Use graph or table values in the derivative

Reading a derivative from a graph

The graph shows a function \(f\) made of straight segments, and the dots mark the values in the table. Let \(F(x)=\int_1^{2x}f(t)\,dt\).\[\begin{array}{c|ccccccc} t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(t) & 3 & 5 & 4 & 1 & -2 & -4 & -1 \end{array}\]

At \(x=2\), the upper edge is at \(2x=4\), where the graph is 2 below the axis, so \(f(4)=-2\). The red parts lie below the axis and count as negative.

The edge moves 2 units for each unit of \(x\). Each unit the edge moves changes the area at the rate \(f(4)=-2\), so \(F'(2)=-2\cdot2=-4\).

Move \(x\). The readouts show where the edge is, the height of the graph there and the rate \(F'(x)\), which needs no area.

Heights at the edges, never areas

A graph or a table gives \(f\) without a formula. That is enough, because the derivative of the accumulation uses \(f\) only at the two edges.

Find where each edge is for the given \(x\). The orange bar at each edge shows the height of the graph there. Multiply each height by the speed of its edge.

Subtract the term of the lower edge from the term of the upper edge.

Move \(x\). The heights at the edges change, but no area is read. If you hide the areas, the two heights remain, and they are all the derivative needs.

Notation

1. The derivative is the height of \(f\) at the upper edge times the speed of that edge, minus the same product for the lower edge.

   \[\begin{aligned} \dfrac{d}{dx}\int_{l(x)}^{u(x)}f(t)\,dt &= \cy{f(u)}\,\cg{u'} \\ &\quad - \cy{f(l)}\,\cg{l'} \end{aligned}\]

2. For a given \(x\), first find the numbers \(u\), \(l\), \(u'\) and \(l'\). Then read \(f(u)\) and \(f(l)\) from the graph or the table.

3. The area does not appear in the derivative, so do not read it.

   A fixed edge does not move, so it adds no term.

When it fails

The table lists \(f\) only at whole numbers. If an edge is between two entries, the table does not give \(f\) there.

At \(x = 2.25\) the edges are at 2.25 and 4.5. From the table alone, you estimate \(f\) with a straight segment between the two neighbors. At 4.5 this gives \(\frac{-2+(-4)}{2}=-3\), so the derivative is an estimate, not an exact value.

Example 1: One edge at 3x

Use the graph and the table of \(f\) below. Let \(G(x) = \displaystyle\int_1^{3x} f(t)\,dt\). Find \(G'(1)\) and \(G'(2)\).\[\begin{array}{c|ccccccc} t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(t) & 3 & 5 & 4 & 1 & -2 & -4 & -1 \end{array}\]

1. The edge is at \(3x\) and moves 3 units for each unit of \(x\). At \(x = 1\) it is at \(t = 3\), and at \(x = 2\) it is at \(t = 6\).

2. Read the heights from the graph or the table: \(f(3) = 1\) and \(f(6) = -1\). Multiply each by the speed 3.

   \[\begin{aligned} G'(1) &= \cy{f(3)}\cdot3 = 3 \\ G'(2) &= \cy{f(6)}\cdot3 = -3 \end{aligned}\]

3. At \(x = 1\) the graph at the edge is above the axis, so a small move of \(x\) adds area and \(G\) increases. The thin green strip at the edge is this area. At \(x = 2\) the graph is below the axis, so the strip is red and \(G\) decreases.

Example 2: Edges at x and x squared

Use the graph and the table of \(f\) below. Let \(K(x) = \displaystyle\int_x^{x^2} f(t)\,dt\). Find \(K'(2)\).\[\begin{array}{c|ccccccc} t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(t) & 3 & 5 & 4 & 1 & -2 & -4 & -1 \end{array}\]

1. At \(x = 2\), the upper edge is at \(x^2 = 4\), and its speed is \(2x = 4\). The lower edge is at 2, and its speed is 1.

2. A tempting wrong move: Read both heights at \(x = 2\), where \(f(2) = 4\).

   \[K'(2) = 4\cdot4 - 4\cdot1 = 12\]

   The upper edge is at \(t = 4\), not at \(t = 2\). The red mark puts the height 4 over \(t = 4\). The graph there is 2 below the axis, so both the size and the sign are wrong.

3. Read the height at each edge: \(f(4) = -2\) at the upper edge and \(f(2) = 4\) at the lower edge.

   \[\begin{aligned} K'(2) &= \cy{f(4)}\cdot4 - \cy{f(2)}\cdot1 \\ &= -8 - 4 = -12 \end{aligned}\]

Practice

Use the graph and the table of \(f\) below. Let \(M(x) = \displaystyle\int_{x+1}^{3x} f(t)\,dt\). Find \(M'(2)\).\[\begin{array}{c|ccccccc} t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(t) & 3 & 5 & 4 & 1 & -2 & -4 & -1 \end{array}\]

Move \(x\) to 2. At each edge, read whether the graph is above or below the axis.

Questions

Questions 1 and 2 use the graph of a function \(g\) made of straight segments. The amount \(H(x)=\int_0^{2x}g(t)\,dt\) is the area between the graph and the t-axis from 0 to the moving edge \(2x\). Where the graph is below the axis, the region is red and counts as negative. After you answer, the edge moves to that \(x\), and a faint mark on the t-axis shows where \(x\) is.

1. At \(x = 2\), what is \(H'(2)\)?

   • 4
   • 2
   • 12
   Answer

   4

2. At \(x = 3\), what is \(H'(3)\)?

   • \(-8\)
   • \(-4\)
   • 8
   Answer

   \(-8\)

3. A table gives values of \(p\). Let \(S(x)=\displaystyle\int_x^{x+2}p(t)\,dt\). What is \(S'(2)\)?\[\begin{array}{c|cccc} t & 1 & 2 & 3 & 4 \\ \hline p(t) & 2 & 5 & -1 & 3 \end{array}\]

   • \(-2\)
   • 8
   • 2
   Answer

   \(-2\)

Summary

You find where each edge is, read \(f\) there from the graph or the table, and multiply by the speed of the edge.

This works because the derivative of an accumulation is made of heights at the edges, never of areas.

You use it when \(f\) is given only as a graph or a table, with no formula to integrate.

This is one of four skills in topic 6.4 of Which & Why: AP Calculus.