Read derivative values from a function graph

Slopes become heights

The top graph shows \(f(x) = \tfrac13 x^3 - 4x\), and the lower graph shows its derivative \(f'\).

The orange line is the tangent. At \(x = 1\), it falls 3 units for each 1 unit to the right, so its slope is \(-3\).

On the lower graph, the orange bar goes down to \(-3\). Its length is the slope of the top curve, not the top curve's height.

Move the point to \(x = 2\). The tangent is horizontal, so the slope is 0 and the bar has no height. At \(x = 3\), the tangent rises 5 units for each 1 unit to the right, so the bar goes up to 5.

As the point moves, the top of the bar traces the graph of \(f'\).

The graph of the slope

The top curve is green where it rises and red where it falls.

Drag the point along the curve. Where the tangent slopes up, the slope is positive and the bar is above the axis.

Where the tangent slopes down, the slope is negative and the bar is below the axis.

A steeper tangent gives a longer bar. At every input, the height of the lower graph is the slope of the top curve.

Notation

1. The derivative \(f'(a)\) is the slope of the tangent to the graph of \(f\) at the input \(a\). On the lower graph, it is the height of the orange bar at \(a\).

   \[\cy{f'(a)} = \text{slope of the tangent at } a\]

2. The slope is the rise over the run. Move 1 unit to the right along the tangent, so the run is 1 and the slope equals the rise.

   \[f'(a) = \dfrac{\text{rise}}{\text{run}}\]

3. For each input \(a\), plot the point \((a, f'(a))\) on the lower graph. These points make the graph of \(f'\).

4. The derivative exists at \(a\) only if the curve has one tangent there, so the slopes from the left and the right must be equal.

When it fails

Drag the sharpness slider to 1. Then move the point to \(x = 0\). The smooth bottom of the curve is now a corner.

At the corner, the red lines are the tangents from the two sides, with slopes \(-1\) and 1. No single line fits both sides, so \(f'(0)\) does not exist, and the lower graph has a gap at 0.

Example 1: a parabola

The graph shows \(g(x) = x^2 + 2x - 3\). Read \(g'\) at \(x = -3\), \(x = -1\) and \(x = 1\).

1. At \(x = -3\), the tangent falls 4 units for each 1 unit to the right, so the slope is \(-4\).

   \[g'(-3) = \cy{-4}\]

2. At \(x = -1\), the tangent is horizontal at the bottom of the curve, so the slope is 0.

   \[g'(-1) = \cy{0}\]

3. At \(x = 1\), the tangent rises 4 units for each 1 unit to the right, so the slope is 4.

   \[g'(1) = \cy{4}\]

4. As the point sweeps across, the top of the bar traces the graph of \(g'\), a straight line.

Example 2: a corner

The graph shows \(h(x) = |x^2 - 4|\). Read \(h'\) at \(x = 1\), \(x = 3\) and \(x = 2\).

1. At \(x = 1\), on the arch between the corners, the tangent falls 2 units for each 1 unit to the right, so the slope is \(-2\).

   \[h'(1) = \cy{-2}\]

2. At \(x = 3\), the tangent rises 6 units for each 1 unit to the right, so the slope is 6.

   \[h'(3) = \cy{6}\]

3. A tempting wrong move: At \(x = 2\), the curve touches the x-axis at the bottom of a V. The tangent is horizontal, so \(h'(2) = 0\).

   \[h'(2) = 0\]

   The red lines are the tangents to the two sides of the V. Both are steep, so a horizontal line fits neither.

4. Read each side separately. The slope from the left is \(-4\) and from the right is 4.

   \[-4 \ne 4\]

5. The two slopes differ, so the curve has no single tangent at \(x = 2\), and \(h'(2)\) does not exist.

Practice

The graph shows \(k(x) = \dfrac{x^4}{4} - x^2\). Read \(k'\) at \(x = -1\), \(x = 0\) and \(x = 2\).

Drag the orange point to each input. Then read how far the tangent rises or falls over 1 unit to the right.

Questions

The graph shows a curve \(m\) and an orange tangent. The curve is a parabola for \(x \le 1\) and a straight line for \(x > 1\).

1. The orange point is at \(x = -1\). What is \(m'(-1)\)?

   • \(-2\)
   • 2
   • 1
   • \(-1\)
   Answer

   \(-2\)

2. At which input does \(m\) have no derivative?

   • \(x = 1\)
   • \(x = 0\)
   • \(x = 2\)
   Answer

   \(x = 1\)

3. A tangent to the graph of \(g\) at \(x = 2\) passes through the points \((2, 5)\) and \((4, 11)\). What is \(g'(2)\)?

   • 3
   • 6
   • \(\dfrac{1}{3}\)
   • 5
   Answer

   3

Summary

The derivative at an input is the slope of the tangent there, which you read as the rise over a run of 1.

At a corner, the slopes from the two sides differ, so no single tangent exists and \(f'\) has no value there.

You use this to turn the graph of \(f\) into the graph of \(f'\), with a gap at each corner.

This is one of five skills in topic 5.8 of Which & Why: AP Calculus.