Consider a function \(F\) and its derivative \(f\) (i.e. \(f = F').\)
Since \(f\) is the derivative of \(F,\)
we could refer to \(F\) as an antiderivative of \(f.\)
And since the operation of “taking the derivative” is referred to as differentiation,
let’s refer to “taking an antiderivative” as antidifferentiation.
A function’s derivative is unique, but it’s antiderivative is not.
A function has infinitely many antiderivatives,
all of them differing by some additive constant \(C.\)
That is, \(F+C\) is the general antiderivative of \(f.\)
Whereas calculating the formula for the derivative of a function
is rather rote and procedural,
calculating a formula for an antiderivative of a function can be much more difficult,
and is sometimes simply impossible.
Determining a function’s antiderivative is simple in the case
when we recognize it as the derivative of a function we know.
\(\displaystyle x^{\color{Maroon}n} = \frac{\mathrm{d}}{\mathrm{d}x}\biggl( {\color{Maroon}\frac{1}{n+1}}x^{\color{Maroon}n+1} + C\biggr) \)
The power rule for \(n \neq -1\)
\(\displaystyle x^{-1} = \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( \ln(x) + C\Bigr) \)
The power rule for \(n = -1\)
\[\begin{align*}
{\color{DarkSlateBlue}\cos}(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( {\color{DarkGoldenRod}\sin}(x) + C \Bigr)
\\[1em]
{\color{DarkGoldenRod}\sin}(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( -{\color{DarkSlateBlue}\cos}(x) + C \Bigr)
\end{align*}\]
\[\begin{align*}
{\color{ForestGreen}\sec}(x){\color{Firebrick}\tan}(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( {\color{ForestGreen}\sec}(x) + C \Bigr)
\\[1em]
{\color{DarkOliveGreen}\csc}(x){\color{Maroon}\cot}(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( -{\color{DarkOliveGreen}\csc}(x) + C \Bigr)
\end{align*}\]
\[\begin{align*}
{\color{ForestGreen}\sec}^2(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( {\color{Firebrick}\tan}(x) + C \Bigr)
\\[1em]
{\color{DarkOliveGreen}\csc}^2(x)
&= \frac{\mathrm{d}}{\mathrm{d}x}\Bigl( -{\color{Maroon}\cot}(x) + C \Bigr)
\end{align*}\]