Wemazi

Topic 2.8

Differentiate a product

Builds on combining derivatives of sine, cosine, exponentials and logarithms. 8 minutes.

A rectangle whose sides both grow

The rectangle has width \(2x\) and height \(x+3\). At \(x=2\) the width is 4 and the height is 5, so the area is 20.

Move the slider h to increase \(x\) from 2 to \(2+h\). The width grows by \(2h\) and the height by \(h\). The dashed line outlines the new rectangle, which adds a green strip on the right, a green strip on top and a small faint corner.

The strip on the right is \(2h\) wide and 5 high. The strip on top is 4 wide and \(h\) high. The corner is \(2h\) wide and \(h\) high.

For \(h=0.1\) the strips add 1.0 and 0.4 and the corner adds 0.02, so the area gains 1.42. This is 14.2 for each unit of \(h\).

If \(h\) is smaller, the corner shrinks much faster than the strips, because its area contains \(h\) twice. The gain for each unit of \(h\) approaches \(2\cdot5+4\cdot1=14\), the rate of the two strips without the corner.

The product rule picture

The width \(u\) and the height \(v\) both change with \(x\), and the area is their product \(uv\). When \(x\) increases by \(h\), the width grows by \(\Delta u\) and the height by \(\Delta v\). The area gains the strip \(\Delta u\cdot v\) on the right, the strip \(u\cdot\Delta v\) on top and the corner \(\Delta u\,\Delta v\).

Press Divide by h. The strip on the right becomes \(u'\) times \(v\), because its width divided by \(h\) approaches the rate \(u'\). The strip on top becomes \(u\) times \(v'\) in the same way.

The corner divided by \(h\) has the height \(\Delta v\div h\), which approaches \(v'\), but its width \(\Delta u\) shrinks toward 0 with \(h\). Only the two strips remain, so the area changes at the rate \(u'v+uv'\).

Notation

  1. This is the product rule for two functions \(u\) and \(v\) of \(x\).

    \[(uv)' = \cy{u'}\,v + u\,\cy{v'}\]

  2. The derivative of a product is the derivative of the first factor times the second, plus the first factor times the derivative of the second.

  3. The rule needs \(u\) and \(v\) to be differentiable at \(x\), because the strips use the rates \(u'\) and \(v'\).

  4. The product \(u'v'\) is not the answer. For \(u=x\) and \(v=x^2\), the product \(x^3\) has the derivative \(3x^2\), but \(u'v'\) is only \(2x\).

    \[(x\cdot x^2)' = 3x^2 \;\neq\; 1\cdot 2x\]

Example 1: two polynomial factors

Find the derivative of \(y=(2x+1)(x^2-3)\). Then check it by multiplying out.

  1. Let \(u=2x+1\) and \(v=x^2-3\). Then \(u'=2\) and \(v'=2x\).

    \[u'=2, \qquad v'=2x\]

  2. Add the two strips \(u'v\) and \(uv'\).

    \[y' = 2(x^2-3) + (2x+1)(2x)\]

  3. Expand and collect like terms.

    \[y' = 6x^2+2x-6\]

  4. To check, multiply out first and differentiate each term. The result is the same.

    \[\begin{aligned} y &= 2x^3+x^2-6x-3 \\ y' &= 6x^2+2x-6 \end{aligned}\]

  5. At \(x=3\) the rectangle is 7 by 6. The strips are \(2\cdot6=12\) and \(7\cdot6=42\), and their sum 54 matches the formula.

Example 2: a product that cannot be multiplied out

Find the slope of \(y=x^2\ln x\) at \(x=2\).

  1. Let \(u=x^2\) and \(v=\ln x\). Then \(u'=2x\) and \(v'=\frac1x\).

    \[u'=2x, \qquad v'=\frac1x\]

  2. A tempting wrong move: Multiply the derivatives of the two factors.

    \[\cr{y' = 2x\cdot\frac1x = 2}\]

    The curve rises from about 2.77 at \(x=2\) to about 3.27 at \(x=2.1\), a rate near 5. The red line of slope 2 is too flat, because \(u'v'\) is not one of the two strips.

  3. Add the two product-rule terms \(u'v\) and \(uv'\). The second one simplifies, because \(x^2\cdot\frac1x=x\).

    \[y' = 2x\ln x + x\]

  4. Substitute \(x=2\). The result is the slope of the orange tangent.

    \[y'(2) = 4\ln 2 + 2\]

  5. As a check, \(4\ln 2+2\approx4.77\), which is close to the slope of the white secant.

Practice

Find the \(x\) where the tangent to \(y=xe^x\) is horizontal.

A horizontal tangent has slope 0. Move the point until the orange bar on the lower graph, which shows the slope, has height 0.

Solution

  1. Let \(u=x\) and \(v=e^x\). Then \(u'=1\) and \(v'=e^x\).

  2. Add the two product-rule terms \(u'v\) and \(uv'\).

    \[y' = 1\cdot e^x + x\,e^x = (1+x)\,e^x\]

  3. The factor \(e^x\) is never 0, so \(y'=0\) only when \(1+x=0\). This gives \(x=-1\), the lowest point \((-1,-\tfrac1e)\).

    \[\cy{x=-1}\]

Questions

Questions 1 and 2 use a rectangle with width \(u=3\) and height \(v=6\). The width grows at the rate \(u'=2\), so the green strip on the right is area gained. The height shrinks at the rate \(v'=-1\), so the red strip on top is area lost. Question 3 needs a calculation.

  1. What is the rate of change of the area \(uv\)?

    • 12
    • 9
    • −2
    • 15
    Answer

    9

  2. Which term of the product rule gives the red strip along the top?

    • u′v
    • u v′
    • u′v′
    Answer

    u v′

  3. What is the derivative of \(x^3\cos x\)?

    • \(-3x^2\sin x\)
    • \(3x^2\cos x+x^3\sin x\)
    • \(3x^2\cos x-x^3\sin x\)
    Answer

    \(3x^2\cos x-x^3\sin x\)

Summary

You differentiate a product by adding two terms, and each term keeps one factor and differentiates the other.

This works because the area of a rectangle gains a strip along each side, and the corner shrinks faster than \(h\), so it drops out.

You use it for a product of two factors that you cannot multiply out, such as \(x^2\ln x\).

This is one of four skills in topic 2.8 of Which & Why: AP Calculus.