Include the chain-rule factor

An edge that moves faster than x

The line is \(y=f(t)=1+\frac t2\). The teal region runs from 0 to the orange edge at \(u=x^2\), and its area is \(A(x)\). A faint \(x\) on the \(t\)-axis marks \(x\) itself, and the edge is farther out.

At \(x=1.5\), the edge is at \(u=2.25\), where the line is \(1+\frac{2.25}{2}=2.125\) high.

Move \(x\) from 1.5 to 1.6. The edge moves from 2.25 to 2.56, about 3 units for each unit of \(x\), because its speed is \(2x=3\). The green strip is the area this move adds.

Each unit that the edge moves adds area at the rate 2.125. The edge moves 3 units for each unit of \(x\), so the area grows at the rate \(2.125\cdot3=6.375\) for each unit of \(x\).

Height at the edge times the speed of the edge

The moving edge is now a function \(u(x)\). The region starts at the dashed line \(t=a\), which does not move, so the area \(A(x)=\int_a^{u(x)}f(t)\,dt\) depends on \(x\) only through the edge.

Two rates multiply. The area grows at the rate \(f(u)\) for each unit that the edge moves, and the edge moves \(u'(x)\) units for each unit of \(x\).

Move \(x\). The faster the edge moves, the faster the area grows, because the same change in \(x\) covers more of the curve. Where the edge stops and turns back, \(u'(x)=0\), so the lower graph is flat there.

The push is a change in \(x\). Narrow it. The faint line through the two ends of the push on the lower graph becomes the orange tangent line. The slope of the tangent is the height at the edge times the speed of the edge.

Notation

1. The curve is \(f\), and the region runs from \(t = a\) to the moving edge at \(t = \cy{u(x)}\).

   \[\cb{A(x) = \int_a^{u(x)} f(t)\,dt}\]

2. For each unit that the edge moves, the area grows at the rate \(\cy{f(u)}\), the height at the edge. This is the rule from the previous skill, with \(u\) as the variable.

   \[\dfrac{dA}{du} = \cy{f(u)}\]

3. A change \(\Delta x\) in \(x\) moves the edge about \(\cg{u'(x)\,\Delta x}\), because \(u'(x)\) is the speed of the edge. The green strip is that wide and \(f(u(x))\) tall.

4. The chain rule multiplies the two rates. The derivative is \(f\) of \(u(x)\) times \(u'(x)\).

   \[\dfrac{dA}{dx} = \dfrac{dA}{du}\cdot\dfrac{du}{dx} = \cy{f(u(x))}\,\cg{u'(x)}\]

   This needs \(f\) continuous on an interval containing \(a\) and \(u(x)\), and \(u\) differentiable at \(x\). If the edge is \(x\) itself, then \(u'(x)=1\) and the formula is the rule from the previous skill.

When it fails

The rule needs \(f\) continuous at the edge. Put a jump in \(f\) and move the edge onto it.

A strip just left of \(c\) and a strip just right of \(c\) have different heights. The lower graph has a corner there, so \(A'\) does not exist.

Example 1: An edge at 3x

Let \(y = \displaystyle\int_0^{3x} \sin\left(t^2\right)dt\). Find \(\dfrac{dy}{dx}\).

1. The edge is \(u = 3x\). It moves 3 units for each unit of \(x\), so \(u' = 3\).

   \[u = 3x, \qquad u' = 3\]

2. The height at the edge is the integrand at \(u\), \(\sin(u^2) = \sin(9x^2)\).

   \[\cy{\sin\left((3x)^2\right)} = \cy{\sin\left(9x^2\right)}\]

3. The derivative is the height at the edge times the speed of the edge.

   \[\dfrac{dy}{dx} = \cy{\sin\left(9x^2\right)}\cdot\cg{3} = 3\sin\left(9x^2\right)\]

Example 2: An edge that moves slowly

Let \(G(x) = \displaystyle\int_0^{\sqrt x} \cos\left(t^2\right)dt\), for \(x > 0\). Find \(G'(x)\) and \(G'(\pi)\).

1. The edge is \(u = \sqrt{x}\), and its speed is \(u' = \dfrac{1}{2\sqrt{x}}\).

2. A tempting wrong move: The height at the edge is \(\cos(u^2) = \cos x\). This is the derivative.

   \[G'(x) = \cos x\]

   At \(x = \pi\), the edge is at about 1.77 and moves only 0.28 units for each unit of \(x\). The red line has slope \(-1\). The orange tangent is much flatter, because the slow edge changes the area slowly.

3. Multiply the height at the edge, \(\cos x\), by the speed of the edge.

   \[G'(x) = \cy{\cos x}\cdot\cg{\dfrac{1}{2\sqrt{x}}}\]

4. At \(x = \pi\), the height is \(\cos\pi = -1\), and the speed is \(\frac{1}{2\sqrt{\pi}}\). The strip is red because the curve is below the axis at the edge, so the slope is negative.

   \[G'(\pi) = -\dfrac{1}{2\sqrt{\pi}}\]

   As a check, \(-\frac{1}{2\sqrt{\pi}} \approx -0.282\).

Practice

Let \(P(x) = \displaystyle\int_1^{x^3} \ln(1 + t)\,dt\). Find \(P'(1)\).

Move x a small step on the figure. The edge moves farther than x, so multiply the height at the edge by the speed of the edge.

Questions

Questions 1 and 2 use the line \(y = f(t) = 6 - t\). The amount \(A(x)\) is the area under the line from 0 to the moving edge \(u = 2x\). The faint \(x\) on the \(t\)-axis shows where \(x\) is.

1. At \(x = 2\), what is the height of the line at the moving edge?

   • 2
   • 4
   • 6
   Answer

   2

2. What is \(A'(2)\)?

   • 4
   • 2
   • 8
   Answer

   4

3. Let \(C(x) = \displaystyle\int_1^{x^2} \sqrt{t + 1}\,dt\). What is \(C'(\sqrt3)\)?

   • \(4\sqrt3\)
   • 2
   • \(2\sqrt3\)
   Answer

   \(4\sqrt3\)

Summary

You differentiate an integral whose upper bound is \(u(x)\) by writing the integrand at \(u(x)\) and multiplying by \(u'(x)\).

This works because two rates multiply: the area grows at the rate \(f(u)\) as the edge moves, and the edge moves at the rate \(u'(x)\).

You use it when the bound of the integral is a function of \(x\), not \(x\) itself.

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