Math Core

Lesson 2.4 · Differentiation: Definition and Basic Rules

The power rule

Computing every derivative from the limit definition would be slow and error-prone. Fortunately, the definition produces patterns, and those patterns become rules you can apply in one line. The first and most-used rule handles powers of xx.

Two rules to start

Constants. The graph of f(x)=cf(x) = c is a horizontal line, so its slope is 00 everywhere. From the definition, c−ch=0\dfrac{c - c}{h} = 0 for every hh, so

ddx[c]=0\frac{d}{dx}[c] = 0

Powers. Look at what the definition gave for small powers:

f(x)f(x)xxx2x^2x3x^3x4x^4
f′(x)f'(x)112x2x3x23x^24x34x^3

The exponent comes down in front, and the new exponent is one less.

The power rule

For any real number nn,

ddx[xn]=nxn−1\frac{d}{dx}\left[x^n\right] = n x^{n - 1}

This holds for positive integers, negative integers, and fractional exponents, wherever xn−1x^{n-1} is defined.

Why it works for whole-number powers

For a positive integer nn, expand (x+h)n(x + h)^n with the binomial theorem:

(x+h)n=xn+nxn−1h+(terms containing h2,h3,…)(x + h)^n = x^n + n x^{n - 1} h + \big(\text{terms containing } h^2, h^3, \ldots\big)

Subtract xnx^n and divide by hh:

(x+h)n−xnh=nxn−1+(terms containing h,h2,…)\frac{(x + h)^n - x^n}{h} = n x^{n - 1} + \big(\text{terms containing } h, h^2, \ldots\big)

As h→0h \to 0, every term that still contains hh vanishes, leaving nxn−1n x^{n - 1}. The proofs for negative and fractional exponents take more work, but the same formula results, and you may use it freely on the AP exam.

Rewrite before you differentiate

The power rule needs the form xnx^n. Radicals and fractions have to be rewritten with exponents first:

written asrewrite as
x\sqrt{x}x1/2x^{1/2}
x23\sqrt[3]{x^2}x2/3x^{2/3}
1x4\dfrac{1}{x^4}x−4x^{-4}
1x\dfrac{1}{\sqrt{x}}x−1/2x^{-1/2}

After differentiating, you can convert back to radicals and positive exponents if you like. Either form is correct.

Worked example: Integer powers

Differentiate (a) y=x7y = x^7 and (b) y=1x3y = \dfrac{1}{x^3}.

(a) dydx=7x6\dfrac{dy}{dx} = 7x^6.

(b) Rewrite as y=x−3y = x^{-3}. Then

dydx=−3x−3−1=−3x−4=−3x4\frac{dy}{dx} = -3x^{-3 - 1} = -3x^{-4} = -\frac{3}{x^4}

Note that −3−1=−4-3 - 1 = -4: subtracting 1 from a negative exponent makes it more negative.

Worked example: Fractional powers

Differentiate (a) f(x)=xf(x) = \sqrt{x} and (b) g(x)=x23g(x) = \sqrt[3]{x^2}.

(a) f(x)=x1/2f(x) = x^{1/2}, so f′(x)=12x−1/2=12xf'(x) = \dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}. This matches what the limit definition gave in the previous lesson, with far less work.

(b) g(x)=x2/3g(x) = x^{2/3}, so

g′(x)=23x2/3−1=23x−1/3=23x3g'(x) = \frac{2}{3} x^{2/3 - 1} = \frac{2}{3} x^{-1/3} = \frac{2}{3\sqrt[3]{x}}

Notice that g′(0)g'(0) is undefined. That's the cusp you saw on the graph of y=x2/3y = x^{2/3} in the differentiability lesson.

Slopes and tangent lines

With the power rule, tangent-line problems become quick.

Worked example: A tangent line in two lines of work

Find the tangent line to y=x3y = x^3 at x=2x = 2.

The slope is y′(2)=3(2)2=12y'(2) = 3(2)^2 = 12, and the point is (2,8)(2, 8). The tangent line is

y−8=12(x−2),ory=12x−16y - 8 = 12(x - 2), \quad \text{or} \quad y = 12x - 16
The tangent line to y = x³ at (2, 8).Open in grapher →

You can also run the question backward: given a slope, find where the curve has it.

Worked example: Where is the slope 12?

At which points on y=x3y = x^3 is the tangent line parallel to y=12x−16y = 12x - 16?

Parallel lines have equal slopes, so set 3x2=123x^2 = 12. Then x2=4x^2 = 4 and x=±2x = \pm 2. The points are (2,8)(2, 8) and (−2,−8)(-2, -8).

Common mistake

The power rule is for a variable base and a constant exponent. It does not apply to 2x2^x (variable in the exponent) and it gives 00, not nxn−1n x^{n-1}, for a constant like π3\pi^3 or 525^2. ddx[π3]=0\dfrac{d}{dx}[\pi^3] = 0 because π3≈31\pi^3 \approx 31 is just a number.

Tip

Before you apply the power rule, ask "Is this exactly xx to a constant power?" If not, rewrite it until it is. Most power-rule mistakes are rewriting mistakes, such as treating 1x3\dfrac{1}{x^3} as x3x^3 or x\sqrt{x} as x2x^2.

Practice

Practice 1

Find dydx\dfrac{dy}{dx} for y=x9y = x^9.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find f′(x)f'(x) for f(x)=1x5f(x) = \dfrac{1}{x^5}.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Find f′(x)f'(x) for f(x)=xxf(x) = x\sqrt{x}, x>0x > 0.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Let f(x)=x3f(x) = \sqrt[3]{x}. Find f′(8)f'(8).

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let g(x)=1xg(x) = \dfrac{1}{\sqrt{x}}. Find g′(4)g'(4).

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find an equation of the tangent line to y=x4y = x^4 at x=−1x = -1.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

What is ddx[π3]\dfrac{d}{dx}\left[\pi^3\right]?

Practice 8

At what point on the graph of y=x3y = x^3 with x>0x > 0 is the tangent line parallel to the line y=27x+5y = 27x + 5?

Enter a point like (2, -3)