Wemazi

Topic 2.5

Evaluate a power-function slope

Builds on differentiating negative and fractional powers. 6 minutes.

The volume of a cube

A cube has an edge of \(s\) centimeters, so its volume is \(V = s^3\) cubic centimeters. The top graph plots \(V\) against \(s\), and the curve is green because the volume rises as the edge grows.

The orange line is the tangent at the orange point. The orange bar on the lower graph is as tall as its slope, in cubic centimeters of volume for each centimeter of edge.

Move the point to \(s = 2\), where the slope is 12. At \(s = 4\) the slope is 48, four times as large, because a thin layer on the larger cube covers faces with four times the area.

The power rule gives the slope at every edge as \(V' = 3s^2\). Substitute 2 to find 12, and substitute 4 to find 48.

Any power function, any point

The top curve is a power function \(f\), and the lower curve is its derivative \(f'\). The second slider changes the exponent.

Move the point to a position \(a\). The orange tangent on the top graph has slope \(f'(a)\), and the orange bar on the lower graph has height \(f'(a)\). These are one number read two ways.

To find the number from a formula, take two steps. First write the formula for \(f'\). Then substitute \(a\) for \(x\).

Notation

  1. The symbol \(f'(a)\) means the derivative \(f'\) evaluated at \(x = a\), which is the slope of the tangent there.

    \[f'(a) = \cy{\text{slope at } x = a}\]

  2. First differentiate with the power rule. Then substitute the input.

    \[\begin{aligned} f(x) = x^3 \;&\Rightarrow\; f'(x) = 3x^2 \\ &\Rightarrow\; f'(2) = 3 \cdot 2^2 = \cy{12} \end{aligned}\]

  3. Do not substitute first. The value \(f(2)\) is a single number, the height of the orange bar on the top graph, and the derivative of a number is 0.

    \[\begin{aligned} f(2) &= 2^3 = 8 \\ \frac{d}{dx}\,8 &= 0 \;\ne\; f'(2) \end{aligned}\]

  4. The slope has units of output per input. If \(x\) is an edge in centimeters and \(f\) is a volume in cubic centimeters, then \(f'(2) = 12\) cubic centimeters per centimeter.

When it fails

The input must be a place where the derivative exists. Take \(f(x) = 5x^{2/3}\). Its derivative \(\tfrac{10}{3}x^{-1/3}\) divides by 0 at \(x = 0\).

Move the point to 0. Just left of 0 the slope is very negative, and just right of 0 it is very positive. The two sides disagree, so \(f'(0)\) does not exist.

Example 1: a negative input

Find the slope of \(y = x^9\) at \(x = -1\).

  1. Differentiate first. The exponent 9 comes down, and the new exponent is 8.

    \[y' = 9x^8\]

  2. Substitute \(x = -1\). An even power of \(-1\) is \(+1\).

    \[y'(-1) = 9\,(-1)^8 = \cy{9}\]

  3. The curve is green at \(x = -1\) because it rises, and the steep orange tangent matches a slope of 9.

Example 2: the slope of x^(3/2)

Find \(f'(4)\) for \(f(x) = x^{3/2}\).

  1. The question asks for \(f'(4)\), the slope of the tangent at \(x = 4\).

  2. A tempting wrong move: Substitute first: \(f(4) = 4^{3/2} = 8\). The derivative of 8 is 0.

    \[f'(4) = 0\]

    A slope of 0 means a flat tangent, shown by the red dashed line. But the curve is green at \(x = 4\), so the tangent rises. The orange bar shows that 8 is the height of the curve, not its slope.

  3. Differentiate first. The exponent \(\tfrac32\) comes down, and the new exponent is \(\tfrac32 - 1 = \tfrac12\).

    \[f'(x) = \tfrac32\,x^{1/2}\]

  4. Then substitute \(x = 4\). The square root of 4 is 2.

    \[f'(4) = \tfrac32 \cdot 2 = \cy{3}\]

Practice

Find the slope of \(y = \dfrac{1}{x^4}\) at \(x = 2\).

Move the point to \(x = 2\). The curve is red there because it falls, so the slope is below zero.

Solution

  1. Write the reciprocal as a power.

    \[y = x^{-4}\]

  2. Differentiate first. The exponent \(-4\) comes down, and the new exponent is \(-5\).

    \[y' = -4x^{-5}\]

  3. Then substitute \(x = 2\).

    \[y'(2) = -4 \cdot 2^{-5} = -\frac{4}{32} = -\frac18\]

Questions

Questions 1 and 2 use the graph of \(y = x^5\) and its slope. Question 3 needs a calculation.

  1. What is the slope of the curve at \(x = 1\)?

    • 1
    • 4
    • 5
    Answer

    5

  2. What is the slope of the curve at \(x = -1\)?

    • −5
    • −1
    • 5
    Answer

    5

  3. Let \(g(x) = x^{5/4}\). Find \(g'(16)\).

    • \(\tfrac52\)
    • 32
    • \(\tfrac58\)
    • 0
    Answer

    \(\tfrac52\)

Summary

You find the slope at a point by differentiating first and then substituting the input into \(f'\).

This works because \(f'(a)\) is the slope of the tangent at \(a\), and the formula for \(f'\) gives that slope at every \(a\).

The answer has units of output per input, and it exists only where \(f'\) is defined.

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