The inverse hyperbolic tangent tanh^(-1)z (Zwillinger 1995, p. 481; Beyer 1987, p. 181), sometimes called the area hyperbolic tangent (Harris and Stocker 1998, p. 267), is the multivalued function that is the inverse function of the hyperbolic tangent.

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Section 3-7 : Derivatives of Inverse Trig Functions

Calculate the derivative of the function (y = arccos x arctan x ) at (x = 0. ) Example 9 Using the chain rule, derive the formula for the derivative of the inverse sine function. Find the Derivative - d/dx y=arctan(1/x) Differentiate using the chain rule, which states that is where. Tap for more steps. To apply the Chain Rule, set as. The derivative of arctan returns an algebraic expression. We can use this expression to find the derivative of inverse trigonometric functions. In our discussion, we’ll understand how to differentiate arctan and use the new derivative rule to differentiate more complex functions. Make sure to have your notes handy!

In this section we are going to look at the derivatives of the inverse trig functions. In order to derive the derivatives of inverse trig functions we’ll need the formula from the last section relating the derivatives of inverse functions. If (fleft( x right)) and (gleft( x right)) are inverse functions then,

[g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}}]

Recall as well that two functions are inverses if (fleft( {gleft( x right)} right) = x) and (gleft( {fleft( x right)} right) = x).

We’ll go through inverse sine, inverse cosine and inverse tangent in detail here and leave the other three to you to derive if you’d like to.

Inverse Sine

Let’s start with inverse sine. Here is the definition of the inverse sine.

[y = {sin ^{ - 1}}xhspace{0.5in} Leftrightarrow hspace{0.5in}sin y = xhspace{0.25in}{mbox{for}}, - frac{pi }{2} le y le frac{pi }{2}]

So, evaluating an inverse trig function is the same as asking what angle (i.e. (y)) did we plug into the sine function to get (x). The restrictions on (y) given above are there to make sure that we get a consistent answer out of the inverse sine. We know that there are in fact an infinite number of angles that will work and we want a consistent value when we work with inverse sine. Using the range of angles above gives all possible values of the sine function exactly once. If you’re not sure of that sketch out a unit circle and you’ll see that that range of angles (the (y)’s) will cover all possible values of sine.

Note as well that since ( - 1 le sin left( y right) le 1) we also have ( - 1 le x le 1).

Let’s work a quick example.

Example 1 Evaluate (displaystyle {sin ^{ - 1}}left( {frac{1}{2}} right)) Show Solution

So, we are really asking what angle (y) solves the following equation.

[sin left( y right) = frac{1}{2}]

and we are restricted to the values of (y) above.

From a unit circle we can quickly see that (y = frac{pi }{6}).

We have the following relationship between the inverse sine function and the sine function.

[sin left( {{{sin }^{ - 1}}x} right) = xhspace{0.5in}{sin ^{ - 1}}left( {sin x} right) = x]

In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,

[fleft( x right) = sin xhspace{0.5in}gleft( x right) = {sin ^{ - 1}}x]

Then,

[g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{cos left( {{{sin }^{ - 1}}x} right)}}]

This is not a very useful formula. Let’s see if we can get a better formula. Let’s start by recalling the definition of the inverse sine function.

[y = {sin ^{ - 1}}left( x right)hspace{0.5in} Rightarrow hspace{0.5in}x = sin left( y right)]

Using the first part of this definition the denominator in the derivative becomes,

[cos left( {{{sin }^{ - 1}}x} right) = cos left( y right)]

Now, recall that

[{cos ^2}y + {sin ^2}y = 1hspace{0.5in} Rightarrow hspace{0.5in}cos y = sqrt {1 - {{sin }^2}y} ]

Using this, the denominator is now,

[cos left( {{{sin }^{ - 1}}x} right) = cos left( y right) = sqrt {1 - {{sin }^2}y} ]

Now, use the second part of the definition of the inverse sine function. The denominator is then,

[cos left( {{{sin }^{ - 1}}x} right) = sqrt {1 - {{sin }^2}y} = sqrt {1 - {x^2}} ]

Putting all of this together gives the following derivative.

[frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }}]

Inverse Cosine

Now let’s take a look at the inverse cosine. Here is the definition for the inverse cosine.

[y = {cos ^{ - 1}}xhspace{0.5in} Leftrightarrow hspace{0.5in}cos y = xhspace{0.25in}{mbox{for}},0 le y le pi ]

As with the inverse sine we’ve got a restriction on the angles, (y), that we get out of the inverse cosine function. Again, if you’d like to verify this a quick sketch of a unit circle should convince you that this range will cover all possible values of cosine exactly once. Also, we also have ( - 1 le x le 1) because ( - 1 le cos left( y right) le 1).

Example 2 Evaluate (displaystyle {cos ^{ - 1}}left( { - frac{{sqrt 2 }}{2}} right)). Show Solution

As with the inverse sine we are really just asking the following.

[cos y = - frac{{sqrt 2 }}{2}]

where (y) must meet the requirements given above. From a unit circle we can see that we must have (y = frac{{3pi }}{4}).

The inverse cosine and cosine functions are also inverses of each other and so we have, [cos left( {{{cos }^{ - 1}}x} right) = xhspace{0.5in}{cos ^{ - 1}}left( {cos x} right) = x]

To find the derivative we’ll do the same kind of work that we did with the inverse sine above. If we start with

[fleft( x right) = cos xhspace{0.5in}gleft( x right) = {cos ^{ - 1}}x]

then,

[g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{ - sin left( {{{cos }^{ - 1}}x} right)}}]

Simplifying the denominator here is almost identical to the work we did for the inverse sine and so isn’t shown here. Upon simplifying we get the following derivative.

[frac{d}{{dx}}left( {{{cos }^{ - 1}}x} right) = - frac{1}{{sqrt {1 - {x^2}} }}]

So, the derivative of the inverse cosine is nearly identical to the derivative of the inverse sine. The only difference is the negative sign.

Inverse Tangent

Here is the definition of the inverse tangent.

[y = {tan ^{ - 1}}xhspace{0.5in} Leftrightarrow hspace{0.5in}tan y = xhspace{0.25in}{mbox{for}}, - frac{pi }{2} < y < frac{pi }{2}]

Again, we have a restriction on (y), but notice that we can’t let (y) be either of the two endpoints in the restriction above since tangent isn’t even defined at those two points. To convince yourself that this range will cover all possible values of tangent do a quick sketch of the tangent function and we can see that in this range we do indeed cover all possible values of tangent. Also, in this case there are no restrictions on (x) because tangent can take on all possible values.

Example 3 Evaluate ({tan ^{ - 1}}1). Show Solution

Here we are asking,

[tan y = 1]

where (y) satisfies the restrictions given above. From a unit circle we can see that (y = frac{pi }{4}).

Because there is no restriction on (x) we can ask for the limits of the inverse tangent function as (x) goes to plus or minus infinity. To do this we’ll need the graph of the inverse tangent function. This is shown below.

From this graph we can see that

[mathop {lim }limits_{x to infty } {tan ^{ - 1}}x = frac{pi }{2}hspace{0.5in}hspace{0.25in}mathop {lim }limits_{x to - infty } {tan ^{ - 1}}x = - frac{pi }{2}]

The tangent and inverse tangent functions are inverse functions so,

[tan left( {{{tan }^{ - 1}}x} right) = xhspace{0.5in}{tan ^{ - 1}}left( {tan x} right) = x]

Therefore, to find the derivative of the inverse tangent function we can start with

[fleft( x right) = tan xhspace{0.5in}gleft( x right) = {tan ^{ - 1}}x]

We then have,

[g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{{{sec }^2}left( {{{tan }^{ - 1}}x} right)}}]

Simplifying the denominator is similar to the inverse sine, but different enough to warrant showing the details. We’ll start with the definition of the inverse tangent.

[y = {tan ^{ - 1}}xhspace{0.5in} Rightarrow hspace{0.5in}tan y = x]

The denominator is then,

[{sec ^2}left( {{{tan }^{ - 1}}x} right) = {sec ^2}y]

Now, if we start with the fact that

[{cos ^2}y + {sin ^2}y = 1]

and divide every term by cos² (y) we will get,

[1 + {tan ^2}y = {sec ^2}y]

The denominator is then,

[{sec ^2}left( {{{tan }^{ - 1}}x} right) = {sec ^2}y = 1 + {tan ^2}y]

Finally using the second portion of the definition of the inverse tangent function gives us,

[{sec ^2}left( {{{tan }^{ - 1}}x} right) = 1 + {tan ^2}y = 1 + {x^2}]

The derivative of the inverse tangent is then,

[frac{d}{{dx}}left( {{{tan }^{ - 1}}x} right) = frac{1}{{1 + {x^2}}}]

There are three more inverse trig functions but the three shown here the most common ones. Formulas for the remaining three could be derived by a similar process as we did those above. Here are the derivatives of all six inverse trig functions.

We should probably now do a couple of quick derivatives here before moving on to the next section.

Example 4 Differentiate the following functions.

1. (fleft( t right) = 4{cos ^{ - 1}}left( t right) - 10{tan ^{ - 1}}left( t right))
2. (y = sqrt z , {sin ^{ - 1}}left( z right))

Show All SolutionsHide All Solutionsa (fleft( t right) = 4{cos ^{ - 1}}left( t right) - 10{tan ^{ - 1}}left( t right)) Show Solution

Not much to do with this one other than differentiate each term.

[f'left( t right) = - frac{4}{{sqrt {1 - {t^2}} }} - frac{{10}}{{1 + {t^2}}}]
b (y = sqrt z , {sin ^{ - 1}}left( z right)) Show Solution

Don’t forget to convert the radical to fractional exponents before using the product rule.

[y' = frac{1}{2}{z^{ - frac{1}{2}}}{sin ^{ - 1}}left( z right) + frac{{sqrt z }}{{sqrt {1 - {z^2}} }}]

Alternate Notation

There is some alternate notation that is used on occasion to denote the inverse trig functions. This notation is,

[begin{array}{ll}{sin ^{ - 1}}x = arcsin x & hspace{1.0in}{cos ^{ - 1}}x = arccos x {tan ^{ - 1}}x = arctan x & hspace{1.0in}{cot ^{ - 1}}x = {mbox{arccot }}x {sec ^{ - 1}}x = {mathop{rm arcsec}nolimits} ,x & hspace{1.0in}{csc ^{ - 1}}x = {mathop{rm arccsc}nolimits} ,xend{array}]

y = arctan x (Figure 2). Note that the function arctan x is dened for all values. We can now use implicit dierentiation to take the derivative of both sides of our original equation to ge Table of Derivatives. (Math | Calculus | Derivatives | Table Of) Derivative of Arctan. Lesson Contents. How do you Differentiate Arctangent? The inverse tangent — known as arctangent or shorthand as arctan, is usually notated as tan-1(some function)

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Table of Derivatives

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Derivative of arctan(x) (Inverse tangent) Detailed Lesso

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Derivative Of Arctan(y/x)

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Derivative Of Arctangent

The derivative of arctan(x) is 1 / (1 + x^2). I coded up a demo using arctan activation and it worked well in some situations, not so well in others. The only implementation detail was that when computing.. L'arcotangente è una funzione goniometrica inversa, indicata con arctan(x), con arctg(x) o talvolta Cerchi un rapido formulario in cui leggere le proprietà della funzione arcotangente y=arctan(x) e.. A derivative is a securitized contract between two or more parties whose value is dependent upon or derived from one or more underlying assets. Its price is determined by fluctuations in that asset.. Do you want to learn more about Derivative Of Arctan? Struggle no more! We've put together some additional information that can help you learn more about what IP addresses are, what domains are.. Struggling with Computing Derivatives? Let us throw some explanations, examples, and practice problems at your problem. Derivatives of Inverse Trigonometric Functions Examples

Derivative Of Arctan U