Higher Order Derivatives The derivative \(f'(x)\) of a differentiable function \(f(x)\) can be thought of as a function in its own right, and if it is differentiable then its derivative—denoted by \(f''(x)\)—is the second derivative of \(f(x)\) (the first derivative being \(f'(x)\)). Likewise, the derivative of \(f''(x)\) would be the third derivative of \(f(x)\), written as \(f'''(x)\). Continuing like this yields the fourth derivative, fifth derivative, and so on. In general the \(\bm{n}\)-th derivative of \(f(x)\) is obtained by differentiating \(f(x)\) a total of \(n\) times. Derivatives beyond the first are called higher order derivatives. For \(f(x) = 3x^4\) find \(f''(x)\) and \(f'''(x)\). Solution: Since \(f'(x) = 12x^3\) then the second derivative \(f''(x)\) is the derivative of \(12x^3\), namely:
\[f''(x) ~=~ 36x^2 \nonumber \]
The third derivative \(f'''(x)\) is then the derivative of \(36x^2\), namely:
\[f'''(x) ~=~ 72x \nonumber \]
Since the prime notation for higher order derivatives can be cumbersome (e.g. writing 50 prime marks for the fiftieth derivative), other notations have been created: Notice that the parentheses around \(n\) in the notation \(f^{(n)}(x)\) indicate that \(n\) is not an exponent—it is the number of derivatives to take. The \(n\) in the Leibniz notation \(\frac{d^ny}{\dx^n}\) indicates the same thing, and in general makes working with higher order derivatives easier:
\[\begin{aligned} \frac{d^2y}{\dx^2} ~&=~ \ddx\,\left(\dydx\right)\
\[4pt] \frac{d^3y}{\dx^3} ~&=~ \ddx\,\left(\frac{d^2y}{\dx^2}\right) ~=~ \frac{d^2}{\dx^2}\,\left(\dydx\right)\\ &\vdots\\ \frac{d^ny}{\dx^n} ~&=~ \ddx\,\left(\frac{d^{n-1}y}{\dx^{n-1}}\right) ~=~ \frac{d^{n-1}}{\dx^{n-1}}\,\left(\dydx\right)\end{aligned} \nonumber \]
A natural question to ask is: what do higher order derivatives represent? Recall that the first derivative \(f'(x)\) represents the instantaneous rate of change of a function \(f(x)\) at the value \(x\). So the second derivative \(f''(x)\) represents the instantaneous rate of change of the function \(f'(x)\) at the value \(x\). In other words, the second derivative is a rate of change of a rate of change. The most famous example of this is for motion in a straight line: let \(s(t)\) be the position of an object at time \(t\) as the object moves along the line. The motion can take two directions, e.g. forward/backward or up/down. Take one direction to represent positive position and the other to represent negative direction, as in the drawing on the right. The (instantaneous) velocity \(v(t)\) of the object at time \(t\) is \(s'(t)\), i.e. the first derivative of \(s(t)\). The acceleration \(a(t)\) of the object at time \(t\) is defined as \(v'(t)\), the instantaneous rate of change of the velocity. Thus, \(a(t) = s''(t)\), i.e. acceleration is the second derivative of position. To summarize:
Example \(\PageIndex{1}\): accel
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Solution
Ignoring wind and air resistance, the position \(s\) of a ball thrown straight up with an initial velocity of 34 m/s from a starting point 2 m off the ground is given by \(s(t) = -4.9t^2 + 34t + 2\) at time \(t\) (measured in seconds) with \(s\) measured in meters. Find the velocity and acceleration of the ball at any time \(t \ge 0\). Solution: The ball moves in a straight vertical line, first straight up then straight down until it hits the ground. Its velocity \(v(t)\) is
\[\begin{aligned} v(t) ~&=~ \dsdt ~=~ -9.8t ~+~ 34 ~~\text{m/s} \intertext{while its acceleration $a(t)$ is} a(t) ~&=~ \frac{d^2s}{\dt^2} ~=~ \ddt\,\left(\dsdt\right) ~=~ \ddt\,(-9.8t ~+~ 34) ~=~ -9.8 ~~\text{m/s}^2\text{,} \end{aligned} \nonumber \]
which is the acceleration due to the force of gravity on Earth. Note that time \(t = 0\) is the time at which the ball was thrown, so that \(v(0)\) is the initial velocity of the ball. Indeed, \(v(0) = -9.8(0) + 34 = 34\) m/s, as expected. Notice in Example
Example \(\PageIndex{1}\): accel
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Solution
that the acceleration of the ball is not only constant but also negative. To see why this makes sense, first consider the case where the ball is moving upward. The ball has an initial upward velocity of 34 m/s then slows down to 0 m/s at the instant it reaches its maximum height above the ground. So the velocity is decreasing, i.e. its rate of change—the acceleration—is negative. The ball reaches its maximum height above the ground when its velocity is zero, that is, when \(v(t) = -9.8t + 34 = 0\), i.e. at time \(t = 34/9.8 = 3.47\) seconds after being thrown (see the picture above). The ball then starts moving downward and its velocity is negative (e.g. at time \(t = 4\) s the velocity is \(v(4) = -9.8(4) + 34 = -5.2\) m/s). Recall that negative velocity indicates downward motion, while positive velocity means the motion is upward (away from the Earth’s center). So in the case where the ball begins moving downward it goes from 0 m/s to a negative velocity, with the ball moving faster towards the ground, which it hits with a velocity of \(-33.43\) m/s (why?). So again the velocity is decreasing, which again means that the acceleration is negative. Common terminology involving motion might cause some confusion with the above discussion. For example, even though the ball’s acceleration is negative as it falls to the ground, it is common to say that the ball is accelerating in that situation, not decelerating (as the ball is doing while moving upward). In general, acceleration is understood to mean that the magnitude (i.e. the absolute value) of the velocity is increasing. That magnitude is called the speed of the object. Deceleration means the speed is decreasing. The first and second derivatives of an object’s position with respect to time represent the object’s velocity and acceleration, respectively. Do the third, fourth, and other higher order derivatives have any physical meanings? It turns out they do. The third derivative of position is called the jerk of the object. It represents the rate of change of acceleration, and is often used in fields such as vehicle dynamics (e.g. minimizing jerk to provide smoother braking). The fourth, fifth, and sixth derivatives of position are called snap, crackle, and pop, respectively.28 The zero-th derivative \(f^{(0)}(x)\) of a function \(f(x)\) is defined to be the function \(f(x)\) itself: \(f^{(0)}(x) = f(x)\). There is a way to define fractional derivatives, e.g. the one-half derivative \(f^{(1/2)}(x)\), which will be discussed in Chapter 6. An immediate consequence of the definition of higher order derivatives is: Recall that the factorial \(n!\) of an integer \(n > 0\) is the product of the integers from 1 to \(n\):
\[n! ~=~ 1 \cdot 2 \cdot 3 \cdot \;\cdots\; \cdot n \nonumber \]
For example:
\[\begin{aligned} {3} 1! ~&=~ 1 \qquad\qquad& 3! ~&=~ 1 \cdot 2 \cdot 3 ~=~ 6\\ 2! ~&=~ 1 \cdot 2 ~=~ 2 \qquad\qquad& 4! ~&=~ 1 \cdot 2 \cdot 3 \cdot 4 ~=~ 24\end{aligned} \nonumber \]
By convention \(0!\) is defined to be 1. The following statement can be proved using induction: Thus,
\[\dfrac{d^{n+1}}{\dx^{n+1}}\,\left(x^n\right) ~=~ \ddx\,\left(\dfrac{d^n}{\dx^n}\,\left(x^n\right)\right) ~=~ \ddx\,(n!) ~=~ 0 \nonumber \]
for all integers \(n \ge 0\), since \(n!\) is a constant. Polynomials are linear combinations of nonnegative powers of a variable (e.g. \(x\)), so the above result combined with the Sum Rule and the Constant Multiple rule—which also hold for higher-order derivatives—yields this important fact: This is the basis for the commonly-used statement that “any polynomial can be differentiated to 0” by taking a sufficient number of derivatives. For example, differentiating the polynomial \(p(x) = 100x^{100} + 50x^{99}\) 101 times would yield 0 (as would differentiating more than 101 times). [sec1dot6] For Exercises 1-6 find the second derivative of the given function. 3 \(f(x) ~=~ x^3 ~+~ x^2 ~+~ x ~+~ 1\) \(f(x) ~=~ x^2\,\sin x\) \(f(x) ~=~ \cos 3x\) 3 \(f(x) ~=~ \dfrac{\sin x}{x}\vphantom{\dfrac{G m_1 m_2}{r^2}}\) \(f(x) ~=~ \dfrac{1}{x}\vphantom{\dfrac{G m_1 m_2}{r^2}}\) \(F(r) ~=~ \dfrac{G m_1 m_2}{r^2}\) Find the first five derivatives of \(f(x) = \sin x\). Use those to find \(f^{(100)}(x)\) and \(f^{(2014)}(x)\). Find the first five derivatives of \(f(x) = \cos x\). Use those to find \(f^{(100)}(x)\) and \(f^{(2014)}(x)\). If an object moves along a straight line such that its position \(s(t)\) at time \(t\) is directly proportional to \(t\) for all \(t\) (written as \(s \propto t\)), then show that the object’s acceleration is always 0. [exer:dnxn] Use induction to show that \(\frac{d^n}{\dx^n}\,\left(x^n\right) ~=~ n!~\) for all integers \(n \ge 1\). Show that for all integers \(n \ge m \ge 1\), \(\frac{d^{m}}{\dx^{m}}\,\left(x^n\right) ~=~ \frac{n!}{(n - m)!}\,x^{n - m} ~\). Find the general expression for the \(n\)-th derivative of \(f(x) = \frac{1}{ax + b} ~\) for all constants \(a\) and \(b\) (\(a \ne 0\)). Show that the function \(y = A \cos\,(\omega t + \phi) ~+~ B \sin\,(\omega t + \phi)\) satisfies the differential equation
\[\frac{d^2y}{\dt^2} ~+~ \omega^2 y ~=~ 0 \nonumber \]
for all constants \(A\), \(B\), \(\omega\), and \(\phi\). If \(s(t)\) represents the position at time \(t\) of an object moving along a straight line, then show that:
\[\begin{aligned} {2} s' ~\text{and}~ s'' ~&\text{have the same sign} \quad&&\Rightarrow\quad \text{the object is accelerating}\\ s' ~\text{and}~ s'' ~&\text{have opposite signs} &&\Rightarrow\quad \text{the object is decelerating}\end{aligned} \nonumber \]
For all twice-differentiable functions \(f\) and \(g\), show that \(\;(f \cdot g)'' = f'' \cdot g ~+~ 2 f' \cdot g' ~+~ f \cdot g''\). Recall that taking a derivative is a way of operating on a function. That is, think of \(\ddx\) as the differentiation operator on the collection of differentiable functions, taking a function \(f(x)\) to its derivative function \(\dfdx\): Likewise, \(\frac{d^2}{d\!x^2}\) is an operator on twice-differentiable functions, taking a function \(f(x)\) to its second derivative function \(\frac{d^2 \negmedspace f}{d\!x^2}\): In general, an eigenfunction of an operator \(A\) is a function \(\phi(x)\) such that \(A(\phi(x)) ~=~ \lambda \cdot \phi(x)\), that is, for all \(x\) in the domain of \(\phi\), for some constant \(\lambda\) called the eigenvalue of the eigenfunction. Show for all constants \(k\) that \(\phi(x) ~=~ \cos\,kx\) is an eigenfunction of the \(\frac{d^2}{d\!x^2}\) operator, and find its eigenvalue. That is, show that \(\frac{d^2}{d\!x^2}(\phi(x)) = \lambda \cdot \phi(x)\) for some constant \(\lambda\) (the eigenvalue). The wave function \(\psi\) for a particle of mass \(m\) moving in a one-dimensional box of length \(L\), given by
\[\psi(x) ~=~ \sqrt{\frac{2}{L}}\;\sin\,\frac{\pi x}{L} \qquad\text{for $~0 \;\le x \;\le L$,} \nonumber \]
is a solution (assuming zero potential energy) of the time-independent Schrödinger equation
\[-\frac{h^2}{8\pi^2 m}\,\frac{d^2 \negmedspace\psi}{d\!x^2} ~=~ E\,\psi(x) \nonumber \]
where \(h\) is Planck’s constant and \(E\) is a constant that represents the total energy of the wave function. Find an expression for the constant \(E\) in terms of the other constants. Notice that this makes \(\psi(x)\) an eigenfunction of the \(\frac{d^2}{d\!x^2}\) operator.
