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Derivatives of the Inverse Trigonometric Functions

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Determining the Derivatives of the Inverse Trigonometric Functions

Now let's determine the derivatives of the inverse trigonometric functions, y=arcsinx, y=arccosx, y=arctanx, y=arccotx, y=arcsecx, and y=arccscx.

One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. An added benefit of this approach is that it will prepare you to be more successful in a future topic called trigonometric substitution.

Example 1: Finding the derivative of y=arcsinx

Find the derivative of y=arcsinx.

Solution:

To find the derivative of y=arcsinx, we will first rewrite this equation in terms of its inverse form. That is, siny=x

Now this equation shows that y can be considered an acute angle in a right triangle with a sine ratio of x1. Since the sine ratio gives us the length of the opposite side over the length of the hypotenuse, this means that the opposite side has a length of x and the hypotenuse has a length of 1. See Figure 1.

inverseTrigSineRtTri1a.png
Figure 1

Now let's differentiate Equation ??? implicitly with respect to x.

cosydydx=1

Then we solve this for dydx.

dydx=1cosy

Looking at Figure 1, we see that cosy=a. Now we use the Pythagorean Theorem to find an expression for a in terms of x using the other sides of the right triangle that we know.

Thus we have:

a2+x2=12a2=1x2a=1x2

inverseTrigSineRtTri1b.png
Figure 2

Now using this result, we see that cosy=1x2

so

dydx=1cosy=11x2

Thus we have found the derivative of y=arcsinx,

ddx(arcsinx)=11x2

Exercise 1

Use the same approach to determine the derivatives of y=arccosx, y=arctanx, and y=arccotx.

Answer

ddx(arccosx)=11x2ddx(arctanx)=11+x2ddx(arccotx)=11+x2

Example 2: Finding the derivative of y=arcsecx

Find the derivative of y=arcsecx.

Solution:

To find the derivative of y=arcsecx, we will first rewrite this equation in terms of its inverse form. That is, secy=x

As before, let y be considered an acute angle in a right triangle with a secant ratio of x1. Since the secant ratio is the reciprocal of the cosine ratio, it gives us the length of the hypotenuse over the length of the adjacent side, so this means that the hypotenuse has a length of x and the adjacent side has a length of 1. See Figure 3.

inverseTrigSecRtTri1a.png
Figure 3

Differentiating Equation ??? implicitly with respect to x, gives us:

secytanydydx=1

Solving this for dydx, we get:

dydx=1secytany

In order to find tany in terms of x, we need to find the length of the opposite side, a, in terms of x. By the Pythagorean Theorem, we have:

12+a2=x2a2=x21a=x21

Figure 4 shows the resulting right triangle.

inverseTrigSecRtTri1b.png
Figure 4

From the right triangle in Figure 4, we can see that tany=x21.

Since secy=x, it appears that

dydx=1secytany=1xx21.

But this is not completely correct, at least not for negative values of x. Considering the graph of y=arcsecx in Figure 5, we see that its slope is always positive. But for negative values of x, the form of the derivative stated above would be negative (and clearly incorrect).

arcsecx2.png
Figure 5

As we'll prove below, the actual derivative formula for this function is:

ddx(arcsecx)=1|x|x21

Consider the domain and range of the original function, y=arcsecx:

Domain: (,1][1,)or|x|1
Range: [0,π2)(π2,π]or0yπ,yπ2

Note that the domain of the derivative is a subset of the domain of the original function, excluding the endpoints, x=1 and x=1.

Now, let's rewrite dydx as:

dydx=1secytany=cosy1cosysiny=cos2ysiny

We see that this function allows y to have all values between 0 and π, except for y=π2, where the original expression 1secytany is undefined. This is a subset of the range of the original function y=arcsecx. Note that in the derivative, y cannot take on the values of the endpoints of this interval, while these were part of the range of y=arcsecx.

Now consider that for all values of y in this range between 0 and π, the expression for dydx is positive, as both

siny>0 and cos2y>0 for all y(0,π).

Therefore, we have that,

ddx(arcsecx)=1|x|x21.

Exercise 2

Use the same approach to determine the derivative of y=arccscx.

Answer

ddx(arccscx)=1|x|x21.

Using the Chain Rule with Inverse Trigonometric Functions

Now let's see how to use the chain rule to find the derivatives of inverse trigonometric functions with more interesting functional arguments.

Example 3:

Find the derivatives for each of the following functions:

  1. y=arcsin(x2)
  2. y=arctan(x3+1)
  3. y=arcsec(ln|x|)
  4. y=arcsin(cosx)

Solution:

  1. Using the chain rule, we see that: ddx(arcsin(x2))=11(x2)2ddx(x2)=2x1x4
  2. Here we have: ddx(arctan(x3+1))=11+(x3+1)2ddx(x3+1)=3x21+(x3+1)2
    Although it would likely be fine as it is, we can simplify it to obtain: ddx(arctan(x3+1))=3x2x6+2x3+2
  3. For y=arcsec(ln|x|), we obtain: ddx(arcsec(ln|x|))=1|ln|x||(ln|x|)21ddx(ln|x|)=1x|ln|x||(ln|x|)21
  4. For y=arcsin(cosx), we obtain: ddx(arcsin(cosx))=11(cosx)2ddx(cosx)=sinxsin2x
    Note that it may look like the denominator should simplify to sinx and the entire derivative to dydx=1. But this is not the case. Remember that
    x2=|x|.
    This means that what we actually obtain here is something more interesting: ddx(arcsin(cosx))=sinx|sinx|
    This function alternates between values of 1 and 1, when sinx is positive and negative, respectively.
    See the graph of the function y=arcsin(cosx) below in Figure 6.

plotArcsin(cosx).png
Figure 6: Graph of y=arcsin(cosx).

Derivative Formulas

In the same way that we can encapsulate the chain rule in the derivative of lnu as ddx(lnu)=uu, we can write formulas for the derivative of the inverse trigonometric functions that encapsulate the chain rule. Note that u represents a function of x in these formulas, and u represents the derivative of u with respect to x.

ddx(arcsinu)=u1u2ddx(arccosu)=u1u2ddx(arctanu)=u1+u2ddx(arccotu)=u1+u2ddx(arcsecu)=u|u|u21ddx(arccscu)=u|u|u21

Contributors

  • Paul Seeburger (Monroe Community College)

This page titled Derivatives of the Inverse Trigonometric Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Paul Seeburger.

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