# Derivative of arcsech

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## Derivative of sech-1(x)

We use the fact from the definition of the inverse that

[ ext{sech}( ext{sech}^{-1} ;x) = x ]

and the fact that

[ ext{sech}', x = - anh (x) ext{sech} (x) ]

Now take the derivative of both sides (using the chain rule on the left hand side) to get

[ - anh ( ext{sech}^{-1} x) ext{sech}( ext{sech}^{-1}, x)( ext{sech}^{-1}, x)' = 1 ]

or

[ -x , anh ( ext{sech}^{-1}x)( ext{sech}^{-1} ,x)' = 1 ag{1}]

We know that

[ cosh^2 x - sinh^2 x = 1]

Dividing by the (cosh^2(x)) gives

[ 1 - anh^2 (x) = ext{sech}^2, x]

or

[ anh x = sqrt{1- ext{sech}^2 ,x}]

so that

[ anh ( ext{sech}^{-1}, x) = sqrt{1- ext{sech}^{-1} , x} = sqrt{1-x^2} ]

Finally substituting into equation 1 gives

[ -xsqrt{1-x^2} ( ext{sech}^{-1}, x) = 1]

[ ext{sech}^{-1} , x = dfrac{-1}{xsqrt{1-x^2}}]

Larry Green (Lake Tahoe Community College)

## Atan2

The function atan2 ⁡ ( y , x ) (y,x)> first appeared in the programming language Fortran (in IBM's implementation FORTRAN-IV in 1961). It was originally intended to return a correct and unambiguous value for the angle θ in converting from cartesian coordinates (x, y) to polar coordinates (r, θ) .

Equivalently, atan2 ⁡ ( y , x ) (y,x)> is the argument (also called phase or angle) of the complex number x + i y .

This only holds when x > 0 . When x < 0 , the angle apparent from the expression above is pointing in the opposite direction of the correct angle, and a value of π (or 180°) must be either added or subtracted from θ to put the Cartesian point (x, y) into the correct quadrant of the Euclidean plane. [1] This requires knowledge of the signs of x and y separately, which is information lost when y is divided by x .

Since any integer multiple of 2π can be added to the angle θ without changing either x or y , implying an ambiguous value for the returned value, the principal value of the angle, in the (left open, right closed) interval (−π, π] is returned. θ is signed, with counterclockwise angles being positive, and clockwise being negative. In other words, atan2 ⁡ ( y , x ) (y,x)> is in the closed interval [0, π] when y ≥ 0 , and in the open interval (−π, 0) when y < 0 .

## Background & Context

ArcSech is the inverse hyperbolic secant function. For a real number , ArcSec [ x ] represents the hyperbolic angle measure such that . ArcSech automatically threads over lists. For certain special arguments, ArcSech automatically evaluates to exact values. When given exact numeric expressions as arguments, ArcSech may be evaluated to arbitrary numeric precision. Operations useful for manipulation of symbolic expressions involving ArcSech include FunctionExpand , TrigToExp , TrigExpand , Simplify , and FullSimplify . ArcSech is defined for complex argument by . ArcSech [ z ] has branch cut discontinuities in the complex plane. Related mathematical functions include Sech , ArcCsch , and ArcSec .

## Complex Inverse Hyperbolic Functions

For inverse hyperbolic functions, the notations sinh -1 and cosh -1 are often used for arcsinh and arccosh , etc. When this notation is used, the inverse functions are sometimes confused with the multiplicative inverses of the functions. The notation using the "arc-" prefix avoids such confusion.

The inverse hyperbolic functions are the multivalued function that are the inverse functions of the hyperbolic functions.

2) Definitions

The inverse hyperbolic functions may be expressed using natural logarithms.

arccosh( z ) = ln( )
For real x > 1, this simplifies to
arccosh( x ) = ln( x + )

arctanh( z ) = [ln(1 + z ) - ln(1 - z ) ]
For real x < 1, this simplifies to
arctanh( x ) = ln( )

arccoth( z ) = [ln(1 + ) - ln(1 - ) ]
For real x < 0 or x >1, this simplifies to
arccoth( x ) = ln( )

arcsech( z ) = ln( )
For real x , it satisfies
arcsech( x ) =

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## Derivative of arcsech

We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest:

Finding the Derivative of Inverse Sine Function, $displaystyle (arcsin x)>$

Suppose $arcsin x = heta$. Then it must be the cases that

Implicitly differentiating the above with respect to $x$ yields

Dividing both sides by $cos heta$ immediately leads to a formula for the derivative.

To be a useful formula for the derivative of $arcsin x$ however, we would prefer that $displaystyle = frac (arcsin x)>$ be expressed in terms of $x$, not $heta$.

Upon considering how to then replace the above $cos heta$ with some expression in $x$, recall the pythagorean identity $cos^2 heta + sin^2 heta = 1$ and what this identity implies given that $sin heta = x$:

So we know either $cos heta$ is then either the positive or negative square root of the right side of the above equation. Since $heta$ must be in the range of $arcsin x$ (i.e., $[-pi/2,pi/2]$), we know $cos heta$ must be positive. Thus,

Finally, plugging this into our formula for the derivative of $arcsin x$, we find

Finding the Derivative of Inverse Cosine Function, $displaystyle (arccos x)>$

The process for finding the derivative of $arccos x$ is almost identical to that used for $arcsin x$:

Suppose $arccos x = heta$. Then it must be the case that

Implicitly differentiating the above with respect to $x$ yields

Dividing both sides by $-sin heta$ immediately leads to a formula for the derivative.

To be a useful formula for the derivative of $arccos x$ however, we would prefer that $displaystyle = frac (arccos x)>$ be expressed in terms of $x$, not $heta$.

Upon considering how to then replace the above $sin heta$ with some expression in $x$, recall the pythagorean identity $cos^2 heta + sin^2 heta = 1$ and what this identity implies given that $cos heta = x$:

So we know either $sin heta$ is then either the positive or negative square root of the right side of the above equation. Since $heta$ must be in the range of $arccos x$ (i.e., $[0,pi]$), we know $sin heta$ must be positive. Thus,

Finally, plugging this into our formula for the derivative of $arccos x$, we find

Finding the Derivative of the Inverse Tangent Function, $displaystyle (arctan x)>$

The process for finding the derivative of $arctan x$ is slightly different, but the same overall strategy is used:

Suppose $arctan x = heta$. Then it must be the case that

Implicitly differentiating the above with respect to $x$ yields

Dividing both sides by $sec^2 heta$ immediately leads to a formula for the derivative.

To be a useful formula for the derivative of $arctan x$ however, we would prefer that $displaystyle = frac (arctan x)>$ be expressed in terms of $x$, not $heta$.

Upon considering how to then replace the above $sec^2 heta$ with some expression in $x$, recall the other pythagorean identity $an^2 heta + 1 = sec^2 heta$ and what this identity implies given that $an heta = x$:

Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of $arccos x$, to find

Finding the Derivative of the Inverse Cotangent Function, $displaystyle ( extrm x)>$

The derivative of $extrm x$ can be found similarly. Suppose $extrm x = heta$. Then $cot heta = x$. Implicitly differentiating with respect to $x$ yields $-csc^2 heta cdot frac = 1$ which implies the following, upon realizing that $cot heta = x$ and the identity $cot^2 heta + 1 = csc^2 heta$ requires $csc^2 heta = 1 + x^2$, $frac = frac<-1> = frac<-1><1+x^2>$ Thus, $frac( extrm x) = frac<-1><1+x^2>$

Finding the Derivative of the Inverse Secant Function, $displaystyle ( extrm x)>$

The derivative of $extrm x$ is more interesting.

Here, we suppose $extrm x = heta$, which means $sec heta = x$. Like before, we differentiate this implicitly with respect to $x$ to find

$sec heta an heta frac = 1$

Solving for $d heta/dx$ in terms of $heta$ we quickly get

This is where we need to be careful. Presuming that the range of the secant function is given by $(0, pi)$, we note that $heta$ must be either in quadrant I or II. In both, the product of $sec heta an heta$ must be positive. This implies

$sec heta an heta = |sec heta|| an heta|$

Of course $|sec heta| = |x|$, and we can use $an^2 heta + 1 = sec^2 heta$ to establish $| an heta| = sqrt$. As such,

## Derivatives Practice Demo

The back of this sheet contains a list of practice problems. It also contains links to solutions generated by Wolfram Alpha and SymboLab.

There are several major computer algebra systems (CAS for short). Many of our undergraduate math classes use a system called Maple. Another major CAS is Mathematica. Mathematica is part of what powers the online computational engine known as Wolfram Alpha.

We have used several demos powered by the SageMath CAS. While SAGE's user interface is not as polished as those of Maple and Mathematica, it is open source and free to use. Even Wolfram Alpha, while available freely, has locked featured behind its paywall.

Of course, Maple, Mathematica, and SAGE are not the only tools available. We have also seen Desmos which creates beautiful graphs (but does not do symbolic calculations). SymboLab provides yet another tool for doing mathematical computations.

SymboLab is much more limited than the CAS mentioned above, but it has an easy user interface and provides step-by-step solutions of basic problems in algebra and calculus including step-by-step derivative calculations. If you're still having trouble with derivative rules, I think you'll find SymboLab to be very helpful.

It is also worth mentioning that SAGE, Desmos, Wolfram Alpha, and SymboLab are all available for free via a web browser. Some of these have accompanying smartphone apps (which generally are not free). For example, Desmos is free while Alpha's is a few dollars. SymboLab's app is free but has locked features which require a payment to unlock. Personally, I'd just access these tools via a browser.

• Maple (a powerful CAS used in many classes at AppState): https://www.maplesoft.com
• Mathematica (a powerful CAS and the machine behind Wolfram Alpha): https://www.wolfram.com/mathematica/
• SAGE (an open source CAS which powers many demos used in our class): http://www.sagemath.org
• Wolfram Alpha (a mostly free online computational engine): http://www.wolframalpha.com
• SymboLab (a free online algebra and calculus calculator): https://www.symbolab.com
• Desmos (a free online graphing utility): https://www.desmos.com

### Practice Problems.

Compute the derivative. These problems begin with basics'' and then start adding rules and new functions. The problems toward the end are generally trickier than those at the beginning.

Try some basic simplification as well. Being able to simplify answers is important in many contexts -- that said -- simplification is not the focus of this exercise so don't waste a lot of time fighting with algebra.

Note: Clicking on "ALPHA" will take you Wolfram Alpha's solution, and clicking on "SYMBO" will take you to SymboLab's solution.

### Step 1: Write (y = arcsin(x))

Start by setting up a variable called (y) equal to (arcsin(x))

### Step 2: Take the (sin) of both sides

Apply the (sin) function to both sides of the equal sign

$fracleft(sin(y) ight) quad = quad frac$

The left-hand side requires the chain rule, while the right-hand side equals 1:

$fracleft(sin(y) ight)frac quad = quad 1$

The left-hand side simplifies to (cos(y)) times the derivative of (y) with respect to (x)

### Step 4: Divide both sides by (cos(y))

Solve (dy/dx) by dividing both sides by (cos(y))

### Step 5: Substitute for (cos(y))

Recall the Pythagorean trigonometric identity:

By subtracting (sin^2(y)) from both sides, then taking the square root of both sides we can isolate the Pythagorean identity for (cos(y)

Recall that (sin(y)=x) from step 2. We can substitute in for (x) in place of (sin(y))

Since (cos(y)) is found to be equal to the square root of 1 minus (x) squared, we can substitute it into our derivative from step 4

Remember that (y) was set equal to (arcsin(x)) in step 1, so the derivative of (y) with respect to (x) is the same as writing the derivative of (arcsin(x))

## Contents

The most common abbreviations are those specified by the ISO 80000-2 standard. They consist of ar- followed by the abbreviation of the corresponding hyperbolic function (e.g., arsinh, arcosh).

However, arc- followed by the corresponding hyperbolic function (e.g., arcsinh, arccosh) is also commonly seen, by analogy with the nomenclature for inverse trigonometric functions. [9] These are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area the hyperbolic functions are not directly related to arcs. [10] [11] [12]

Other authors prefer to use the notation argsinh, argcosh, argtanh, and so on, where the prefix arg is the abbreviation of the Latin argumentum. [13] In computer science, this is often shortened to asinh.

The notation sinh −1 (x) , cosh −1 (x) , etc., is also used, [14] [15] [16] [17] despite the fact that care must be taken to avoid misinterpretations of the superscript −1 as a power, as opposed to a shorthand to denote the inverse function (e.g., cosh −1 (x) versus cosh(x) −1 ).

Since the hyperbolic functions are rational functions of e x whose numerator and denominator are of degree at most two, these functions may be solved in terms of e x , by using the quadratic formula then, taking the natural logarithm gives the following expressions for the inverse hyperbolic functions.

For complex arguments, the inverse hyperbolic functions, the square root and the logarithm are multi-valued functions, and the equalities of the next subsections may be viewed as equalities of multi-valued functions.

For all inverse hyperbolic functions (save the inverse hyperbolic cotangent and the inverse hyperbolic cosecant), the domain of the real function is connected.

### Inverse hyperbolic sine Edit

Inverse hyperbolic sine (a.k.a. area hyperbolic sine) (Latin: Area sinus hyperbolicus): [14] [15]

### Inverse hyperbolic cosine Edit

Inverse hyperbolic cosine (a.k.a. area hyperbolic cosine) (Latin: Area cosinus hyperbolicus): [14] [15]

### Inverse hyperbolic tangent Edit

Inverse hyperbolic tangent (a.k.a. area hyperbolic tangent) (Latin: Area tangens hyperbolicus): [15]

### Inverse hyperbolic cotangent Edit

Inverse hyperbolic cotangent (a.k.a., area hyperbolic cotangent) (Latin: Area cotangens hyperbolicus):

The domain is the union of the open intervals (−∞, −1) and (1, +∞) .

### Inverse hyperbolic secant Edit

Inverse hyperbolic secant (a.k.a., area hyperbolic secant) (Latin: Area secans hyperbolicus):

The domain is the semi-open interval (0, 1] .

### Inverse hyperbolic cosecant Edit

Inverse hyperbolic cosecant (a.k.a., area hyperbolic cosecant) (Latin: Area cosecans hyperbolicus):

The domain is the real line with 0 removed.

For an example differentiation: let θ = arsinh x, so (where sinh 2 θ = (sinh θ) 2 ):

Expansion series can be obtained for the above functions:

Asymptotic expansion for the arsinh x is given by

As functions of a complex variable, inverse hyperbolic functions are multivalued functions that are analytic, except at a finite number of points. For such a function, it is common to define a principal value, which is a single valued analytic function which coincides with one specific branch of the multivalued function, over a domain consisting of the complex plane in which a finite number of arcs (usually half lines or line segments) have been removed. These arcs are called branch cuts. For specifying the branch, that is, defining which value of the multivalued function is considered at each point, one generally define it at a particular point, and deduce the value everywhere in the domain of definition of the principal value by analytic continuation. When possible, it is better to define the principal value directly—without referring to analytic continuation.

For example, for the square root, the principal value is defined as the square root that has a positive real part. This defines a single valued analytic function, which is defined everywhere, except for non-positive real values of the variables (where the two square roots have a zero real part). This principal value of the square root function is denoted x >> in what follows. Similarly, the principal value of the logarithm, denoted Log > in what follows, is defined as the value for which the imaginary part has the smallest absolute value. It is defined everywhere except for non-positive real values of the variable, for which two different values of the logarithm reach the minimum.

For all inverse hyperbolic functions, the principal value may be defined in terms of principal values of the square root and the logarithm function. However, in some cases, the formulas of § Definitions in terms of logarithms do not give a correct principal value, as giving a domain of definition which is too small and, in one case non-connected.

### Principal value of the inverse hyperbolic sine Edit

The principal value of the inverse hyperbolic sine is given by

The argument of the square root is a non-positive real number, if and only if z belongs to one of the intervals [i, +i∞) and (−i∞, −i] of the imaginary axis. If the argument of the logarithm is real, then it is positive. Thus this formula defines a principal value for arsinh, with branch cuts [i, +i∞) and (−i∞, −i] . This is optimal, as the branch cuts must connect the singular points i and −i to the infinity.

### Principal value of the inverse hyperbolic cosine Edit

The formula for the inverse hyperbolic cosine given in § Inverse hyperbolic cosine is not convenient, since similar to the principal values of the logarithm and the square root, the principal value of arcosh would not be defined for imaginary z . Thus the square root has to be factorized, leading to

The principal values of the square roots are both defined, except if z belongs to the real interval (−∞, 1] . If the argument of the logarithm is real, then z is real and has the same sign. Thus, the above formula defines a principal value of arcosh outside the real interval (−∞, 1] , which is thus the unique branch cut.

### Principal values of the inverse hyperbolic tangent and cotangent Edit

for the definition of the principal values of the inverse hyperbolic tangent and cotangent. In these formulas, the argument of the logarithm is real if and only if z is real. For artanh, this argument is in the real interval (−∞, 0] , if z belongs either to (−∞, −1] or to [1, ∞) . For arcoth, the argument of the logarithm is in (−∞, 0] , if and only if z belongs to the real interval [−1, 1] .

Therefore, these formulas define convenient principal values, for which the branch cuts are (−∞, −1] and [1, ∞) for the inverse hyperbolic tangent, and [−1, 1] for the inverse hyperbolic cotangent.

### Principal value of the inverse hyperbolic cosecant Edit

For the inverse hyperbolic cosecant, the principal value is defined as

It is defined when the arguments of the logarithm and the square root are not non-positive real numbers. The principal value of the square root is thus defined outside the interval [−i, i] of the imaginary line. If the argument of the logarithm is real, then z is a non-zero real number, and this implies that the argument of the logarithm is positive.

Thus, the principal value is defined by the above formula outside the branch cut, consisting of the interval [−i, i] of the imaginary line.

For z = 0 , there is a singular point that is included in the branch cut.

### Principal value of the inverse hyperbolic secant Edit

Here, as in the case of the inverse hyperbolic cosine, we have to factorize the square root. This gives the principal value

If the argument of a square root is real, then z is real, and it follows that both principal values of square roots are defined, except if z is real and belongs to one of the intervals (−∞, 0] and [1, +∞) . If the argument of the logarithm is real and negative, then z is also real and negative. It follows that the principal value of arsech is well defined, by the above formula outside two branch cuts, the real intervals (−∞, 0] and [1, +∞) .

For z = 0 , there is a singular point that is included in one of the branch cuts.

### Graphical representation Edit

In the following graphical representation of the principal values of the inverse hyperbolic functions, the branch cuts appear as discontinuities of the color. The fact that the whole branch cuts appear as discontinuities, shows that these principal values may not be extended into analytic functions defined over larger domains. In other words, the above defined branch cuts are minimal.

## Inverse Secant Calculator | Arcsecant Calculation

The Arcsec Calculator is also called the Inverse Secant Calculator, as it is the inverse of the secant value. The term Arcsec is short form of the term 'Arc Secant'. It can also be termed as asec. It is just the inverse function of sec(x). Make use of the below calculator to find the arc secant values in degrees & radians. Enter a value, and click calculate to find the corresponding arc secant radians and degrees. This arcsecant calculator will find the inverse trigonometric values of secant.

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