Wemazi

Topic 2.9

Check a simplified derivative

Builds on using a quotient derivative at a point. 7 minutes.

Two routes to one derivative

The function is \(f(x) = \dfrac{x^2 - 9}{x - 3}\). Drag \(x\), and the orange line rises to the curve. At \(x = 3\) the bottom is 0, so \(f\) has no value there, and the open circle marks the missing point.

Route 1 uses the quotient rule. The top of the rule is \(2x(x-3) - (x^2-9) = x^2 - 6x + 9\), which is \((x-3)^2\), so \(f' = (x-3)^2/(x-3)^2 = 1\).

Route 2 simplifies first. The top is \((x-3)(x+3)\), so \(f(x) = x + 3\) for \(x \ne 3\). The first button draws the orange dashed line \(y = x + 3\), which lies on the curve and has the slope 1.

Both routes give 1. The second button shows the slope graph, a line at height 1 with an open circle at \(x = 3\), where \(f\) has no slope.

Two forms of one function

Two expressions are equivalent when they have the same value at every input where the original function has a value.

Drag \(x\). The orange dashed curve lies on the chalk curve at every input except the open circle, where a cancelled factor is 0.

The original function has no value at the open circle, so the simplified form does not apply there, although it has a value.

Notation

  1. First list the inputs that the original function excludes. The bottom of \(f\) is 0 at \(x = 3\).

    \[f(x) = \frac{x^2 - 9}{x - 3}, \qquad x \ne 3\]

  2. Route 1 applies the quotient rule and simplifies the top. The figure now shows the graph of \(f'\), a line at height 1.

    \[\begin{aligned}f'(x) &= \frac{2x(x-3) - (x^2-9)(1)}{(x-3)^2} \\ &= \frac{(x-3)^2}{(x-3)^2}\end{aligned}\]

  3. Route 2 factors the top and cancels \(x - 3\). The cancel is valid only where \(x - 3\) is not 0.

    \[\begin{aligned}f(x) &= \frac{(x-3)(x+3)}{x-3} \\ &= x + 3 \quad (x \ne 3)\end{aligned}\]

  4. The derivative of \(x + 3\) is 1, the same as route 1, so the answer is confirmed. Keep the exclusion in the answer.

    \[f'(x) = \cy{1} \quad (x \ne 3)\]

    The constant 1 has a value at \(x = 3\), but \(f'(3)\) does not exist, because \(f(3)\) has no value.

When it fails

At \(x = 3\), the orange line reaches the dashed line at height 6, but the original function has no value there. The orange point becomes a ring around the empty circle.

The simplified form is correct for \(x \ne 3\) only. A formula that has a value at an excluded input does not restore that input.

Example 1: divide first

Find \(f'(x)\) for \(f(x) = \dfrac{x^3 + x^2}{x}\) by two routes. State where the answer holds.

  1. In route 1, the top of the quotient rule is \((3x^2 + 2x)\,x - (x^3 + x^2)(1)\).

    \[= 2x^3 + x^2\]

  2. The bottom is \(x^2\). Divide the top and the bottom by \(x^2\), for \(x \ne 0\).

    \[\begin{aligned}f'(x) &= \frac{2x^3 + x^2}{x^2} \\ &= 2x + 1 \quad (x \ne 0)\end{aligned}\]

  3. Route 2 divides each term of the top by \(x\) first. Then \(f(x) = x^2 + x\) for \(x \ne 0\), so the derivative needs no quotient rule.

    \[f'(x) = 2x + 1 \quad (x \ne 0)\]

  4. The routes agree. Route 2 is shorter, because the bottom divides every term of the top.

  5. The original function has no value at 0, so \(f'(0)\) does not exist, although \(2x + 1\) is 1 at 0.

Example 2: a hidden exclusion

Find \(f'(x)\) for \(f(x) = \dfrac{x^2 - 1}{x^2 - x}\). Then decide whether \(f'(1)\) exists.

  1. The top factors as \((x-1)(x+1)\) and the bottom as \(x(x-1)\). The bottom is 0 at \(x = 0\) and at \(x = 1\), so \(f\) excludes both.

    \[x \ne 0, \quad x \ne 1\]

  2. Cancel \(x - 1\), which is allowed for \(x \ne 1\). Then \(f(x) = (x+1)/x = 1 + 1/x\).

    \[f(x) = 1 + \frac{1}{x}\]

  3. The derivative of \(1/x\) is \(-1/x^2\).

    \[f'(x) = -\frac{1}{x^2}\]

    The quotient rule on the original expression gives a top of \(-(x-1)^2\) over \(x^2(x-1)^2\), which cancels to the same result.

  4. A tempting wrong move: The formula is defined at \(x = 1\) and gives \(-1\). So \(f'(1) = -1\).

    \[f'(1) = -\frac{1}{1^2} = -1\]

    The graph of \(f\) has an open circle at \(x = 1\), so there is no point, no tangent line and no slope there.

  5. Keep both exclusions. The answer is \(f'(x) = -1/x^2\) for \(x \ne 0, 1\), and \(f'(1)\) does not exist.

Practice

For \(g(x) = \dfrac{x^3 - 8}{x - 2}\), show that the quotient rule and simplifying first give the same derivative. State where it holds.

Factor the top as \((x-2)(x^2 + 2x + 4)\). The graph of \(g\) has an open circle at \(x = 2\).

Solution

  1. For route 2, cancel \(x - 2\), which is allowed for \(x \ne 2\). Then \(g(x) = x^2 + 2x + 4\).

    \[g'(x) = 2x + 2 \quad (x \ne 2)\]

  2. For route 1, substitute into the quotient rule. Then multiply out the top.

    \[\begin{aligned}&3x^2(x-2) - (x^3-8) \\ &= 2x^3 - 6x^2 + 8 \\ g'(x) &= \frac{2x^3 - 6x^2 + 8}{(x-2)^2}\end{aligned}\]

  3. Factor the top. It is \(2(x-2)^2(x+1)\), because \((x-2)^2(x+1) = x^3 - 3x^2 + 4\).

    \[\begin{aligned}g'(x) &= \frac{2(x-2)^2(x+1)}{(x-2)^2} \\ &= 2x + 2 \quad (x \ne 2)\end{aligned}\]

  4. The routes agree for \(x \ne 2\). The expression \(2x + 2\) is 6 at \(x = 2\), but \(g'(2)\) does not exist.

Questions

The figure shows \(h(x) = \dfrac{x^3 - x}{x + 1}\) and its simplified form, with a tangent at the marked point. The point moves to the input that each question names. Questions 1 and 2 use the graph, and question 3 needs a calculation.

  1. At the marked point \(x = 2\), the tangent in the figure has slope 3. Which expression for \(h'(x)\) gives 3 at \(x = 2\)?

    • \(2x - 1\)
    • \(3x^2 - 1\)
    • \(2x + 1\)
    • \(x^2 - x\)
    Answer

    \(2x - 1\)

  2. The marked point is now the open circle at \(x = -1\). What is \(h'(-1)\)?

    • It does not exist, because \(h(-1)\) has no value
    • \(-3\), the value of \(2x - 1\) at \(-1\)
    • 0
    Answer

    It does not exist, because \(h(-1)\) has no value

  3. Which statement is correct for \(k(x) = \dfrac{4x^2 - 9}{2x + 3}\)?

    • \(k'(x) = 2\) for every \(x \ne -\tfrac32\)
    • \(k'(x) = 2\) for every \(x\)
    • \(k'(x) = 8x\)
    • \(k'(x) = 2x - 3\)
    Answer

    \(k'(x) = 2\) for every \(x \ne -\tfrac32\)

Summary

You find the same derivative by the quotient rule and by simplifying first, and you compare the two results as expressions.

The comparison holds only where the original function has a value, so the answer excludes every input that the original excludes.

You use this when an answer looks different from the key, or when a cancelled factor could be 0.

This is one of four skills in topic 2.9 of Which & Why: AP Calculus.