DIFFERENTIATION OF INVERSE TRIGONOMETRIC
FUNCTIONS
None of the six basic trigonometry functions is a one-to-one function. However, in the
following list, each trigonometry function is listed with an appropriately restricted
domain, which makes it one-to-one.
1.
for
2.
for
3.
for
4.
for
5.
for
6.
for
, except
, except x = 0
Because each of the above-listed functions is one-to-one, each has an inverse
function. The corresponding inverse functions are
1.
for
2.
for
3.
for
4.
arc
for
5.
arc
for
6.
arc
for
, except
, except y = 0
In the following discussion and solutions the derivative of a function h(x) will be
denoted by
or h'(x) . The derivatives of the above-mentioned inverse
trigonometric functions follow from trigonometry identities, implicit differentiation,
and the chain rule. They are as follows.
1.
2.
3.
4.
arc
1
5.
arc
6.
arc
In the list of problems which follows, most problems are average and a few are
somewhat challenging
SOLUTIONS TO DIFFERENTIATION OF INVERSE
TRIGONOMETRIC FUNCTIONS
SOLUTION 1 : Differentiate
. Apply the product rule. Then
(Factor an x from each term.)
.
SOLUTION 2 : Differentiate
. Apply the quotient rule. Then
2
.
SOLUTION 3 : Differentiate
arc
arc
arc
arc
arc
arc
arc
arc
= ( arc
SOLUTION 4 : Let
differentiating f . Then
. Apply the product rule. Then
arc
.
arc
. Solve f'(x) = 0 for x . Begin by
(Get a common denominator and subtract fractions.)
3
.
(It is a fact that if
, then A = 0 .) Thus,
2(x - 2)(x+2) = 0 .
(It is a fact that if AB = 0 , then A = 0 or B=0 .) It follows that
x-2 = 0 or x+2 = 0 ,
that is, the only solutions to f'(x) = 0 are
x = 2 or x = -2 .
SOLUTION 5 : Let
. Show that f'(x) = 0 . Conclude that
. Begin by differentiating f . Then
.
If f'(x) = 0 for all admissable values of x , then f must be a constant function, i.e.,
for all admissable values of x ,
i.e.,
for all admissable values of x .
In particular, if x = 0 , then
4
i.e.,
.
Thus,
and
for all admissable values of x .
SOLUTION 6 : Evaluate
. It may not be obvious,
but this problem can be viewed as a derivative problem. Recall that
(Since h approaches 0 from either side of 0, h can be either a positve or a negative
number. In addition,
is equivalent to
. This explains the following
equivalent variations in the limit definition of the derivative.)
.
If
, then
, and letting
5
, it follows that
.
The following problems require use of the chain rule.
SOLUTION 7 : Differentiate
. Use the product rule first. Then
(Apply the chain rule in the first summand.)
(Factor out
. Then get a common denominator and add.)
.
6
SOLUTION 8 : Differentiate
(Recall that
. Apply the chain rule twice. Then
.)
.
SOLUTION 9 : Differentiate
(Recall that
. Apply the chain rule twice. Then
.)
.
7
SOLUTION 10 : Determine the equation of the line tangent to the graph of
at x = e . If x = e , then
the line passes through the point
derivative (Apply the chain rule.)
, so that
. The slope of the tangent line follows from the
.
The slope of the line tangent to the graph at x = e is
.
Thus, an equation of the tangent line is
.
8