Wemazi

Topic 2.5

Differentiate positive integer powers

Builds on reading the derivative of a function as the slope of its graph. 6 minutes.

Slopes of x², x³ and x⁴

The top graph shows \(y = x^2\). The orange line touches the curve at the orange point, and its steepness is the slope of the curve there.

The curve is green where it rises and red where it falls. The lower graph plots the slope, so the height of the orange bar is the slope at the point.

The lower graph is the graph of \(2x\), so at \(x = 2\) the slope is 4.

Show \(x^3\). At \(x = 2\) the slope is 12, and the lower graph is \(3x^2\). Show \(x^4\). At \(x = 2\) the slope is 32, and the lower graph is \(4x^3\).

In each case, the exponent of the curve comes down in front, and the exponent on \(x\) is one smaller.

Any whole-number power

Slide the exponent \(n\) from 1 to 6. The top curve changes shape, and the point and the bar stay at the same \(x\).

The first number below the graph is the slope that the graph shows, and the second is \(n\) times \(x^{n-1}\). The two numbers are equal for every exponent and every point.

Notation

  1. The slope of \(x^n\) is its derivative, where \(n\) is a constant positive whole number. The goal is a formula for it.

    \[\frac{d}{dx}\,x^n = \;?\]

  2. To find the slope at \(x\), step a short distance \(h\) to the right. The secant through the two points has slope equal to the rise divided by \(h\).

    \[\frac{(x+h)^n - x^n}{h}\]

  3. The power \((x+h)^n\) is a product of \(n\) factors \(x+h\). Its terms with one \(h\) take \(h\) from one factor and \(x\) from the rest, so there are \(n\) of them, each equal to \(x^{n-1}h\).

    \[\begin{aligned} (x+h)^n &= x^n + \cy{n\,x^{n-1}}\,h \\ &\quad + \cq{(\text{higher powers of } h)} \end{aligned}\]

    For \(n = 3\): \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\).

  4. Subtract \(x^n\) and divide by \(h\). Every term that still contains \(h\) goes to 0 as \(h\) goes to 0, so the secant turns into the tangent.

    \[\begin{aligned} \frac{(x+h)^n - x^n}{h} &= \cy{n\,x^{n-1}} \\ &\quad + \cq{(\text{terms in } h)} \\ &\to \cy{n\,x^{n-1}} \end{aligned}\]

  5. The exponent \(n\) comes down in front of \(x\), and the new exponent on \(x\) is \(n - 1\).

    \[\frac{d}{dx}\,x^n = \cy{n}\,x^{\cy{n-1}}\]

When it fails

The rule needs a constant exponent. Change \(x^3\) into \(2^x\), where \(x\) is in the exponent.

If the rule applied, it would give \(x \cdot 2^{x-1}\), which is 12 at \(x = 3\), the slope of the red dashed line. The orange tangent has slope about 5.55.

Example 1: a high power

Find the derivative of \(y = x^6\).

  1. The exponent 6 comes down in front of \(x\).

    \[y' = 6\,x^{?}\]

  2. The new exponent on \(x\) is \(6 - 1 = 5\). The lower graph shows this formula.

    \[y' = \cy{6}\,x^{\cy{5}}\]

  3. At \(x = 1\) the slope is \(6 \cdot 1^5 = 6\). At \(x = -1\) the slope is \(6(-1)^5 = -6\), and the curve is red there because it falls.

Example 2: the sign of the slope

Find the derivative of \(y = x^5\). Then say whether the curve rises or falls at \(x = -1\).

  1. The exponent 5 comes down in front of \(x\).

  2. A tempting wrong move: Keep the exponent 5 on \(x\).

    \[y' = 5x^5\]

    At \(x = -1\) this formula gives \(-5\), so the tangent would fall, as the red dashed line shows. But the curve is green there, so it rises.

  3. The exponent on \(x\) must be one smaller, so the new exponent is \(5 - 1 = 4\).

    \[y' = 5x^4\]

  4. At \(x = -1\) the slope is \(5(-1)^4 = 5\). The slope is positive, so the curve rises.

Practice

Find the derivative of \(y = x^8\).

Move the point to \(x = 1\). The slope there is the number that comes down in front.

Solution

  1. The exponent 8 comes down in front of \(x\).

  2. The new exponent on \(x\) is \(8 - 1 = 7\).

    \[y' = 8x^7\]

  3. At \(x = 1\) the formula gives 8, which matches the orange bar.

Questions

Questions 1 and 2 use the graph of \(y = x^4\) and its slope. Question 3 needs a calculation.

  1. At \(x = -1\), is the slope of the curve positive, negative or zero?

    • Positive
    • Negative
    • Zero
    Answer

    Negative

  2. At \(x = 2\), the orange bar on the lower graph reaches 32. Which formula gives the height of the lower graph?

    • \(4x^3\)
    • \(4x^4\)
    • \(x^3\)
    • \(3x^4\)
    Answer

    \(4x^3\)

  3. A student differentiates \(x^n\) and writes \(20x^{19}\). What is \(n\)?

    • 19
    • 20
    • 21
    • 380
    Answer

    20

Summary

You differentiate \(x^n\) by bringing the exponent down in front and lowering the exponent on \(x\) by 1.

The rule works because \((x+h)^n\) contains \(n\) copies of \(x^{n-1}h\), and the terms with more \(h\) go to 0 in the limit.

Use the rule when the base is \(x\) and the exponent is a constant whole number.

This is one of four skills in topic 2.5 of Which & Why: AP Calculus.