15.1 Higher Order Derivatives

15.1   Definition (Higher order derivatives.) Let $f$ be a function whose domain is a subset of R. We define a function $f'$ (called the derivative of $f$) by

$$\mbox{domain}(f') = \{ x \in \mbox{{\rm dom}}(f): f'(x) \mbox{ exists}\}.$$

and for all $x \in \mbox{{\rm dom}}(f)$, the value of $f'$ at $x$ is the derivative $f'(x)$. We may also write $f^{(1)}$ for $f'$. Since $f'$ is itself a function, we can calculate its derivative: this derivative is denoted by $f''$ or $f^{(2)}$, and is called the second derivative of $f$. For integers $n \geq 2$ we define

$$f^{(n+1)} = (f^{(n)})'.$$ (15.2)

and we call $f^{(n)}$ the $n$th derivative of $f$. We also define

$$f^{(0)} = f.$$

In Leibniz's notation we write

$${d^nf \over dx^n} = f^{(n)}, \mbox{ or }{d^n \over dx^n} f = ... ...)}f(x) = f^{(n)}(x) \mbox{ or }{d^nf \over dx^n} = f^{(n)}(x),$$

so that equation (15.2) becomes

$${d^{n+1} f \over dx^{n+1} } = {d\over dx}\left( {d^n f \over dx^n} \right).$$

If $a$ and $b$ are real numbers, and $f$ and $g$ are functions then from known properties of the derivative we can show that

$$(af +bg)^{(n)} = af^{(n)} + bg^{(n)} \mbox{ on } \mbox{{\rm dom}}(f^{(n)}) \cap \mbox{{\rm dom}}(g^{(n)}).$$

or

$${d^n \over dx^n}(af+bg) = a {d^{n}f \over dx^n} + b{ d^n f \over dx^n}.$$

15.3   Examples. If $h(x) = \sin(\omega x)$, where $\omega \in \mbox{{\bf R}}$, then

$$\begin{eqnarray*} h'(x) &=& \omega \cos(\omega x),\\ h''(x) &=& -\omega^2\sin(\... ...ga x),\\ h^{(4)}(x) &=& \omega^4\sin(\omega x) = \omega^4 h(x). \end{eqnarray*}$$

It should now be apparent that

$$h^{(4n+k)}(x) = \omega^{4n}h^{(k)}(x) \mbox{ for } k = 0,1,2,3.$$

so that

$$h^{(98)}(x) = h^{(4\cdot 24 + 2)}(x) = \omega^{96}h^{(2)}(x) = -\omega^{98}\sin(\omega x).$$

If

$$g(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$$

then

$$\begin{eqnarray*} g'(x) &=& 1 + x + \frac{x^2}{2!},\\ g''(x) &=& 1 + x,\\ g^{(... ...\\ g^{(n)}(x) &=& 0 \mbox{ for } n \in \mbox{{\bf Z}}_{\geq 4}. \end{eqnarray*}$$

If $y = \ln(x)$ then

$$\begin{eqnarray*} {dy \over dx} &=& {1 \over x},\\ {d^2y \over dx^2} &=& -{1 \over x^2},\\ {d^3y \over dx^3} &=& {2 \over x^3}. \end{eqnarray*}$$

15.4   Exercise. Calculate $g^{(5)}(t)$ if $g(t) = t^4\ln(t)$.

15.5   Exercise. A Let $g(t) = tf(t)$. Calculate $g'(t)$, $g''(t)$, $g^{(3)}(t)$ and $g^{(4)}(t)$ in terms of $f(t)$, $f'(t)$, $f''(t)$, $f^{(3)}(t)$ and $f^{(4)}(t)$. What do you think is the formula for $g^{(n)}(t)$?

15.6   Exercise. Find $${ {d^2y \over dx^2} }$$ if $${ y = {1/(x^2-1)}}$$.

15.7   Exercise. Find $f''(x)$ if $${ f(x) = e^{ {1 \over x^2} } = \exp\left({1\over x^2}\right) }$$.

15.8   Exercise. A Suppose $f''(x) = 0$ for all $x \in \mbox{{\bf R}}$. What can you say about $f$?

15.9   Exercise. A Let $f$ and $g$ be functions such that $f^{(2)}$ and $g^{(2)}$ are defined on all of $\mbox{{\bf R}}$. Show that

$$(fg)^{(2)} = fg^{(2)} + 2f^{(1)}g^{(1)} + f^{(2)}g.$$

Find a similar function for $(fg)^{(3)}$ (assuming that $f^{(3)}$ and $g^{(3)}$ are defined.)

In Leibniz's calculus, $d^2f$ or $ddf$ was actually an infinitely small quantity that was so much smaller than $dx$ that the quotient $${d^2f \over dx}$$ was zero, and $${ d^2f\over dx^2}$$ was obtained by multiplying $dx$ by itself and then dividing the result into $d^2f$. Leibniz also used notations like $${ddy \over ddx}$$ and $${dxds\over ddy}$$ for which our modern notation has no counterparts. Leibniz considered the problem of defining a meaning for $${d^{1\over 2}f}$$, but he did not make much progress on this problem. Today there is considerable literature on fractional derivatives. A brief history of the subject can be found in [36, ch I and ch VIII].

15.10   Exercise. Let $a$ be a real number. Show that for $k = 0,1,2,3$

$${d^k\over dx^k} e^{ax} = a^k e^{ax}.$$ (15.11)

After doing this it should be clear that equation (15.11), in fact holds for all $n\in\mbox{{\bf Z}}_{\geq 0}$ (this can be proved by induction). Now suppose that $a>0$ and we will define

$${d^r\over dx^r} e^{ax} = a^r e^{ax} \mbox{ for all }r \in \mbox{{\bf R}}.$$ (15.12)

Show that then for all $p$ and $q$ in $\mbox{{\bf R}}$,

$$\left({d\over dx}\right)^p \left(\left( {d\over dx}\right)^q( e^{ax}) \right) = \left( {d \over dx}\right)^{p+q} (e^{ax}).$$

Find $${ \left( {d\over dx}\right)^{1\over 2} e^{3x} }$$ and $${ \left( {d \over dx}\right)^{1\over 2}e^{5x} }$$. What do you think $${ \left( {d\over dx}\right)^{1\over 2} \left(3e^{3x} + 4e^{5x}\right)}$$ should be?

Equation (15.12) was the starting point from which Joseph Liouville (1809-1882) developed a theory of fractional calculus[36, pp 4-6].

15.13   Exercise. A Let $a$ and $b$ be real numbers. Show that for $k = 0,1,2,3$

$$\left({d \over dx}\right)^{k} \cos(ax + b) = a^k\cos(ax + b + {k\pi \over 2}).$$ (15.14)

After doing this exercise it should be clear that in fact equation (15.14) holds for all $k \in \mbox{{\bf Z}}_{\geq 0}$ (this can be proved by induction). Now suppose that $a>0$, and we will define

$$\left({d\over dx}\right)^r \cos(ax+b) = a^r\cos(ax + b +{r\pi \over 2}) \mbox{ for all }r \in \mbox{{\bf R}}.$$ (15.15)

Show that for all $p$ and $q$ in $\mbox{{\bf R}}$

$$\left( {d\over dx}\right)^p \left( \left( {d\over dx} \right... ...os(ax + b) \right) = \left({d\over dx}\right)^{p+q} \cos(ax+b).$$

Equation (15.15) was used as the starting point for a definition of fractional derivatives for general functions, by Joseph Fourier (1768-1830)[36, page 3].