Basic arithmetic
Addition and subtraction
Given \(a\left( t \right) = b\left( t \right) \pm c\left( t \right)\), trivially
(3)\[a^{\left[ n \right]}\left( t \right) = b^{\left[ n \right]}\left( t \right) \pm c^{\left[ n \right]}\left( t \right).\]
Multiplication
Given \(a\left( t \right) = b\left( t \right) c\left( t \right)\), the derivative \(a^{\left[ n \right]}\left( t \right)\)
is given directly by the application of the general Leibniz rule (2).
Division
Given \(a\left( t \right) = \frac{b\left( t \right)}{c\left( t \right)}\), we can write
(4)\[a\left( t \right) c\left( t \right) = b\left( t \right).\]
We can now apply the normalised derivative of order \(n\) to both sides, use (2) and re-arrange to obtain:
(5)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{c^{\left[ 0 \right]}\left( t \right)}\left[ b^{\left[ n \right]}\left( t \right) - \sum_{j=1}^n a^{\left[ n - j \right]}\left( t \right) c^{\left[ j \right]}\left( t \right)\right].\]
Square root
Given \(a\left( t \right) =\sqrt{b\left( t \right)}\), we can write
(7)\[a\left( t \right)^2 = b\left( t \right).\]
We can apply the normalised derivative of order \(n\) to both sides, and, with the help of (6), we obtain:
(8)\[\begin{split}b^{\left[ n \right]}\left( t \right) =
\begin{cases}
2\sum_{j=0}^{\frac{n}{2}-1} a^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) + \left( a^{\left[ \frac{n}{2} \right]}\left( t \right) \right)^2 \mbox{ if $n$ is even}, \\
2\sum_{j=0}^{\frac{n-1}{2}} a^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \mbox{ if $n$ is odd}.
\end{cases}\end{split}\]
We can then isolate \(a^{\left[ n \right]}\left( t \right)\) to obtain:
(9)\[\begin{split}a^{\left[ n \right]}\left( t \right) =
\begin{cases}
\frac{1}{2a^{\left[ 0 \right]}\left( t \right)} \left[ b^{\left[ n \right]}\left( t \right) - 2\sum_{j=1}^{\frac{n}{2}-1} a^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) - \left( a^{\left[ \frac{n}{2} \right]}\left( t \right) \right)^2 \right] \mbox{ if $n$ is even}, \\
\frac{1}{2a^{\left[ 0 \right]}\left( t \right)} \left[ b^{\left[ n \right]}\left( t \right) - 2\sum_{j=0}^{\frac{n-1}{2}} a^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right] \mbox{ if $n$ is odd}.
\end{cases}\end{split}\]
Exponentiation
Given \(a\left( t \right) = b\left( t \right)^\alpha\), with \(\alpha \neq 0\), we have
(10)\[a^\prime\left( t \right) = \alpha b\left( t \right)^{\alpha - 1} b^\prime\left( t \right).\]
By multiplying both sides by \(b\left( t \right)\) we obtain
(11)\[\begin{split}\begin{aligned}
b\left( t \right) a^\prime\left( t \right) & = b\left( t \right) \alpha b\left( t \right)^{\alpha - 1} b^\prime\left( t \right) \\
& = \alpha b^\prime\left( t \right) a\left( t \right).
\end{aligned}\end{split}\]
We can now apply the normalised derivative of order \(n-1\) to both sides, use (2) and re-arrange to obtain,
for \(n > 0\):
(12)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n b^{\left[ 0 \right]}\left( t \right)} \sum_{j=0}^{n-1} \left[ n\alpha - j \left( \alpha + 1 \right) \right] b^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right).\]
Exponentials
Natural exponential
Given \(a\left( t \right) = e^{b\left( t \right)}\), we have
(13)\[a^\prime\left( t \right) = e^{b\left( t \right)}b^\prime\left( t \right) = a\left( t \right) b^\prime\left( t \right).\]
We can now apply the normalised derivative of order \(n-1\) to both sides, use (1) and (2)
and obtain, for \(n > 0\):
(14)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n} \sum_{j=1}^{n} j a^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right).\]
Standard logistic function
Given \(a\left( t \right) = \operatorname{sig} {b\left( t \right)}\), where \(\operatorname{sig}\left( x \right)\)
is the standard logistic function
(15)\[\operatorname{sig} \left( x \right) = \frac{1}{1+e^{-x}},\]
we have
(16)\[a^\prime\left( t \right) = \operatorname{sig}{b\left( t \right)} \left[1 - \operatorname{sig}{b\left( t \right)} \right] b^\prime\left( t \right) = a\left( t \right) \left[1 - a\left( t \right) \right] b^\prime\left( t \right),\]
which, after the introduction of the auxiliary function
(17)\[c\left( t \right) = a^2\left( t \right) ,\]
becomes
(18)\[a^\prime\left( t \right) = \left[ a\left( t \right) - c\left( t \right) \right] b^\prime\left( t \right).\]
After applying the normalised derivative of order \(n-1\) to both sides, we can use (1),
(2) and (3) to obtain, for \(n > 0\):
(19)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n}\sum_{j=1}^{n} j \left[ a^{\left[ n - j \right]} \left( t \right)- c^{\left[ n - j \right]}\left( t \right)\right] b^{\left[ j \right]}\left( t \right).\]
Trigonometric functions
Tangent
Given \(a\left( t \right) = \tan b\left( t \right)\), we have
(23)\[a^\prime\left( t \right) = \left[ \tan^2 b\left( t \right) + 1 \right] b^\prime\left( t \right) = a^2\left( t \right)b^\prime\left( t \right) + b^\prime\left( t \right),\]
which, after the introduction of the auxiliary function
(24)\[c\left( t \right) = a^2\left( t \right) ,\]
becomes
(25)\[a^\prime\left( t \right) = c\left( t \right) b^\prime\left( t \right) + b^\prime\left( t \right).\]
After applying the normalised derivative of order \(n-1\) to both sides, we can use (1),
(2) and (3) to obtain, for \(n > 0\):
(26)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n}\sum_{j=1}^{n} j c^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right) + b^{\left[ n \right]}\left( t \right).\]
Inverse trigonometric functions
Inverse sine
Given \(a\left( t \right) = \arcsin b\left( t \right)\), we have
(27)\[a^\prime\left( t \right) = \frac{b^\prime\left( t \right)}{\sqrt{1 - b^2\left( t \right) }},\]
or, equivalently,
(28)\[a^\prime\left( t \right) \sqrt{1 - b^2\left( t \right) } = b^\prime\left( t \right).\]
We introduce the auxiliary function
(29)\[c\left( t \right) = \sqrt{1 - b^2\left( t \right) },\]
so that (28) can be rewritten as
(30)\[a^\prime\left( t \right) c\left( t \right) = b^\prime\left( t \right).\]
Applying the normalised derivative of order \(n-1\) to both sides yields, via (1):
(31)\[\left[a^\prime\left( t \right) c\left( t \right)\right]^{\left[ n - 1 \right]} = n b^{\left[ n \right]} \left( t \right).\]
We can now apply the general Leibniz rule (2) to the left-hand side and re-arrange
the terms to obtain, for \(n > 0\):
(32)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n c^{\left[ 0 \right]}\left( t \right)}\left[ n b^{\left[ n \right]}\left( t \right) - \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right].\]
Inverse cosine
The derivation is identical to the inverse sine, apart from a sign change.
Given \(a\left( t \right) = \arccos b\left( t \right)\),
the final result is, for \(n > 0\):
(33)\[a^{\left[ n \right]}\left( t \right) = -\frac{1}{n c^{\left[ 0 \right]}\left( t \right)}\left[ n b^{\left[ n \right]}\left( t \right) + \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right],\]
with \(c\left( t \right)\) defined as:
(34)\[c\left( t \right) = \sqrt{1 - b^2\left( t \right) }.\]
Inverse tangent
Given \(a\left( t \right) = \arctan b\left( t \right)\), we have
(35)\[a^\prime\left( t \right) = \frac{b^\prime\left( t \right)}{1 + b^2\left( t \right) },\]
or, equivalently,
(36)\[a^\prime\left( t \right) \left[1 + b^2\left( t \right) \right] = b^\prime\left( t \right).\]
We introduce the auxiliary function
(37)\[c\left( t \right) = b^2\left( t \right),\]
so that (36) can be rewritten as
(38)\[a^\prime\left( t \right) + a^\prime\left( t \right) c\left( t \right) = b^\prime\left( t \right).\]
Applying the normalised derivative of order \(n-1\) to both sides yields, via (1) and (3):
(39)\[n a^{\left[ n \right]} \left( t \right) + \left[a^\prime\left( t \right) c\left( t \right)\right]^{\left[ n - 1 \right]} = n b^{\left[ n \right]} \left( t \right).\]
With the help of the general Leibniz rule (2), after re-arranging we obtain, for \(n > 0\):
(40)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n \left[ c^{\left[ 0 \right]}\left( t \right) + 1 \right]}\left[ n b^{\left[ n \right]}\left( t \right) - \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right].\]
Two-argument inverse tangent
Given \(a\left( t \right) = \operatorname{arctan2}\left( b\left( t \right), c\left( t \right) \right)\), we have
(41)\[a^\prime\left( t \right) = \frac{c\left( t \right) b^\prime\left( t \right)-b\left( t \right)c^\prime \left( t \right)}
{b^2\left( t \right)+c^2\left( t \right)}.\]
After the introduction of the auxiliary function
(42)\[d\left( t \right) = b^2\left( t \right)+c^2\left( t \right),\]
(41) can be rewritten as
(43)\[d\left( t \right)a^\prime\left( t \right) = c\left( t \right) b^\prime\left( t \right)-b\left( t \right)c^\prime \left( t \right).\]
We can now apply the normalised derivative of order \(n-1\) to both sides, and, via (2), obtain, for \(n > 0\):
(44)\[\begin{split}\begin{aligned}
a^{\left[ n \right]}\left( t \right) &= \frac{1}{nd^{\left[ 0 \right]}\left( t \right)}\left[\vphantom{\sum_{j=1}^{n-1}j\left( \right)}
n\left( c^{\left[ 0 \right]}\left( t \right) b^{\left[ n \right]}\left( t \right) - b^{\left[ 0 \right]}\left( t \right) c^{\left[ n \right]}\left( t \right)\right) \right.\\
&\left. + \sum_{j=1}^{n-1}j\left( c^{\left[ n-j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right) -
b^{\left[ n-j \right]}\left( t \right) c^{\left[ j \right]}\left( t \right) -
d^{\left[ n-j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right) \right].
\end{aligned}\end{split}\]
Hyperbolic functions
Hyperbolic sine
Given \(a\left( t \right) = \sinh b\left( t \right)\), we have
(45)\[a^\prime\left( t \right) = b^\prime\left( t \right) \cosh b\left( t \right).\]
We introduce the auxiliary function
(46)\[c\left( t \right) = \cosh b\left( t \right),\]
so that (45) can be rewritten as
(47)\[a^\prime\left( t \right) = c\left( t \right) b^\prime\left( t \right).\]
We can now apply the normalised derivative of order \(n-1\) to both sides, and, via (2), obtain, for \(n > 0\):
(48)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n} \sum_{j=1}^{n} j c^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right).\]
Hyperbolic cosine
Given \(a\left( t \right) = \cosh b\left( t \right)\), the process of deriving \(a^{\left[ n \right]}\left( t \right)\) is
identical to the hyperbolic sine. After the definition of the auxiliary function
(49)\[s\left( t \right) = \sinh b\left( t \right),\]
the final result, for \(n > 0\), is:
(50)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n} \sum_{j=1}^{n} j s^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right).\]
Hyperbolic tangent
Given \(a\left( t \right) = \tanh b\left( t \right)\), the process of deriving \(a^{\left[ n \right]}\left( t \right)\) is
identical to the tangent, apart from a sign change. After the definition of the auxiliary function
(51)\[c\left( t \right) = a^2\left( t \right),\]
the final result, for \(n > 0\), is:
(52)\[a^{\left[ n \right]}\left( t \right) = b^{\left[ n \right]}\left( t \right) - \frac{1}{n}\sum_{j=1}^{n} j c^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right).\]
Inverse hyperbolic functions
Inverse hyperbolic sine
Given \(a\left( t \right) = \operatorname{arsinh} b\left( t \right)\), the process of deriving \(a^{\left[ n \right]}\left( t \right)\) is
identical to the inverse sine, apart from a sign change. After the definition of the auxiliary function
(53)\[c\left( t \right) = \sqrt{1 + b^2\left( t \right) },\]
the final result, for \(n > 0\), is:
(54)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n c^{\left[ 0 \right]}\left( t \right)}\left[ n b^{\left[ n \right]}\left( t \right) - \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right].\]
Inverse hyperbolic cosine
Given \(a\left( t \right) = \operatorname{arcosh} b\left( t \right)\), the process of deriving \(a^{\left[ n \right]}\left( t \right)\) is
identical to the inverse hyperbolic sine, apart from a sign change. After the definition of the auxiliary function
(55)\[c\left( t \right) = \sqrt{b^2\left( t \right) - 1 },\]
the final result, for \(n > 0\), is:
(56)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n c^{\left[ 0 \right]}\left( t \right)}\left[ n b^{\left[ n \right]}\left( t \right) - \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right].\]
Inverse hyperbolic tangent
Given \(a\left( t \right) = \operatorname{artanh} b\left( t \right)\), the process of deriving \(a^{\left[ n \right]}\left( t \right)\) is
identical to the inverse tangent, apart from a sign change. After the definition of the auxiliary function
(57)\[c\left( t \right) = b^2\left( t \right),\]
the final result, for \(n > 0\), is:
(58)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n \left[1 - c^{\left[ 0 \right]}\left( t \right) \right]}\left[ n b^{\left[ n \right]}\left( t \right) + \sum_{j=1}^{n-1} j c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right) \right].\]
Special functions
Error function
Given \(a\left( t \right) = \operatorname{erf} b\left( t \right)\), we have
(59)\[a^\prime\left( t \right) = \frac 2{\sqrt\pi} \exp{\left[-b^2\left( t \right)\right]} b^\prime\left( t \right),\]
which, after the introduction of the auxiliary function
(60)\[c\left( t \right) = \exp{\left[ -b^2\left( t \right)\right]} ,\]
becomes
(61)\[a^\prime\left( t \right) = \frac 2{\sqrt\pi}c\left( t \right) b^\prime\left( t \right).\]
After applying the normalised derivative of order \(n-1\) to both sides, we can use (1)
and (2) to obtain, for \(n > 0\):
(62)\[a^{\left[ n \right]}\left( t \right) = \frac 1n \frac 2{\sqrt\pi}\sum_{j=1}^{n} j c^{\left[ n - j \right]}\left( t \right) b^{\left[ j \right]}\left( t \right).\]
Celestial mechanics
Kepler’s eccentric anomaly
The eccentric anomaly is the bivariate function \(E = E\left( e, M \right)\) implicitly defined by the
trascendental equation
(63)\[M = E - e \sin E,\]
with \(e \in \left[ 0, 1 \right)\). Given \(a\left( t \right) = E\left( e\left( t \right), M \left( t \right) \right)\), we have
(64)\[a^\prime\left( t \right) = \frac{\partial E}{\partial e}e^\prime\left( t \right) + \frac{\partial E}{\partial M}M^\prime\left( t \right),\]
where the partial derivatives are
(65)\[\begin{split}\begin{cases}
\frac{\partial E}{\partial e} = \frac{\sin E}{1-e\cos E}, \\
\frac{\partial E}{\partial M} = \frac{1}{1-e\cos E}. \\
\end{cases}\end{split}\]
Expanding the partial derivatives yields
(66)\[a^\prime\left( t \right) = \frac{e^\prime \left( t \right)\sin a\left(t \right) + M^\prime\left( t \right)}{1-e\left( t \right)\cos a\left(t \right)},\]
or, equivalently,
(67)\[a^\prime\left( t \right) - a^\prime\left( t \right) e \left( t \right) \cos a\left(t \right) = e^\prime \left( t \right)\sin a\left(t \right) + M^\prime\left( t \right).\]
We can now introduce the auxiliary functions
(68)\[\begin{split}\begin{cases}
c\left( t \right) = e\left( t \right) \cos a\left(t \right), \\
d\left( t \right) = \sin a\left(t \right), \\
\end{cases}\end{split}\]
so that (67) can be rewritten as
(69)\[a^\prime\left( t \right) - a^\prime\left( t \right) c\left( t \right) = e^\prime \left( t \right)d\left(t \right) + M^\prime\left( t \right).\]
After applying the normalised derivative of order \(n-1\) to both sides, we can use (1)
and (2) and re-arrange to obtain, for \(n > 0\):
(70)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n \left(1 - c^{\left[ 0 \right]}\left( t \right)\right)}
\left[
n\left( e^{\left[ n \right]}\left( t \right) d^{\left[ 0 \right]}\left( t \right) + M^{\left[ n \right]}\left( t \right)\right) +
\sum_{j=1}^{n-1}j\left( c^{\left[ n - j \right]}\left( t \right) a^{\left[ j \right]}\left( t \right)+
d^{\left[ n - j \right]}\left( t \right)e^{\left[ j \right]}\left( t \right)
\right)
\right].\]
Eccentric longitude
The eccentric longitude
is the trivariate function \(F = F\left( h, k, \lambda \right)\) implicitly defined by the
trascendental equation
(71)\[\lambda = F + h\cos F - k\sin F,\]
with \(h^2+k^2 < 1\). Given \(a\left( t \right) = F\left( h\left( t \right), k \left( t \right), \lambda \left( t \right) \right)\),
we have
(72)\[a^\prime\left( t \right) = \frac{k^\prime \left( t \right)\sin a\left(t \right) - h^\prime \left( t \right)\cos a\left(t \right)
+ \lambda^\prime\left( t \right)}{1-h\left( t \right)\sin a\left(t \right) -k\left( t \right)\cos a\left(t \right)}.\]
After the introduction of the auxiliary functions
(73)\[\begin{split}\begin{cases}
c\left( t \right) = h\left( t \right) \sin a\left(t \right), \\
d\left( t \right) = k\left( t \right) \cos a\left(t \right), \\
e\left( t \right) = \sin a\left(t \right), \\
f\left( t \right) = \cos a\left(t \right), \\
\end{cases}\end{split}\]
we can then proceed in the same way as explained for the eccentric anomaly.
The final result, for \(n > 0\), is:
(74)\[a^{\left[ n \right]}\left( t \right) = \frac{1}{n \left(1 - c^{\left[ 0 \right]}\left( t \right) -d^{\left[ 0 \right]}\left( t \right)\right)}
\left\{
n\left( k^{\left[ n \right]}\left( t \right) e^{\left[ 0 \right]}\left( t \right)
- h^{\left[ n \right]}\left( t \right) f^{\left[ 0 \right]}\left( t \right)
+ \lambda^{\left[ n \right]}\left( t \right)\right) +
\sum_{j=1}^{n-1}j\left[
a^{\left[ j \right]}\left( t \right) \left(
c^{\left[ n - j \right]}\left( t \right) + d^{\left[ n - j \right]}\left( t \right)
\right)
+ k^{\left[ j \right]}\left( t \right) e^{\left[ n - j \right]}\left( t \right)
- h^{\left[ j \right]}\left( t \right) f^{\left[ n - j \right]}\left( t \right)
\right]
\right\}.\]
Time functions
Time polynomials
Given the time polynomial of order \(n\)
(75)\[p_n\left( t \right) = \sum_{i=0}^n a_i t^i,\]
its derivative of order \(j\) is
(76)\[\left(p_n\left( t \right)\right)^{\left( j \right)} = \sum_{i=j}^n \left( i \right)_j a_i t^{i - j},\]
where \(\left( i \right)_j\) is the falling factorial.
The normalised derivative of order \(j\) is
(77)\[\left(p_n\left( t \right)\right)^{\left[ j \right]} = \frac{1}{j!}\sum_{i=j}^n \left( i \right)_j a_i t^{i - j},\]
which, with the help of elementary relations involving factorials and after re-arranging the indices, can be rewritten as
(78)\[\left(p_n\left( t \right)\right)^{\left[ j \right]} = \sum_{i=0}^{n-j} {i+j \choose j} a_{i+j} t^i.\]
