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The power rule

Change and motion · level 7, lesson 4 of 6

A question to start

Find the derivative of \(x^{100}\).

From first principles, you would have to expand \((x + h)^{100}\), which has 101 terms, some with numbers more than 25 digits long. Nobody wants to do that. Yet by the end of this lesson you will be able to write the answer in one line, \(100x^{99}\), and be completely sure it is right, without expanding anything.

A second surprise is waiting in the last lesson's results. The derivatives of \(\frac{1}{x}\) and \(\sqrt{x}\) look nothing like the derivative of \(x^3\). But they all follow the same rule.

The idea

The pattern

Here is everything we have found so far, plus the two simplest cases (a constant is a flat line, slope 0; the line \(y = x\) has slope 1):

\(f(x)\) \(1\) \(x\) \(x^2\) \(x^3\) \(x^4\) \(\frac{1}{x} = x^{-1}\) \(\sqrt{x} = x^{1/2}\)
\(f'(x)\) \(0\) \(1\) \(2x\) \(3x^2\) \(4x^3\) \(-\frac{1}{x^2} = -1\,x^{-2}\) \(\frac{1}{2\sqrt{x}} = \frac12 x^{-1/2}\)

Every column follows one rule: bring the power down in front, then lower the power by one.

\[\frac{d}{dx}\left(x^n\right) = n\,x^{n-1}.\]

This is the power rule. Check the last two columns: for \(n = -1\), the rule gives \(-1 \cdot x^{-2}\); for \(n = \frac12\), it gives \(\frac12 x^{-1/2}\). Both match the answers we worked out the long way. Even \(x^0 = 1\) fits: \(0 \cdot x^{-1} = 0\).

Why it works for whole numbers

Proof for \(n = 1, 2, 3, \ldots\) Multiply out \((x + h)^n = (x + h)(x + h)\cdots(x + h)\), with \(n\) brackets. Each term of the answer comes from choosing either \(x\) or \(h\) from every bracket and multiplying.

  • Choose \(x\) every time: you get \(x^n\). There is one way to do this.
  • Choose \(h\) from exactly one bracket and \(x\) from all the others: you get \(x^{n-1}h\). There are \(n\) ways (which bracket gives the \(h\)?), so together these make \(n\,x^{n-1}h\).
  • Every other choice takes \(h\) from two or more brackets, giving a term with \(h^2\), \(h^3\), and so on.

So

\[(x + h)^n = x^n + n\,x^{n-1}h + (\text{terms with } h^2 \text{ or higher}).\]

Subtract \(x^n\) and divide by \(h\):

\[\frac{(x + h)^n - x^n}{h} = n\,x^{n-1} + (\text{terms with } h \text{ or higher}) \;\longrightarrow\; n\,x^{n-1}.\]

That is the power rule. The \(n\) in front is just the number of ways to pick which bracket supplies the \(h\): a counting fact (see Counting). For \(x^{100}\), there are 100 brackets, so the answer is \(100x^{99}\), with no need to know any of the other 100 terms.

Why it works for other powers

The rule holds for every real power \(n\): negative, fractions, even \(x^{\pi}\). You have checked \(n = -1\) and \(n = \frac12\) by hand. A proof for all negative whole numbers uses the trick from \(\frac{1}{x}\) (try \(x^{-2}\) in the exercises), and a proof for every power at once comes in level 8, using logarithms and the chain rule. From now on, use it for any power.

Three rules that let us do polynomials

The constant rule. If \(f(x) = c\), then \(f'(x) = 0\). Every difference quotient is \(\frac{c - c}{h} = 0\). A flat line has slope 0.

The constant multiple rule. \((c\,f)' = c\,f'\). A number in front stays in front:

\[\frac{c\,f(x + h) - c\,f(x)}{h} = c \cdot \frac{f(x + h) - f(x)}{h} \;\longrightarrow\; c\,f'(x).\]

In a picture, multiplying \(f\) by 3 stretches the graph upwards by 3, so every rise is 3 times bigger while the runs stay the same, and every slope is 3 times bigger.

The sum rule. \((f + g)' = f' + g'\), and likewise for differences. The difference quotient of \(f + g\) splits into the difference quotient of \(f\) plus that of \(g\), and the sum law for limits finishes it. In words: if two amounts are changing, their total changes at the sum of the two rates. If your savings grow by 30 dollars a week and your friend's by 20 dollars a week, your combined savings grow by 50 dollars a week.

With these, any polynomial can be differentiated term by term:

\[\frac{d}{dx}\left(4x^5 - 3x^2 + 7x - 9\right) = 20x^4 - 6x + 7.\]

Rewrite first, then differentiate

The power rule needs powers of \(x\). So rewrite roots and fractions as powers before you start.

Written as Rewrite as Derivative
\(\dfrac{1}{x^3}\) \(x^{-3}\) \(-3x^{-4} = -\dfrac{3}{x^4}\)
\(\sqrt[3]{x}\) \(x^{1/3}\) \(\frac13 x^{-2/3}\)
\(\dfrac{5}{\sqrt{x}}\) \(5x^{-1/2}\) \(-\frac52 x^{-3/2}\)
\(\dfrac{x^2 + 3}{x}\) \(x + 3x^{-1}\) \(1 - 3x^{-2}\)
\((x + 2)^2\) \(x^2 + 4x + 4\) \(2x + 4\)

A rule that is not there

It is tempting to guess that the derivative of a product is the product of the derivatives. Test it on \(x^2 = x \cdot x\). The guess gives \(1 \cdot 1 = 1\), but the truth is \(2x\). Wrong. There is a product rule, but it is different, and it is the first lesson of level 8. Until then, multiply out first.

Tangent and normal lines

The tangent at \(x = a\) has slope \(m = f'(a)\). The normal is the line through the same point at right angles to the tangent. Perpendicular lines have slopes that multiply to \(-1\), so the normal's slope is \(-\frac{1}{m}\) (when \(m \ne 0\)). Normals point straight out from a curve, the way light bounces off a curved mirror.

Worked examples

1. Roots and fractions. Differentiate \(y = 3\sqrt{x} - \dfrac{2}{x^2} + \dfrac{1}{4x}\).

  1. Rewrite: \(y = 3x^{1/2} - 2x^{-2} + \frac14 x^{-1}\).
  2. Power rule on each term: \(y' = \frac32 x^{-1/2} + 4x^{-3} - \frac14 x^{-2}\).
  3. Tidy up: \(y' = \dfrac{3}{2\sqrt{x}} + \dfrac{4}{x^3} - \dfrac{1}{4x^2}\).

2. Tangent and normal. Find the tangent and normal to \(y = x^3 - 2x\) at \(x = 1\).

  1. The point: \(y = 1 - 2 = -1\), so \((1, -1)\).
  2. Slope: \(y' = 3x^2 - 2\), so at \(x = 1\) the slope is \(1\).
  3. Tangent: \(y = -1 + 1(x - 1)\), that is \(y = x - 2\).
  4. Normal: slope \(-\frac{1}{1} = -1\), so \(y = -1 - (x - 1)\), that is \(y = -x\).

3. A tangent with a given slope. Where on \(y = x^3\) is the tangent parallel to the line \(y = 12x + 1\)?

  1. Parallel means the same slope, 12. So \(3x^2 = 12\), \(x^2 = 4\), \(x = 2\) or \(x = -2\).
  2. The points are \((2, 8)\) and \((-2, -8)\). Two answers: the curve climbs at the same steepness at both.

4. The tangent that meets the curve again. Level 6 noted that the tangent to \(y = x^3\) at \((1, 1)\) meets the curve again at \((-2, -8)\). Now we can prove it.

  1. Slope at \(x = 1\): \(3 \times 1^2 = 3\). Tangent: \(y = 1 + 3(x - 1) = 3x - 2\).
  2. Where does it meet the curve? Solve \(x^3 = 3x - 2\), that is \(x^3 - 3x + 2 = 0\).
  3. We know \(x = 1\) is a solution, and in fact a double one (a tangent touches), so \((x - 1)^2\) is a factor: \(x^3 - 3x + 2 = (x - 1)^2(x + 2)\). Check by expanding: \((x^2 - 2x + 1)(x + 2) = x^3 - 3x + 2\). ✓
  4. The other solution is \(x = -2\), giving the point \((-2, -8)\). A tangent touches the curve at its point, but it can cross the curve somewhere else.
-2-112-8-44xy (1, 1) (−2, −8) tangent y = 3x − 2

Try it

  1. Differentiate: (a) \(x^7\); (b) \(5x^4\); (c) \(-2x\); (d) \(8\).
  2. For \(f(x) = x^3 - 6x^2 + 9x - 4\), find \(f'(x)\), \(f'(0)\) and \(f'(2)\). What do the signs tell you?
  3. Differentiate \(y = 6\sqrt{x} + \dfrac{4}{x}\).
  4. Differentiate \(y = (2x + 3)^2\) (multiply out first).
  5. Differentiate \(y = \dfrac{x^3 + 2x}{x}\).
  6. Find the equation of the tangent to \(y = x^2 - 4x + 1\) at \(x = 3\).
  7. Find the points on \(y = x^3 - 3x\) where the tangent is horizontal.
  8. (a) Show that the tangent to \(y = x^2\) at \(x = a\) crosses the \(y\)-axis at \(-a^2\). (b) Use the method of the last lesson to prove the power rule for \(x^{-2}\) from first principles.
Answers
  1. (a) \(7x^6\); (b) \(20x^3\); (c) \(-2\); (d) \(0\).
  2. \(f'(x) = 3x^2 - 12x + 9\). \(f'(0) = 9\) (going up steeply at \(x = 0\)) and \(f'(2) = 12 - 24 + 9 = -3\) (going down at \(x = 2\)).
  3. \(y = 6x^{1/2} + 4x^{-1}\), so \(y' = 3x^{-1/2} - 4x^{-2} = \dfrac{3}{\sqrt{x}} - \dfrac{4}{x^2}\).
  4. \(y = 4x^2 + 12x + 9\), so \(y' = 8x + 12\).
  5. For \(x \ne 0\), \(y = x^2 + 2\), so \(y' = 2x\).
  6. \(y' = 2x - 4 = 2\) at \(x = 3\), and \(y = 9 - 12 + 1 = -2\). Tangent: \(y = -2 + 2(x - 3)\), so \(y = 2x - 8\).
  7. \(y' = 3x^2 - 3 = 0\) gives \(x = 1\) or \(x = -1\). The points are \((1, -2)\) and \((-1, 2)\).
  8. (a) Tangent: \(y = a^2 + 2a(x - a) = 2ax - a^2\). At \(x = 0\), \(y = -a^2\). (b) \(\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2} = \dfrac{x^2 - (x + h)^2}{x^2(x + h)^2} = \dfrac{-2xh - h^2}{x^2(x + h)^2}\). Divide by \(h\): \(\dfrac{-2x - h}{x^2(x + h)^2} \to \dfrac{-2x}{x^4} = -\dfrac{2}{x^3} = -2x^{-3}\), as the power rule says.

Common mistake

Lowering a negative power the wrong way. The derivative of \(x^{-2}\) is \(-2x^{-3}\), not \(-2x^{-1}\). "Lower by one" means subtract 1: \(-2 - 1 = -3\).

Differentiating \(\frac{1}{x^3}\) as \(\frac{1}{3x^2}\). The power rule works on powers of \(x\), not on the bottom of a fraction. Rewrite as \(x^{-3}\) first: the answer is \(-3x^{-4}\).

Where it leads