Wemazi

Topic 3.5

Choose the outermost derivative rule

Builds on using the product, quotient and chain rules one at a time. 7 minutes.

The last operation sets the rule

The figure computes \(p(x) = x^2\sqrt{x+1}\) one operation at a time, starting at \(x = 3\).

One branch squares \(x\), which gives 9. The other branch adds 1 and takes the square root, which gives 2.

The last operation multiplies 9 by 2 to give 18. This last operation is called the outermost operation, and the figure outlines it in orange.

Drag \(x\). The numbers change, but the last operation stays a multiplication, so \(p\) is a product and the product rule comes first.

Inside the second factor, \(\sqrt{x+1}\) is a composition, which is a function placed inside another function, so it needs the chain rule second.

Four possible last operations

The letters \(u\) and \(v\) now stand for any two functions. The last operation can add them, multiply them, divide them, or put one inside the other.

Press each button to mark a row of the table, which names the rule for that last operation.

Whatever the functions are, the rule for the last operation comes first, and the pieces inside then use their own rules.

Notation

  1. If the last operation is a sum, differentiate each part.

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

  2. If it is a product, add two terms. Each term differentiates one factor and keeps the other. For \(x^3 = x\cdot x^2\), the curve in the figure, the terms add to the slope 3 at \(x = 1\).

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

    Both terms appear because both factors change.

  3. If it is a quotient, the numerator is the derivative of the top times the bottom, minus the top times the derivative of the bottom. Divide it by the bottom squared.

    \[\cy{\left(\dfrac uv\right)'} = \dfrac{u'v - uv'}{v^2}\]

  4. If it is a composition, differentiate the outer function at the inside, then multiply by the derivative of the inside.

    \[\cy{\big(f(g(x))\big)'} = f'(g(x))\,g'(x)\]

  5. Each rule needs its functions to be differentiable at \(x\). A quotient also needs \(v(x) \ne 0\), because you cannot divide by 0.

When it fails

A common mistake is to multiply the derivatives, which would make \((uv)'\) equal to \(u'v'\). Test it with \(u = x\) and \(v = x^2\), so \(uv = x^3\), which has slope 3 at \(x = 1\) (the orange tangent).

Multiplying the derivatives gives \(1\cdot 2x = 2x\), which is 2 at \(x = 1\), so the red line cuts across the curve. The product rule gives \(1\cdot x^2 + x\cdot 2x = 3x^2\), which is 3.

Example 1: a quotient

Find \(g'(x)\) for \(g(x) = \dfrac{x^2 - 4}{x + 3}\).

  1. The last operation divides \(x^2 - 4\) by \(x + 3\), so the quotient rule applies. The top is \(u = x^2 - 4\) with \(u' = 2x\), and the bottom is \(v = x + 3\) with \(v' = 1\).

  2. Substitute the pieces into the rule, which holds for \(x \ne -3\), where the bottom is not 0.

    \[g'(x) = \dfrac{2x(x+3) - (x^2 - 4)\cdot 1}{(x+3)^2}\]

  3. Expand the top: \(2x^2 + 6x - x^2 + 4\) is \(x^2 + 6x + 4\).

    \[g'(x) = \dfrac{x^2 + 6x + 4}{(x+3)^2}\]

  4. At \(x = 1\), the top is 11 and the bottom is 16, so the orange tangent has slope \(\tfrac{11}{16}\).

    \[g'(1) = \dfrac{11}{16}\]

Example 2: divide first

Find \(h'(x)\) for \(h(x) = \dfrac{x^3 + 2x^2}{x}\), where \(x \ne 0\).

  1. The last operation is a division. Before you use the quotient rule, check whether the top divides by the bottom, because a simpler form has an easier derivative.

  2. The top is \(x^2(x+2)\). For \(x \ne 0\), dividing by \(x\) leaves \(x^2 + 2x\), so the graph of \(h\) is this parabola with a hole at \(x = 0\).

  3. A tempting wrong move: Differentiate the top and the bottom separately. Then divide the results.

    \[h'(x) = \dfrac{3x^2 + 4x}{1}\]

    At \(x = 1\) this gives 7. The parabola \(x^2 + 2x\) has slope 4 at \(x = 1\), so the red line is too steep. The derivative of a quotient is not the quotient of the derivatives.

  4. Differentiate the simpler form instead. The quotient rule gives the same result, because its numerator simplifies to \(2x^3 + 2x^2\).

    \[\begin{aligned} h'(x) &= 2x + 2 \\ \dfrac{2x^3 + 2x^2}{x^2} &= 2x + 2 \end{aligned}\]

Practice

Find \(p'(x)\) for \(p(x) = (x^2 + 1)\sin x\). Name the last operation first.

Move the slider to x = 2 and read the orange slope. Your formula must give that value.

Solution

  1. The last operation is a product. Let \(u = x^2 + 1\) and \(v = \sin x\), so \(u' = 2x\) and \(v' = \cos x\).

  2. Apply the product rule: \(u'v + uv'\).

    \[p'(x) = 2x\sin x + (x^2 + 1)\cos x\]

  3. As a check, \(p'(2) \approx 1.556\), which matches the orange slope at \(x = 2\).

Questions

The figure shows \(y = x(4 - x)\) with a tangent at the point each question names.

  1. The tangent at \(x = 2\) is horizontal. Which calculation of the derivative of \(x(4-x)\) at \(x = 2\) gives that slope?

    • \(u'v + uv' = 1\cdot 2 + 2\cdot(-1)\)
    • \(u'v' = 1\cdot(-1)\)
    • \(u'v - uv' = 1\cdot 2 - 2\cdot(-1)\)
    • \(uv' = 2\cdot(-1)\)
    Answer

    \(u'v + uv' = 1\cdot 2 + 2\cdot(-1)\)

  2. At \(x = 3\) the curve falls. The function \(x(4-x)\) is also \(4x - x^2\). What is the slope at \(x = 3\)?

    • −2
    • 2
    • −1
    • 4
    Answer

    −2

  3. What is the derivative of \(\dfrac{e^x}{x}\) at \(x = 2\)?

    • \(\dfrac{e^2}{4}\)
    • \(-\dfrac{e^2}{4}\)
    • \(e^2\)
    • \(\dfrac{3e^2}{4}\)
    Answer

    \(\dfrac{e^2}{4}\)

Summary

The last operation in an expression decides which rule you apply first.

This works because each rule treats its pieces as whole functions, and each piece then takes its own rule.

Name the last operation at the start of every derivative that mixes operations.

This is one of four skills in topic 3.5 of Which & Why: AP Calculus.