1?

More practice: on your own

Differentiate y = (x³ − 4x)/x² for x ≠ 0.

Differentiate f(x) = √x (x + 1).

Find the point on y = √x where the tangent has slope 1/10.

Find the equation of the tangent to y = 4/x at x = 2.

Is there a point on y = 1/x where the tangent is horizontal? Explain.

Challenge

Show that every tangent to y = 1/x (x > 0) forms a triangle with the coordinate axes of area 2.

Find the equation of the normal to y = ∛x at x = 8.

For f(x) = x^(4/3), find f′ and f″. Is f′ differentiable at x = 0?

Check your understanding

Quiz

What is the derivative of y = √x?

What is the derivative of f(x) = 1/x²?

Rewrite 5/∛x as a power of x.

What is d/dx (x^(3/4))?

Where is y = x^(1/3) not differentiable?

What is the slope of the tangent to y = 2√x at x = 16?

For parents and teachers

Big idea (A3.4): the power rule d/dx (xⁿ) = nxⁿ⁻¹ also works for negative and fractional exponents. The key skill is rewriting roots and reciprocals as powers before differentiating.

How to help: Write expressions such as √x, 1/x³ and 2/∛x and ask your teen to rewrite each as a power of x before doing anything else. Then ask what 1/2 − 1 and −1/3 − 1 equal; fraction arithmetic with negatives is where most errors happen.

Common difficulties:

Writing 1/x² as x² or x^(−1/2).
Subtracting 1 from the wrong part of a fraction: 2/3 − 1 = −1/3, not 1/3 or 2/2.
Moving a constant into the exponent: 3/x² = 3x⁻², not x^(−2/3) or (3x)⁻².
Splitting fractions with more than one term in the denominator, which is not allowed: (x + 1)/(x + 2) ≠ x/x + 1/2.