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Lecture 12: Higher Order Derivatives

GOAL: Compute derivatives of derivatives. Interpret the second and third derivative in a physics context.

Summary:

\[ \begin{gathered} \frac{d^n f}{dx^n} = \frac{d}{dx}\left(\frac{d^{n-1} f}{dx^{n-1}}\right) \\[0.5em] f^{(n)}(x) = [f^{(n-1)}(x)]' \\[0.5em] a(t) = v'(t) = s''(t) \quad\\[0.5em] j(t) = a'(t) = v''(t) = s'''(t) \end{gathered} \]

Recall that the height of a ball tossed straight into the air can be modeled using a parabola. Say

\[ s(t) = -4.9t^2 + 7t + 6 \]

Then the velocity of the ball is given by the derivative.

\[ v(t) = s'(t) = -9.8t + 7 \]

\( v(t) \) is a differentiable function, so we can continue this process

\[ v'(t) = -9.8 \]
Hand-drawn graph of \(s(t)\) rising from origin, peaking, then falling toward the horizontal axis.
Figure 1
A simple hand-drawn sketch on a coordinate axis system. A vertical line represents the y-axis and a horizontal line represents the t-axis, intersecting near the lower left. A small circle marks the origin point where the curve begins, labeled with the number 0. From this point, a curve labeled \(s(t)\) rises steeply, reaches a rounded peak near the top of the image, then descends in a curved arc toward the right side. The curve crosses the horizontal axis and continues slightly below it, ending with an arrowhead pointing downward, indicating the function continues to decrease past the axis. The overall shape resembles a single hump, illustrating the function \(s(t)\) increasing from zero, reaching a maximum, and then decreasing below the horizontal axis.
Hand-drawn graph of \(v(t)\) as a line starting at a dot on the vertical axis, sloping down through the horizontal axis.
Figure 2
A simple hand-drawn diagram with a vertical axis (y-axis) and a horizontal axis (x-axis) intersecting near the bottom left. On the vertical axis, above the horizontal axis, there is a small filled dot marking a starting point. From this dot, a straight diagonal line, labeled \(v(t)\), descends downward and to the right, crossing the horizontal axis, and ending in an arrowhead pointing further down and to the right below the horizontal axis. This suggests the line represents a quantity \(v(t)\) that starts at a positive value and decreases over time, eventually becoming negative.”},
Hand-drawn diagram with a vertical axis, a horizontal line above, and an arrow labeled \(a(t)\) from a marked origin point.
Figure 3
A simple hand-sketched diagram on a white background. A vertical curved line runs from top to bottom of the image, crossed near the top by a slightly wavy horizontal line extending to the right, forming a plus-like intersection. Below this intersection, near the bottom of the vertical line, there is a small filled circle marking a point of origin. From this circle, a horizontal arrow extends to the right, ending in an arrowhead, indicating direction or motion. Just above this arrow, the label \(a(t)\) is written in cursive-style handwriting, suggesting the arrow represents a time-dependent quantity such as acceleration as a function of time \(t\).

Definition The quantity

\[ \begin{gathered} \frac{d}{dx}\left(\frac{df}{dx}\right) = \frac{d^2 f}{dx^2} \\[0.5em] [f'(x)]' = f''(x) \end{gathered} \]

is called the second derivative of \( f \) with respect to \( x \).

Definition The acceleration of an object, whose position at time \( t \) is given by \( s(t) \), is defined by

\[ a(t) \stackrel{\text{def}}{=} v'(t) = s''(t) \]

Example 1

Find \( f''(x) \) if \( f(x) = 2\sin x + 3\cos x \)

\[ \begin{aligned} f(x) &= 2\sin x + 3\cos x \\ f'(x) &= 2\cos x - 3\sin x \\ f''(x) &= -2\sin x - 3\cos x \end{aligned} \]

NOTE

This is a fun fact (ie, you do not need to know this). If \( A \) and \( B \) are constants, then any function of the form \( f(x) = A\sin x + B\cos x \) satisfies the equation

\[ \frac{d^2 f}{dx^2} + f(x) = 0 \]

Example 2

A particle is traveling along a straight line; its position at time \( t \) is given by

\[ s(t) = t^3 - 6t^2 + 9t \]

  1. Find the velocity and acceleration of the particle at time \( t \).

    \[ \begin{gathered} v(t) = s'(t) = 3t^2 - 12t + 9 \\[0.5em] a(t) = v'(t) = 6t - 12 \end{gathered} \]

  2. If \( t \) is measured in seconds and \( s(t) \) is measured in meters, find \( v(4) \) and \( a(4) \).

\[ \begin{aligned} v(4) &= 3(4)^2 - 12(4) + 9 \\ &= 48 - 48 + 9 \\ &= 9\ \frac{m}{s} \end{aligned} \] \[ \begin{aligned} a(t) &= 6(4) - 12 \\ &= 12\ \frac{m/s}{s} \\ &= 12\ \frac{m}{s^2} \end{aligned} \]

  1. What is the acceleration when the velocity is \( 24\ \frac{m}{s} \)?

    Step 1 Find \( t \) when \( v(t) = 24 \)

    \[ \begin{aligned} v(t) = 3t^2 - 12t + 9 &\stackrel{\text{set}}{=} 24 \\ 3t^2 - 12t - 15 &= 0 \\ t^2 - 4t - 5 &= 0 \\ (t+1)(t-5) &= 0 \end{aligned} \]

    Thus, \( t = -1 \) or \( t = 5 \). As \( t \) represents time, we get \( v(t) = 24 \) when \( t = 5 \).

    Step 2 Find \( a(5) \)

    \[ \begin{aligned} a(5) &= 6(5) - 12 \\ &= 30 - 12 \\ &= 18\ \frac{m}{s^2} \end{aligned} \]

Example 3

Suppose the position of a particle is given by

\[ s(t) = \frac{t}{t^2+1} \qquad (t \ge 0) \]

When is the acceleration \( 0 \)?

\[ \begin{aligned} v(t) &= s'(t) \\ &= \frac{(1)(t^2+1) - t(2t)}{(t^2+1)^2} \end{aligned} \]

NOTE:

When you need to take multiple derivatives, it is generally a good idea to simplify as much as possible before you continue.

\[ \begin{aligned} v(t) &= \frac{t^2+1 - 2t^2}{(t^2+1)^2} \\ &= \frac{1-t^2}{(t^2+1)^2} \end{aligned} \] \[ \begin{aligned} a(t) &= \frac{(-2t)(t^2+1)^2 - (1-t^2)(2)(t^2+1)(2t)}{(t^2+1)^4} \\ &= \frac{-2t(t^2+1)\left[(t^2+1) + (1-t^2)(2)\right]}{(t^2+1)^4} \\ &= \frac{-2t(t^2+1)\left[3-t^2\right]}{(t^2+1)^4} \stackrel{\text{set}}{=} 0 \end{aligned} \]

The denominator is never \( 0 \), so we only need to check when the numerator is \( 0 \). We also disregard any negative solutions as \( t \ge 0 \).

\[ -2t = 0 \quad \text{OR} \quad t^2+1 = 0 \quad \text{OR} \quad 3-t^2 = 0 \] \[ \boxed{t = 0} \qquad \text{Never} \qquad \begin{aligned} &t^2 = 3 \\ &t = \pm\sqrt{3}\\ &\boxed{t =~ \sqrt{3}} \end{aligned} \]

Higher Order Derivatives

As long as the derivative is differentiable, there is nothing stopping us from taking more derivatives.

NOTATION:

3rd Derivative:

\[ \begin{gathered} \frac{d^3 f}{dx^3} \stackrel{\text{def}}{=\!=} \frac{d}{dx}\left( \frac{d^2 f}{dx^2} \right) \\[0.5em] f'''(x) = \left[ f''(x) \right]' \end{gathered} \]

4th Derivative:

\[ \begin{gathered} \frac{d^4 f}{dx^4} = \frac{d}{dx}\left( \frac{d^3 f}{dx^3} \right) \\[0.5em] f''''(x) = \left[ f'''(x) \right]' \end{gathered} \]

nth Derivative:

\[ \begin{gathered} \frac{d^n f}{dx^n} = \frac{d}{dx}\left( \frac{d^{n-1} f}{dx^{n-1}} \right) \\[0.5em] f^{(n)}(x) = \left[ f^{(n-1)}(x) \right]' \end{gathered} \]

NOTE:

In a physics context, the 3rd Derivative of a position function \( s(t) \) is called the jerk at time \( t \), denoted by:

\[ j(t) \stackrel{\text{def}}{=\!=} a'(t) = v''(t) = s'''(t) \]

Example 4

If \( f^{(9)}(x) = 6\sec(2x-5) \), find \( f^{(10)}(x) \).

\[ \begin{aligned} f^{(10)}(x) = \left[ f^{(9)}(x) \right]' &= 6\sec(2x-5)\tan(2x-5)\,(2) \\ &= 12\sec(2x-5)\tan(2x-5) \end{aligned} \]

Example 5

Examine the various derivatives of \( f(x) = \sin x \)

\[ \begin{aligned} f(x) &= \sin x \\ f'(x) &= \cos x \\ f''(x) &= -\sin x \\ f'''(x) &= -\cos x \\ f^{(4)}(x) &= \sin x \end{aligned} \]

So we get a loop:

Circular loop diagram showing sin x to cos x to minus sin x to minus cos x and back, labeled d/dx
Figure 4
A hand-drawn circular loop diagram illustrating the cycle of derivatives of the sine function. Four expressions are arranged around a circle: \( \sin x \) at the top, \( \cos x \) at the right, \( -\sin x \) at the bottom, and \( -\cos x \) at the left. Curved arrows connect them clockwise: from \( \sin x \) to \( \cos x \), from \( \cos x \) to \( -\sin x \), from \( -\sin x \) to \( -\cos x \), and from \( -\cos x \) back up to \( \sin x \). The arrow from \( \sin x \) to \( \cos x \) is labeled \( \frac{d}{dx} \), indicating that each arrow represents taking a derivative.

Example 6

Repeat for the function \( f(x) = x^4 + x^3 + x^2 + x + 1 \)

\[ \begin{aligned} f(x) &= x^3 + x^2 + x + 1 \\ f'(x) &= 3x^2 + 2x + 1 \\ f''(x) &= 6x + 2 \\ f'''(x) &= 6 \\ f^{(4)}(x) &= 0 \end{aligned} \]

So \(f\) (and polynomials in general) go to \( 0 \) if you take enough (1 + the degree of the polynomial) derivatives

Chain diagram of boxed polynomial derivatives ending at 0 with a self-loop labeled d/dx
Figure 5
A vertical flow diagram of successive derivatives. At the top is a boxed expression \( x^3 + x^2 + x + 1 \). A downward arrow labeled \( \frac{d}{dx} \) leads to a boxed \( 3x^2 + 2x + 1 \). Another downward arrow labeled \( \frac{d}{dx} \) leads to a boxed \( 6x + 2 \). Another downward arrow labeled \( \frac{d}{dx} \) leads to a boxed \( 6 \). A final downward arrow labeled \( \frac{d}{dx} \) leads to a boxed \( 0 \). From the box containing 0, a curved loop leaves and returns to the same box with an arrowhead pointing back into it, labeled \( \frac{d}{dx} \), indicating that differentiating zero repeatedly gives zero.

Example 7

Repeat for the function \( f(x) = e^{2x} \)

\[ \begin{aligned} f(x) &= e^{2x} \\ f'(x) &= e^{2x}(2) = 2e^{2x} \\ f''(x) &= 2e^{2x}(2) = 2^2 e^{2x} \\ f'''(x) &= 2^2 e^{2x}(2) = 2^3 e^{2x} \\ &\;\;\vdots \\ f^{(n)}(x) &= 2^{n-1} e^{2x}(2) = 2^n e^{2x} \end{aligned} \]

So we get a never ending chain

Vertical chain of boxed expressions from e to the 2x down to 2 to the n+1 times e to the 2x, each arrow labeled d/dx
Figure 6
A hand-drawn vertical flow chart of successive derivatives. At the top is a box containing \( e^{2x} \). A downward arrow labeled \( \frac{d}{dx} \) leads to a box containing \( 2e^{2x} \). Another arrow labeled \( \frac{d}{dx} \) leads to a box containing \( 2^2 e^{2x} \). Another arrow labeled \( \frac{d}{dx} \) leads to a box containing \( 2^3 e^{2x} \). A further arrow labeled \( \frac{d}{dx} \) leads to a vertical ellipsis of three dots, followed by another arrow labeled \( \frac{d}{dx} \) leading to a box containing \( 2^n e^{2x} \). An arrow labeled \( \frac{d}{dx} \) leads to a box containing \( 2^{n+1} e^{2x} \), and a final arrow labeled \( \frac{d}{dx} \) points down to a vertical ellipsis, indicating the chain continues forever.

Supplemental Examples from the Homework

Example 8 (HW 12, #2)

Find \( g''(x) \) if \(g(x) = 4e^{3x}\cos(4x)\)

\begin{align*} g(x) = 4e^{3x} \cos(4x) \end{align*} \begin{align*} g'(x) =&~ 4e^{3x}(3) \cos(4x) + 4e^{3x} (-\sin(4x))(4)\\[0.5em] =&~ 12e^{3x} \cos(4x) - 16 e^{3x}\sin(4x)\\[0.5em] =&~ 4e^{3x}[3 \cos(4x) - 4\sin(4x)] \end{align*} \begin{align*} g''(x) =&~ 12e^{3x} [3\cos(4x) - 4\sin(4x)] + 4e^{3x}[-12\sin(4x) - 16\cos(4x)]\\[0.5em] =&~ (36 - 64)e^{3x}\cos(4x) + (-48 - 48) e^{3x}\sin(4x)\\[0.5em] =&~ -28e^{3x} \cos(4x) - 96e^{3x}\sin(4x) \end{align*}