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Derivative of arctan. What is the derivative of the arctangent function of x? The derivative of the arctangent function of x is equal to 1 divided by (1+x 2). Derivative of arctan(x) Let’s use our formula for the derivative of an inverse function to find the deriva tive of the inverse of the tangent function: y = tan−1 x = arctan x. We simplify the equation by taking the tangent of both sides.
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Section 3-7 : Derivatives of Inverse Trig Functions
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, Download graphicconverter 10 5 5.
[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
![]()
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.
[begin{array}{ll}displaystyle frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }} & hspace{1.0in}displaystyle frac{d}{{dx}}left( {{{cos }^{ - 1}}x} right) = - frac{1}{{sqrt {1 - {x^2}} }} displaystyle frac{d}{{dx}}left( {{{tan }^{ - 1}}x} right) = frac{1}{{1 + {x^2}}} & hspace{1.0in}displaystyle frac{d}{{dx}}left( {{{cot }^{ - 1}}x} right) = - frac{1}{{1 + {x^2}}} displaystyle frac{d}{{dx}}left( {{{sec }^{ - 1}}x} right) = frac{1}{{left| x right|sqrt {{x^2} - 1} }} & hspace{1.0in}displaystyle frac{d}{{dx}}left( {{{csc }^{ - 1}}x} right) = - frac{1}{{left| x right|sqrt {{x^2} - 1} }}end{array}]
We should probably now do a couple of quick derivatives here before moving on to the next section.
Example 4 Differentiate the following functions.
![]()
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}]
An Approach
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13
IT IS NOT NECESSARY to memorize the derivatives of this Lesson. Rather, the student should know now to derive them.
In Topic 19 of Trigonometry, we introduced the inverse trigonometric functions. According to the inverse relations:
y = arcsin x implies sin y = x.
And similarly for each of the inverse trigonometric functions.
Problem 1. If y = arcsin x, show:
To see the answer, pass your mouse over the colored area.
To cover the answer again, click 'Refresh' ('Reload'). Do the problem yourself first!
Begin:
We take the positive sign because cos y is positive for all values of y in the range of y = arcsin x, which is the 1st and 4th quadrants. (Topic 19 of Trigonometry.)
Problem 2. If y = arcsec x, show:
Begin:
The derivative of y = arcsin x
The derivative of the arcsine with respect to its argument
is equal to 1 over the square root of 1 minus the square of the argument.
Here is the proof:
according to Problem 1.
That is what we wanted to prove.
Note: We could have used the theorem of Lesson 8 directly:
We will use that theorem in the proofs that follow.
Problem 3. Calculate these derivatives. [In parts a) and b), use the chain rule.]
The derivative of y = arccos x
The derivative of arccos x is the negative of the derivative
of arcsin x. That will be true for the inverse of each pair of cofunctions.
The derivative of arccot x will be the negative
of the derivative of arctan x.
The derivative of arccsc x will be the negative
of the derivative of arcsec x.
For, beginning with arccos x:
The angle whose cosine is x is the complement
of the angle whose sine is x.
Problem 4. Calculate these derivatives.
The derivative of y = arctan x
First,
y = arctan x implies tan y = x.
Therefore, according to the theorem of Lesson 8:
Which is what we wanted to prove.
Therefore, the derivative of arccot x is its negative:
Problem 5. Calculate these derivatives.
Derivative Of Arctanh
*
The remaining derivatives come up rarely in calculus. Nevertheless, here are the proofs.
The derivative of y = arcsec x
Again,
y = arcsec x implies sec y = x.
Therefore, according to the theorem of Lesson 8:
Now, according to the theorem of Topic 19 of Trigonometry: that product is never negative. Therefore to ensure that, rather than replacing sec y with x, we will replace it with |x|. And in Problem 2 we will take only the positive root of tan y.
Derivative Of Arctan 1
Therefore,
Which is what we wanted to prove.
If we took the range of arcsec x to be a third quadrant angle between −π and −π/2, when x is negative, then we would not need to write the absolute value, and the proof would be straightforward. We would simply replace sec y with x, and take the positive root of tan y, because tan y is positive in the first and third quadrants. In the graph of y = arcsec x with that range, the slope for negative x is negative. The disadvantage of taking that range is that, when x is negative, arcsec x will not equal arccos 1/x, because arccos 1/x will be a 2nd quadrant angle. But then in the proof we have to write the absolute value.
The derivative, therefore, of arccsc x is its negative:
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