Solutions to Differentiation of Inverse Trigonometric Functions

8/7/2019 Solutions to Differentiation of Inverse Trigonometric Functions

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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

.


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SOLUTION 3 : Differentiate arc arc . Apply the product rule. Then

arc arc arc arc

arc arc

= ( arc arc .

SOLUTION 4 : Let arc . Solve f'(x) = 0 forx . Begin by

differentiating f. Then

(Get a common denominator and subtract fractions.)

.


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(It is a fact that if , then A = 0 .) Thus,

2(x - 2)(x+2) = 0 .

(It is a fact that ifAB = 0 , then A = 0 orB=0 .) It follows that

x-2 = 0 orx+2 = 0 ,

that is, the only solutions to f'(x) = 0 are

x = 2 orx = -2 .

SOLUTION 5 : Let . Show that f'(x) = 0 . Conclude

that . Begin by differentiating f. Then

.

Iff'(x) = 0 for all admissable values ofx , then fmust be a constant function, i.e.,

for all admissable values ofx ,

i.e.,

for all admissable values ofx .

In particular, ifx = 0 , then

i.e.,

.

Thus, and for all admissable values ofx .


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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 , it follows that


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.

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.)

.


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SOLUTION 8 : Differentiate . Apply the chain rule twice. Then

(Recall that .)

.

SOLUTION 9 : Differentiate . Apply the chain rule twice. Then

(Recall that .)

.


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SOLUTION 10 : Determine the equation of the line tangent to the graph

of at x = e . Ifx = e , then , so that

the line passes through the point . The slope of the tangent line follows from

the derivative (Apply the chain rule.)

.

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

.

Thus, an equation of the tangent line is

.

SOLUTION 11 : Differentiate arc . What conclusion can be

drawn from your answer about function y ? What conclusion can be drawn about

functions arc and ? First, differentiate, applying the chain rule to

the inverse cotangent function. Then


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= 0 .

Ify' = 0 for all admissable values ofx , then y must be a constant function, i.e.,

for all admissable values ofx ,

i.e.,

arc for all admissable values of x .

In particular, ifx = 1 , then

arc

i.e.,

.

Thus, c = 0 and arc for all admissable values of x . We

conclude that

arc .

Note that this final conclusion follows even more simply and directly from the

definitions of these two inverse trigonometric functions.

SOLUTION 12 : Differentiate . Begin by applying the

product rule to the first summand and the chain rule to the second summand. Then


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.

SOLUTION 13 : Find an equation of the line tangent to the graph

of at x=2 . Ifx = 2 ,

then , so that the line passes through thepoint . The slope of the tangent line follows from the derivative

(Recall that when dividing by a fraction, one must invert and multiply by the

reciprocal. That is .)


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.

The slope of the line tangent to the graph at x = 2 is

.

Thus, an equation of the tangent line is

or

or

.

SOLUTION 14 : Evaluate . Since and , it

follows that takes the indeterminate form `` zero over zero.'' Thus,

we can apply L'H pital's Rule. Begin by differentiating the numerator and

denominator separately. DO NOT apply the quotient rule ! Then

=

=

(Recall that when dividing by a fraction, one must invert and multiply by the

reciprocal. That is .)


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=

=

= .

SOLUTION 15 : A movie screen on the front wall in your classroom is 16 feet high

and positioned 9 feet above your eye -level. How far away from the front of the

room should you sit in order to have the ``best" view ? Begin by introducingvariables x and . (See the diagram below.)

From trigonometry it follows that

,

so that

.

In addition,


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so that

.

It follows that

,

that is, angle is explicitly represented as a function of distance x . Now find the

value ofx which maximizes the value of function . Begin by differentiating

function and setting the derivative equal to zero. Then

.

.

Now solve this equation forx . Then

if

if

if


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if

if

feet .

(Use the first or second derivative test (The first derivative test is easier.) to verify

that this value ofx determines a maximum value for .)

Thus, the ``best'' view is found x=15 feet from the front of the room.

Solutions to Differentiation of Inverse Trigonometric Functions

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