DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS



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. [pic]for [pic]

2. [pic]for [pic]

3. [pic]for [pic]

4. [pic]for [pic], except [pic]

5. [pic]for [pic], except x = 0

6. [pic]for [pic]

Because each of the above-listed functions is one-to-one, each has an inverse function. The corresponding inverse functions are

1. [pic]for [pic]

2. [pic]for [pic]

3. [pic]for [pic]

4. [pic]arc[pic] for [pic], except [pic]

5. [pic]arc[pic] for [pic], except y = 0

6. [pic]arc[pic] for [pic]

In the following discussion and solutions the derivative of a function h(x) will be denoted by [pic]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. [pic]

2. [pic]

3. [pic]

4. [pic]arc[pic]

5. [pic]arc[pic]

6. [pic]arc[pic]

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 [pic]. Apply the product rule. Then

[pic]

[pic]

(Factor an x from each term.)

[pic].

SOLUTION 2 : Differentiate [pic]. Apply the quotient rule. Then

[pic]

[pic]

[pic]

[pic]

[pic].

SOLUTION 3 : Differentiate [pic]arc[pic]arc[pic] . Apply the product rule. Then

[pic]arc[pic]arc[pic]arc[pic]arc[pic]

[pic]arc[pic]arc[pic]

= ( arc[pic]arc[pic] .

SOLUTION 4 : Let [pic]arc[pic] . Solve f'(x) = 0 for x . Begin by differentiating f . Then

[pic]

[pic]

(Get a common denominator and subtract fractions.)

[pic]

[pic]

[pic]

[pic]

[pic].

(It is a fact that if [pic], 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 [pic]. Show that f'(x) = 0 . Conclude that [pic]. Begin by differentiating f . Then

[pic].

If f'(x) = 0 for all admissable values of x , then f must be a constant function, i.e.,

[pic]for all admissable values of x ,

i.e.,

[pic]for all admissable values of x .

In particular, if x = 0 , then

[pic]

i.e.,

[pic].

Thus, [pic]and [pic]for all admissable values of x .

SOLUTION 6 : Evaluate [pic]. It may not be obvious, but this problem can be viewed as a derivative problem. Recall that

[pic]

(Since h approaches 0 from either side of 0, h can be either a positve or a negative number. In addition, [pic]is equivalent to [pic]. This explains the following equivalent variations in the limit definition of the derivative.)

[pic]

[pic].

If [pic], then [pic], and letting [pic], it follows that

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic].

The following problems require use of the chain rule.

SOLUTION 7 : Differentiate [pic]. Use the product rule first. Then

[pic]

(Apply the chain rule in the first summand.)

[pic]

[pic]

(Factor out [pic]. Then get a common denominator and add.)

[pic]

[pic]

[pic].

SOLUTION 8 : Differentiate [pic]. Apply the chain rule twice. Then

[pic]

[pic]

(Recall that [pic].)

[pic]

[pic].

SOLUTION 9 : Differentiate [pic]. Apply the chain rule twice. Then

[pic]

(Recall that [pic].)

[pic]

[pic]

[pic].

SOLUTION 10 : Determine the equation of the line tangent to the graph of [pic]at x = e . If x = e , then [pic], so that the line passes through the point [pic]. The slope of the tangent line follows from the derivative (Apply the chain rule.)

[pic]

[pic]

[pic].

The slope of the line tangent to the graph at x = e is

[pic]

[pic]

[pic].

Thus, an equation of the tangent line is

[pic].

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download