Wemazi

Topic 3.6

Use higher-derivative notation

Builds on finding a second derivative. 6 minutes.

Counting the differentiations

The ladder starts with \(f(x) = x^5\). Press Differentiate once. The orange arrow shows one differentiation, and the orange marker moves down to its result.

After one press the marked row reads \(f'(x) = 5x^4\), and after two presses it reads \(f''(x) = 20x^3\).

The order of a derivative is the number of times you differentiate, and the name of the row counts it. The exponent of \(x\) is a different number, because it belongs to the formula and falls by 1 at each press.

After five presses the formula is the constant 120, and a sixth press gives 0.

Two numbers that look alike

The ladder now shows only names, three for each row, so \(f''\), \(f^{(2)}\) and \(\dfrac{d^2y}{dx^2}\) all name the second derivative.

Press Differentiate once to move down a row. In every name, the primes, the number in parentheses or the 2 in \(d^2y\) and \(dx^2\) give the order, which is the number of differentiations.

An exponent outside the name is a power. Press Show the powers to see \(f(x)^2\) and \(\left(\dfrac{dy}{dx}\right)^2\) in the faint dashed row. They multiply a value by itself, so they are not rows of the ladder.

Notation

  1. Prime marks count the differentiations, so the row \(f''\) is the derivative of \(f'\), and \(f'''\) is the derivative of \(f''\).

    \[f'' = (f')', \qquad f''' = (f'')'\]

  2. For a higher order, write the order in parentheses. The symbol \(f^{(n)}\) means differentiate \(n\) times.

    \[\cy{f^{(4)}} = (f''')'\]

    Without parentheses, \(f(x)^2\) is a power, the square of the value.

  3. In Leibniz notation, the 2 in \(d^2y\) and in \(dx^2\) counts the differentiations. It is not the square of \(\dfrac{dy}{dx}\).

    \[\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right)\]

  4. To evaluate \(f^{(n)}(a)\), differentiate \(n\) times while \(x\) is still a variable. Then substitute \(x = a\). Each derivative up to the order \(n\) must exist at \(x = a\).

When it fails

Take \(f(x) = x^{8/3}\). Then \(f'(x) = \tfrac83x^{5/3}\) and \(f''(x) = \tfrac{40}{9}x^{2/3}\), and both are defined at \(x = 0\).

The third derivative is \(f'''(x) = \tfrac{80}{27}x^{-1/3}\). It is not defined at \(x = 0\), so \(f'''(0)\) does not exist. The red outlines mark the formulas that are not defined at 0.

Example 1: order and power

Let \(y = x^3 - 4x\). Find \(\dfrac{d^2y}{dx^2}\) at \(x = 2\), and compare it with \(\left(\dfrac{dy}{dx}\right)^2\) at \(x = 2\).

  1. Differentiate once. Then substitute \(x = 2\).

    \[\frac{dy}{dx} = 3x^2 - 4, \qquad 3(2)^2 - 4 = \cy{8}\]

  2. Differentiate \(3x^2 - 4\) again. Then substitute \(x = 2\).

    \[\frac{d^2y}{dx^2} = 6x, \qquad 6(2) = \cy{12}\]

  3. The faint dashed row shows the square of the first derivative, \(8^2 = 64\). This is not 12, so \(\dfrac{d^2y}{dx^2}\) is not \(\left(\dfrac{dy}{dx}\right)^2\).

    \[\left(\frac{dy}{dx}\right)^2 = 64\]

Example 2: when to substitute

Let \(f(x) = \tfrac12x^4 - 3x\). Find \(f''(1)\).

  1. Differentiate once. The top graph shows the result, \(f'\), which is a curve, and the bottom graph shows the slope of that curve.

    \[f'(x) = 2x^3 - 3\]

  2. A tempting wrong move: Substitute \(x = 1\) now. The value \(f'(1) = -1\) is a constant, and a constant has derivative 0.

    \[f''(1) = (-1)' = \cr{0}\]

    The flat red line shows the slope this move claims, but the orange tangent to the curve at \(x = 1\) has slope 6. Substituting first replaces the curve by the single number \(-1\), which says nothing about how \(f'\) changes.

  3. Differentiate \(f'\) while \(x\) is still a variable. Then substitute \(x = 1\), where the bottom graph is at 6.

    \[f''(x) = 6x^2, \qquad f''(1) = \cy{6}\]

Practice

Let \(f(x) = x^6 - x^2\). Find \(f^{(4)}(1)\).

Count four arrows on the ladder before you substitute \(x = 1\).

Solution

  1. Differentiate four times, one row at a time.

    \[\begin{aligned} f' &= 6x^5 - 2x, \quad f'' = 30x^4 - 2, \\ f''' &= 120x^3, \quad f^{(4)} = 360x^2 \end{aligned}\]

  2. Substitute \(x = 1\) into the fourth derivative.

    \[f^{(4)}(1) = 360\]

Questions

The figure shows the ladder for \(f(x) = x^5\) and \(y = f(x)\).

  1. The marker is on the row \(60x^2\). Which name belongs to this row?

    • \(f'''(x)\)
    • \(f(x)^3\)
    • \(f^{(60)}(x)\)
    Answer

    \(f'''(x)\)

  2. What is \(\left(\dfrac{dy}{dx}\right)^2\)?

    • \(25x^8\)
    • \(20x^3\)
    • \(5x^8\)
    Answer

    \(25x^8\)

  3. Let \(g(x) = x^4 - 3x^3\). What is \(g'''(1)\)?

    • 6
    • \(-6\)
    • 24
    • \(-125\)
    Answer

    6

Summary

The order of a derivative counts differentiations, so \(f''\), \(f^{(2)}\) and \(\dfrac{d^2y}{dx^2}\) all name the row of order 2, \(20x^3\).

An exponent outside the name, as in \(\left(\dfrac{dy}{dx}\right)^2 = 25x^8\), is a power of the value and not a row of the ladder.

To evaluate a derivative of order \(n\), you differentiate \(n\) times while \(x\) is a variable, and then you substitute.

This is one of four skills in topic 3.6 of Which & Why: AP Calculus.