For all these trigonometric functions, we will also have their respective reciprocal function.
Between a function and its reciprocal function, we do have the following relation:
$$ f \circ f^{-1} = id$$
Here is an example with \(sin(x)\) and \(arcsin(x)\) :
$$ \Biggl \{ \begin{align*} f : x \longmapsto sin(x), \hspace{3.1em} \mathbb{R } \longmapsto [-1, \enspace 1] \\ f^{-1} : x \longmapsto arcsin(x), \enspace [-1, \enspace 1] \longmapsto \mathbb{R } \end{align*} $$
$$ arcsin(sin(x)) = x \Longleftrightarrow sin(arcsin(x)) = x $$
Be careful not to be confused with the notation "\( f^{-1} \)" of reciprocal functions with that of the secant.
Indeed, we note "\( cos^{-1}, \ sin^{-1}, \ tan^{-1}... \)" for trigonometric reciprocal functions \( (arcsin, \ arccos, \ arctan...) \), but this is a different notation from "\( f^{-1} \)" which generally means the inverse function \( (x^{-1} = \frac{1}{x}) \).
$$ cos^2(x) = cos(x)cos(x) $$
$$ (but) $$
$$ \Biggl[ cos^{-1}(x) = arccos(x) \Biggr] \ \neq \ \Biggl[ \Bigl(cos(x)\Bigr)^{-1} = \frac{1}{cos(x)} = sec(x) \Biggr] $$
The basic trigonometric functions\(:sin(x), cos(x), tan(x)\)
Applying the Thales' theorem, we definitely see the following relation:
$$ \frac{cos(\theta)}{1} = \frac{sin(\theta)}{tan(\theta)} \Longleftrightarrow tan(\theta) = \frac{sin(\theta)}{cos(\theta)} $$
The sines function\(: sin(x)\)
The \( sin(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sin(x) $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ sin(x)' = cos(x) $$
The cosines function\(: cos(x)\)
The \( cos(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cos(x) $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ cos(x)' = -sin(x) $$
The tangent function\(: tan(x)\)
The \( tan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = tan(x) = \frac{sin(x)}{cos(x)} $$
Its derivative is:
$$ \forall k \in \mathbb{Z}, \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr] $$
$$ tan(x)' = 1 + tan^2(x) = \frac{1}{cos^2(x)}= sec^2(x) $$
The basic trigonometric reciprocal functions: \(arcsin(x), arccos(x), arctan(x)\)
The arcsines function\(: arcsin(x)\)
The \( arcsin(x) \) is the reciprocal function of the \( sin(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arcsin(x) = sin^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-1 ,\hspace{0.2em} 1[, $$
$$ arcsin(x)' = \frac{1}{\sqrt{1 - x^2}} $$
The arccosines function\(: arccos(x)\)
The \( arccos(x) \) function is the reciprocal function of the \( cos(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arccos(x) = cos^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-1 , \hspace{0.2em}1[, $$
$$ arccos(x)' = -\frac{1}{\sqrt{1 - x^2}} $$
The arctangent function\(: arctan(x)\)
The \( arctan(x) \) function is the reciprocal function of the \( tan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arctan(x) = tan^{-1}(x) $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ arctan(x)' = \frac{1}{1 + x^2} $$
The secant trigonometric functions: \(cosec(x), sec(x), cotan(x)\)
The three secant trigonometric functions are the \( cosec(x), sec(x) \) and \( cotan(x) \) functions.
They are respectively the inverses of the \( sin(x), cos(x) \) and \( tan(x) \) functions.
Applying the Thales' theorem, we definitely see the following relation:
$$ \left \{ \begin{align*} \frac{cosec(\theta)}{1} = \frac{1}{sin(\theta)} \Longleftrightarrow cosec(\theta) = \frac{1}{sin(\theta)} \\ \frac{sec(\theta)}{1} = \frac{1}{cos(\theta)} \Longleftrightarrow sec(\theta) = \frac{1}{cos(\theta)} \\ \frac{cotan(\theta)}{1} = \frac{cosec(\theta)}{sec(\theta)} = \frac{1}{tan(\theta)} \Longleftrightarrow cotan(\theta) = \frac{1}{tan(\theta)} \end{align*} \right \} $$
The cosecant function\(: cosec(x)\)
The \( cosec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], \enspace f(x) = cosec(x) = \frac{1}{sin(x)} $$
Its derivative is:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ cosec(x)' = - cosec^2(x)cos(x) = -cosec(x)cotan(x) $$
Furthermore, we do notice that:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ \frac{cosec'(x)}{cosec(x)} = -cosec(x)cos(x) = -tan(x)$$
The secant function\(: sec(x)\)
The \( sec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = sec(x) = \frac{1}{cos(x)} $$
Its derivative is:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ sec(x)' = sec^2(x) sin(x) = sec(x)tan(x) $$
Furthermore, we do notice that:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ \frac{sec'(x)}{sec(x)} = sec(x)sin(x) = tan(x)$$
The cotangent function\(: cotan(x)\)
The \( cotan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr] , \enspace f(x) = cotan(x) = \frac{cosec(x)}{sec(x)} = \frac{1}{tan(x)} $$
Its derivative is:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ cotan(x)' = -(1 + cotan^2(x)) = - cosec^2(x) $$
The secant trigonometric reciprocal functions: \(arccosec(x)\), \(arcsec(x)\), \( arccotan(x)\)
The arccosecant function\(: arccosec(x)\)
The \( arccosec(x) \) is the reciprocal function of the \( cosec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccosec(x) = cosec^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ arccosec(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$
The arcsecant function\(: arcsec(x)\)
The \( arcsec(x) \) is the reciprocal function of the \( sec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arcsec(x) = sec^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ arcsec(x)' = \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$
The arccotangent function\(: arccotan(x)\)
The \( arccotan(x) \) is the reciprocal function of the \( cotan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R} , \enspace f(x) = arccotan(x) = cotan^{-1}(x) $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ arccotan(x)' = - \frac{1}{ 1 + x^2} $$
The hyperbolic functions: \(sinh(x), cosh(x), tanh(x)\)
The three hyperbolic functions are the \( sinh(x), cosh(x) \) and \( tanh(x) \) functions.
They are the twins of the \( sin(x), cos(x) \) and \( tan(x) \) functions, and notably concerning their properties.
The hyperbolic sines function\(: sinh(x)\)
The \( sinh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sinh(x) = \frac{e^x - e^{-x} }{2} $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ sinh(x)' = cosh(x) $$
The hyperbolic cosines function\(: cosh(x)\)
The \( cosh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cosh(x) = \frac{e^x + e^{-x} }{2} $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ cosh(x)' = sinh(x) $$
The hyperbolic tangent function\(: tanh(x)\)
The \( tanh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ tanh(x)' = 1 - tanh^2(x) = sech^2(x) $$
The hyperbolic reciprocal functions: \( arcsinh(x)\), \(arccosh(x)\), \( arctanh(x)\)
The hyperbolic arcsines function\(: arcsinh(x)\)
The \( arcsinh(x) \) is the reciprocal function of the \( sinh(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arcsinh(x)= sinh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \mathbb{R},$$
$$ arcsinh(x) = ln \left|x + \sqrt{x^2 + 1}\right| $$
(\(\Longrightarrow\) see demonstration of it)
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ arcsinh(x)' = \frac{1}{\sqrt{1 + x^2}} $$
The hyperbolic arccosines function\(: arccosh(x)\)
The \( arccosh(x) \) is the reciprocal function of the \( cosh(x) \) function, it is defined as follows:
$$ \forall x \in [1, \hspace{0.1em} +\infty[, \enspace f(x) = arccosh(x) = cosh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x) = ln \Bigl| x + \sqrt{x^2 - 1}\Bigr| $$
(\(\Longrightarrow\) see demonstration of it)
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x)' = \frac{1}{\sqrt{x^2 -1}} $$
The hyperbolic arctangent function\(: arctanh(x)\)
The \( arctanh(x) \) is the reciprocal function of the \( tanh(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \enspace f(x) = arctanh(x) = tanh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$
$$ arctanh(x) = \frac{1}{2} ln \left| \frac{1 + x}{1 - x} \right| $$
(\(\Longrightarrow\) see demonstration of it)
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$
$$ arctanh(x)' = \frac{1}{1 - x^2} $$
The hyperbolic secant functions: \(cosech(x), sech(x), cotanh(x)\)
The three hyperbolic secant functions are the \( cosech(x), sech(x) \) and \( cotanh(x) \) functions.
They are respectively the inverses of the \( sinh(x), cosh(x) \) and \( tanh(x) \) functions.
The hyperbolic cosecant function\(: cosech(x)\)
The \( cosech(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cosech(x) = \frac{1}{sinh(x)} $$
Its derivative is:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ cosech(x)' = - cosech^2(x) cosh(x) = -cosech(x)cotanh(x) $$
Furthermore, we do notice that:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ \frac{cosech'(x)}{cosech(x)} = -cosech(x)cosh(x) = -cotanh(x)$$
The hyperbolic secant function\(: sech(x)\)
The \( sech(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sech(x) = \frac{1}{cosh(x)} $$
Its derivative is:
$$ \forall x \in \mathbb{R}, $$
$$ sech(x)' = -sech^2(x)sinh(x) = -sech(x)tanh(x) $$
The hyperbolic cotangent function\(: cotanh(x)\)
The \( cotanh(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cotanh(x) = \frac{1}{tanh(x)} $$
Its derivative is:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ cotanh(x)' = 1 - cotan^2(x) = -cosech^2(x)$$
The hyperbolic secant reciprocal functions: \(arccosech(x)\), \(arcsech(x)\), \( arccotanh(x)\)
The hyperbolic arccosecant function\(: arccosech(x)\)
The \( arccosech(x) \) is the reciprocal function of the \( cosech(x) \) function, it is defined as follows:
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , \enspace f(x) = arccosech(x) = cosech^{-1}(x) $$
Its derivative is:
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , $$
$$ arccosech(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 + \frac{1}{ x^2}}} $$
The hyperbolic arcsecant function\(: arcsech(x)\)
The \( arcsech(x) \) is the reciprocal function of the \( sech(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1] , \enspace f(x) = arcsech(x) = sech^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.1em} ]0, \hspace{0.1em} 1], $$
$$ arcsech(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{\frac{1}{ x^2} - 1}} $$
The hyperbolic arccotangent function\(: arccotanh(x)\)
The \( arccotanh(x) \) is the reciprocal function of the \( cotanh(x) \) function, it is defined as follows:
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccotanh(x) =cotanh^{-1}(x) $$
Its derivative is:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[, $$
$$ arccotanh(x)' = \frac{1}{ 1 - x^2} $$
Recap table of the trigonometric functions derivatives
Click on the title to access to the recap table.
The \( sin(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sin(x) $$
With the definition of the derivative, we do have:
$$ sin(x)' = lim_{h \to 0 } \enspace \frac{ sin(x + h) - sin(x)}{h} $$
We know from the trigonometric addition formulas that:
$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(\alpha + \beta) = sin(\alpha) cos(\beta) + cos(\alpha) sin(\beta) $$
So:
$$ sin(x)' = lim_{h \to 0 } \enspace \frac{ sin(x) cos(h) + cos(x) sin(h) - sin(x)}{h} $$
When \( h \to 0\), \( cos(h) \to 1\) and \( sin(h) \to h\).
Then,
$$ sin(x)' = lim_{h \to 0 } \enspace \frac{ sin(x) + cos(x). h - sin(x)}{h} $$
$$ sin(x)' = lim_{h \to 0 } \enspace \frac{cos(x). h }{h} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ sin(x)' = cos(x) $$
The \( cos(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cos(x) $$
With the definition of the derivative, we do have:
$$cos(x)' = lim_{h \to 0 } \enspace \frac{ cos(x + h) - cos(x)}{h} $$
We know from the trigonometric addition formulas that:
$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ cos(\alpha + \beta) = cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta) $$
So:
$$cos(x)' = lim_{h \to 0 } \enspace \frac{ cos(x) cos(h) - sin(x) sin(h) - cos(x)}{h} $$
When \( h \to 0\), \( cos(h) \to 1\) and \( sin(h) \to h\).
Then,
$$cos(x)' = lim_{h \to 0 } \enspace \frac{ cos(x) - sin(x). h - cos(x)}{h} $$
$$cos(x)' = lim_{h \to 0 } \enspace \frac{ - sin(x). h }{h} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ cos(x)' = -sin(x) $$
The \( tan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = tan(x) = \frac{sin(x)}{cos(x)} $$
By definition:
$$ tan(x)' = \left( \frac{sin(x)}{cos(x)} \right)' $$
We know from the derivative of a quotient that:
$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$
$$ \left ( f \over g \right)' = \frac{f'g - g'f}{g^2} $$
So in our case:
$$ tan(x)' = \frac{cos(x)cos(x) + sin(x)sin(x)}{cos^2(x)} $$
$$ tan(x)' = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} $$
$$ tan(x)' = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} $$
And finally we do have,
$$ \forall k \in \mathbb{Z}, \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr] $$
$$ tan(x)' = 1 + tan^2(x) = \frac{1}{cos^2(x)} = sec^2(x) $$
The \( arcsin(x) \) is the reciprocal function of the \( sin(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arcsin(x) = sin^{-1}(x) $$
We can calculate this derivative using the derivative of a reciprocal function :
$$ ( f^{-1} )'= \frac{1}{ (f' \circ f^{-1})} $$
$$ with \ \Biggl \{ \begin{align*} f(x) = sin(x) \\ f'(x) = cos(x) \\ f^{-1}(x) = arcsin(x) \end{align*} $$
Consequently,
$$ arcsin(x)' = \frac{1}{cos(arcsin(x))} $$
Furthermore,
$$ cos^2(x) + sin^2(x) = 1$$
$$ cos^2(x) = 1 - sin^2(x) $$
$$ cos(x) = \sqrt{1 - sin^2(x)} $$
Thus,
$$ arcsin(x)' = \frac{1}{\sqrt{1 - sin^2(arcsin(x))}} $$
And as a result,
$$ \forall x \in \hspace{0.05em} ]-1 ,\hspace{0.2em} 1[, $$
$$ arcsin(x)' = \frac{1}{\sqrt{1 - x^2}} $$
The \( arccos(x) \) function is the reciprocal function of the \( cos(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arccos(x) = cos^{-1}(x) $$
By the same reasoning as before with \(arcsin(x)'\) calculation:
$$ arccos(x)' = - \frac{1}{\sqrt{1 - cos^2(arccos(x))}} $$
And finally,
$$ \forall x \in \hspace{0.05em} ]-1 , \hspace{0.2em}1[, $$
$$ arccos(x)' = -\frac{1}{\sqrt{1 - x^2}} $$
The \( arctan(x) \) function is the reciprocal function of the \( tan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arctan(x) = tan^{-1}(x) $$
By the same reasoning as before with \(arcsin(x)'\) calculation:
$$ arctan(x)' = \frac{1}{1 + tan^2(arctan(x))} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ arctan(x)' = \frac{1}{1 + x^2} $$
The \( cosec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z},\ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], \enspace f(x) = cosec(x) = \frac{1}{sin(x)} $$
By definition:
$$ cosec(x)' = \biggl(\frac{1}{sin(x)} \biggr)' $$
We know how to calculate the inverse of a function's derivative:
$$ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$
$$ \left ( 1 \over g \right)' = -\frac{ g' }{g^2}$$
So in our case,
$$ cosec(x)' = -\frac{cos(x)}{sin^2(x)} $$
And finally,
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ cosec(x)' = - cosec^2(x)cos(x) = -cosec(x)cotan(x)$$
Furthermore, we do notice that:
$$ \frac{cosec'(x)}{cosec(x)} = \frac{-cosec^2(x)cos(x)}{cosec(x)} $$
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ \frac{cosec'(x)}{cosec(x)} = -cosec(x)cos(x) = -tan(x)$$
The \( sec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = sec(x) = \frac{1}{cos(x)} $$
By definition:
$$ sec(x)' = \biggl(\frac{1}{cos(x)} \biggr)' $$
We apply again the inverse of a function's derivative:
$$ sec(x)' = \frac{sin(x)}{cos^2(x)} $$
And finally,
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ sec(x)' = sec^2(x) sin(x) = sec(x)tan(x) $$
Furthermore, we do notice that:
$$ \frac{sec'(x)}{sec(x)} = \frac{sec^2(x)tan(x)}{sec(x)} $$
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ \frac{sec'(x)}{sec(x)} = sec(x)sin(x) = tan(x)$$
The \( cotan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr] , \enspace f(x) = cotan(x) = \frac{cosec(x)}{sec(x)} = \frac{1}{tan(x)} $$
By definition:
$$ cotan(x)' = \biggl(\frac{1}{tan(x)} \biggr)' $$
We apply again the inverse of a function's derivative:
$$ cotan(x)' = - \frac{1 + tan^2(x)}{tan^2(x)} $$
And finally,
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$
$$ cotan(x)' = -(1 + cotan^2(x)) = - cosec^2(x) $$
The \( arccosec(x) \) is the reciprocal function of the \( cosec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccosec(x) = cosec^{-1}(x) $$
We can calculate this derivative using the derivative of a reciprocal function:
$$ ( f^{-1} )'= \frac{1}{ (f' \circ f^{-1})} $$
$$ with \ \Biggl \{ \begin{align*} f(x) = cosec(x) \\ f'(x) = - cosec^2(x)cos(x) \\ f^{-1}(x) = arccosec(x) \end{align*} $$
$$ arccosec(x)' = -\frac{1}{cosec^2(arccosec(x)) \times cos(arccosec(x))} $$
Furthermore,
$$ cos^2(x) + sin^2(x) = 1$$
$$ cos^2(x) = 1 - sin^2(x) $$
$$ cos(x) = \sqrt{1 - sin^2(x)} $$
$$ arccosec(x)' = -\frac{1}{cosec^2(arccosec(x)) \times \sqrt{1 - sin^2(arccosec(x))}} $$
But:
$$ cosec(x) = \frac{1}{sin(x)} \Longleftrightarrow sin(x) = \frac{1}{cosec(x)} $$
So,
$$ arccosec(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{cosec^2(arccosec(x))}}} $$
Soit finalement,
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ arccosec(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$
The \( arcsec(x) \) is the reciprocal function of the \( sec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arcsec(x) = sec^{-1}(x) $$
By the same reasoning as before with \(arccosec(x)'\) calculation:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ arcsec(x)' = \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$
The \( arccotan(x) \) is the reciprocal function of the \( cotan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R} , \enspace f(x) = arccotan(x) = cotan^{-1}(x) $$
By the same reasoning as before with \(arccosec(x)'\) calculation:
$$ \forall x \in \mathbb{R}, $$
$$ arccotan(x)' = - \frac{1}{ 1 + x^2} $$
The \( sinh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sinh(x) = \frac{e^x - e^{-x} }{2} $$
Here, we will use a chain derivation to derivate exponentials:
$$ sinh(x)' = \biggl(\frac{e^x - e^{-x}}{2} \biggr)' $$
$$ sinh(x)' = \frac{1}{2} \bigl( e^x + e^{-x} \bigr) $$
$$ sinh(x)' = \biggl(\frac{e^x + e^{-x}}{2} \biggr) $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ sinh(x)' = cosh(x) $$
The \( cosh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cosh(x) = \frac{e^x + e^{-x} }{2} $$
By the same reasoning as before with \(sinh(x)'\) calculation:
$$ cosh(x)' = \biggl(\frac{e^x + e^{-x}}{2} \biggr)' $$
$$ cosh(x)' = \frac{1}{2} \bigl( e^x - e^{-x} \bigr) $$
$$ cosh(x)' = \biggl(\frac{e^x - e^{-x}}{2} \biggr) $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ cosh(x)' = sinh(x) $$
The \( tanh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
By definition:
$$ tanh(x)' = \biggl(\frac{e^x - e^{-x}}{e^x + e^{-x}} \biggr)' $$
Let us apply the derivative of a quotient:
$$ tanh(x)' = \frac{(e^x + e^{-x}) (e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2} $$
$$ tanh(x)' = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2} $$
$$ tanh(x)' = 1 - \frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ tanh(x)' = 1 - tanh^2(x) = sech^2(x) $$
The \( arcsinh(x) \) is the reciprocal function of the \( sinh(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arcsinh(x)= sinh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \mathbb{R}, $$
$$ arcsinh(x) = ln \left|x + \sqrt{x^2 + 1}\right| $$
(\(\Longrightarrow\) see demonstration of it)
We can calculate this derivative using the derivative of a reciprocal function:
$$ ( f^{-1} )'= \frac{1}{ (f' \circ f^{-1})} $$
$$ with \ \Biggl \{ \begin{align*} f(x) = sinh(x) \\ f('x) = cosh(x) \\ f^{-1}(x) = arcsinh(x) \end{align*} $$
$$ arcsinh(x)' = \frac{1}{cosh(arcsinh(x))} $$
Furthermore,
$$ cosh^2(x) -sinh^2(x) = 1$$
$$ cosh^2(x) = 1 + sinh^2(x) $$
$$ cosh(x) = \sqrt{1 +sinh^2(x)} $$
So,
$$ arcsinh(x)' = \frac{1}{\sqrt{1 + sinh^2(arcsinh(x))}} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ arcsinh(x)' = \frac{1}{\sqrt{1 + x^2}} $$
The \( arccosh(x) \) is the reciprocal function of the \( cosh(x) \) function, it is defined as follows:
$$ \forall x \in [1, \hspace{0.1em} +\infty[, \enspace f(x) = arccosh(x) = cosh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x) = ln \Bigl| x + \sqrt{x^2 - 1}\Bigr| $$
(\(\Longrightarrow\) see demonstration of it)
By the same reasoning as before with \(arcsinh(x)'\) calculation:
$$ \forall x \in \hspace{0.05em} ]1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x)' = \frac{1}{\sqrt{x^2 -1}} $$
The \( arctanh(x) \) is the reciprocal function of the \( tanh(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \enspace f(x) = arctanh(x) = tanh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ \ arctanh(x) = \frac{1}{2} ln \left| \frac{1 + x}{1 - x} \right| $$
(\(\Longrightarrow\) see demonstration of it)
By the same reasoning as before with \(arcsinh(x)'\) calculation:
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$
$$ arctanh(x)' = \frac{1}{1 - x^2} $$
The \( cosech(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cosech(x) = \frac{1}{sinh(x)} $$
By definition:
$$ cosech(x)' = \biggl(\frac{1}{sinh(x)} \biggr)' $$
We know how to calculate the inverse of a function's derivative:
$$ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$
$$ \left ( 1 \over g \right)' = -\frac{ g' }{g^2}$$
So in our case,
$$ cosech(x)' = -\frac{cosh(x)}{sinh^2(x)} $$
And finally,
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ cosech(x)' = - cosech^2(x) cosh(x) = -cosech(x)cotanh(x) $$
Furthermore, we do notice that:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ \frac{cosech'(x)}{cosech(x)} = -cosech(x)cosh(x) = -cotanh(x)$$
The \( sech(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sech(x) = \frac{1}{cosh(x)} $$
By definition:
$$ sech(x)' = \biggl(\frac{1}{cosh(x)} \biggr)' $$
We apply again the inverse of a function's derivative:
$$ sech(x)' = -\frac{sinh(x)}{cosh^2(x)} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ sech(x)' = -sech^2(x)sinh(x) = -sech(x)tanh(x) $$
Furthermore, we do notice that:
$$ \forall x \in \mathbb{R}, $$
$$ \frac{sech'(x)}{sech(x)} = -sech(x)sinh(x) = -tanh(x)$$
The \( cotanh(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cotanh(x) = \frac{1}{tanh(x)} $$
By definition:
$$ cotanh(x)' = \biggl(\frac{1}{tanh(x)} \biggr)' $$
We apply again the inverse of a function's derivative:
$$ cotanh(x)' = - \frac{1 - tanh^2(x)}{tanh^2(x)} $$
And as a result,
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], $$
$$ cotanh(x)' = 1 - cotan^2(x) = -cosech^2(x)$$
The \( arccosech(x) \) is the reciprocal function of the \( cosech(x) \) function, it is defined as follows:
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , \enspace f(x) = arccosech(x) = cosech^{-1}(x) $$
We can calculate this derivative using the derivative of a reciprocal function:
$$ ( f^{-1} )'= \frac{1}{ (f' \circ f^{-1})} $$
$$ with \ \Biggl \{ \begin{align*} f(x) = cosech(x) \\ f'(x) = - cosech^2(x) \ cosh(x) \\ f^{-1}(x) = arccosech(x) \end{align*} $$
$$ arccosech(x)' = \frac{1}{-cosech^2(arccosech(x)) \times \hspace{0.2em} cosh(arccosech(x))} $$
Furthermore,
$$ cosh^2(x) -sinh^2(x) = 1$$
$$ cosh^2(x) = 1 + sinh^2(x) $$
$$ cosh(x) = \sqrt{1 +sinh^2(x)} $$
So,
$$ arccosech(x)' = -\frac{1}{cosech^2(arccosech(x)) \times \sqrt{1 +sinh^2(arccosech(x))}} $$
But:
$$ cosech(x) = \frac{1}{sinh(x)} \Longleftrightarrow sinh(x) = \frac{1}{cosech(x)} $$
Soit,
$$ arccosech(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1+ \frac{1}{cosech^2(arccosech(x))}}} $$
And as a result,
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , $$
$$ arccosech(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 + \frac{1}{ x^2}}} $$
The \( arcsech(x) \) is the reciprocal function of the \( sech(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1] , \enspace f(x) = arcsech(x) = sech^{-1}(x) $$
By the same reasoning as before with \(arccosech(x)'\) calculation:
$$ \forall x \in \hspace{0.1em} ]0, \hspace{0.1em} 1], $$
$$ arcsech(x)' = - \frac{1}{ x^2} \times \frac{1}{ \sqrt{\frac{1}{ x^2} - 1}} $$
The \( arccotanh(x) \) is the reciprocal function of the \( cotanh(x) \) function, it is defined as follows:
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccotanh(x) =cotanh^{-1}(x) $$
By the same reasoning as before with \(arccosech(x)'\) calculation:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[, $$
$$ arccotanh(x)' = \frac{1}{ 1 - x^2} $$
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