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.125truein ^^^^''0 0^''

SOLUTION 1: Differentiate tex2html_wrap_inline416 . Apply the product rule. Then

tex2html_wrap_inline418

tex2html_wrap_inline420

(Factor an x from each term.)

tex2html_wrap_inline424 .

SOLUTION 2: Differentiate tex2html_wrap_inline426 . Apply the quotient rule. Then

tex2html_wrap_inline428

tex2html_wrap_inline430

tex2html_wrap_inline432

tex2html_wrap_inline434

tex2html_wrap_inline436 .

SOLUTION 3: Differentiate tex2html_wrap_inline438arctex2html_wrap_inline440arctex2html_wrap_inline442 . Apply the product rule. Then

tex2html_wrap_inline444arctex2html_wrap_inline446arctex2html_wrap_inline448arctex2html_wrap_inline450arctex2html_wrap_inline442

tex2html_wrap_inline454arctex2html_wrap_inline456arctex2html_wrap_inline442

= ( arctex2html_wrap_inline462arctex2html_wrap_inline464 .

SOLUTION 4: Let tex2html_wrap_inline466arctex2html_wrap_inline468 . Solve f'(x) = 0 for x . Begin by differentiating f . Then

tex2html_wrap_inline476

tex2html_wrap_inline478

(Get a common denominator and subtract fractions.)

tex2html_wrap_inline480

tex2html_wrap_inline482

tex2html_wrap_inline484

tex2html_wrap_inline486

tex2html_wrap_inline488 .

(It is a fact that if tex2html_wrap_inline490 , 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 tex2html_wrap_inline512 . Show that f'(x) = 0 . Conclude that tex2html_wrap_inline516. Begin by differentiating f . Then

tex2html_wrap_inline520 .

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

tex2html_wrap_inline528 for all admissable values of x ,

i.e.,

tex2html_wrap_inline532 for all admissable values of x .

In particular, if x = 0 , then

tex2html_wrap_inline538

i.e.,

tex2html_wrap_inline540 .

Thus, tex2html_wrap_inline542 and tex2html_wrap_inline516 for all admissable values of x .

SOLUTION 6: Evaluate tex2html_wrap_inline548 . It may not be obvious, but this problem can be viewed as a derivative problem. Recall that

tex2html_wrap_inline550

(Since h approaches 0 from either side of 0, h can be either a positve or a negative number. In addition, tex2html_wrap_inline556 is equivalent to tex2html_wrap_inline558. This explains the following equivalent variations in the limit definition of the derivative.)

tex2html_wrap_inline560

tex2html_wrap_inline562 .

If tex2html_wrap_inline564 , then tex2html_wrap_inline566, and letting tex2html_wrap_inline568, it follows that

tex2html_wrap_inline570

tex2html_wrap_inline572

tex2html_wrap_inline574

tex2html_wrap_inline576

tex2html_wrap_inline578

tex2html_wrap_inline580

tex2html_wrap_inline582

tex2html_wrap_inline584

tex2html_wrap_inline586 .

The following problems require use of the chain rule.

SOLUTION 7: Differentiate tex2html_wrap_inline588 . Use the product rule first. Then

tex2html_wrap_inline590

(Apply the chain rule in the first summand.)

tex2html_wrap_inline592

tex2html_wrap_inline594

(Factor out tex2html_wrap_inline596. Then get a common denominator and add.)

tex2html_wrap_inline598

tex2html_wrap_inline600

tex2html_wrap_inline602 .

SOLUTION 8: Differentiate tex2html_wrap_inline604 . Apply the chain rule twice. Then

tex2html_wrap_inline606

tex2html_wrap_inline608

(Recall that tex2html_wrap_inline610.)

tex2html_wrap_inline612

tex2html_wrap_inline614 .

SOLUTION 9: Differentiate tex2html_wrap_inline616 . Apply the chain rule twice. Then

tex2html_wrap_inline618

(Recall that tex2html_wrap_inline620.)

tex2html_wrap_inline622

tex2html_wrap_inline624

tex2html_wrap_inline626 .

SOLUTION 10: Determine the equation of the line tangent to the graph of tex2html_wrap_inline628 at x = e . If x = e , then tex2html_wrap_inline634, so that the line passes through the point tex2html_wrap_inline636. The slope of the tangent line follows from the derivative (Apply the chain rule.)

tex2html_wrap_inline638

tex2html_wrap_inline640

tex2html_wrap_inline642 .

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

tex2html_wrap_inline646

tex2html_wrap_inline648

tex2html_wrap_inline650 .

Thus, an equation of the tangent line is

tex2html_wrap_inline652 .

SOLUTION 11: Differentiate tex2html_wrap_inline654arctex2html_wrap_inline656 . What conclusion can be drawn from your answer about function y ? What conclusion can be drawn about functions arctex2html_wrap_inline662 and tex2html_wrap_inline596 ? First, differentiate, applying the chain rule to the inverse cotangent function. Then

tex2html_wrap_inline666

tex2html_wrap_inline668

tex2html_wrap_inline670

= 0 .

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

tex2html_wrap_inline680 for all admissable values of x ,

i.e.,

arctex2html_wrap_inline686 for all admissable values of x .

In particular, if x = 1 , then

arctex2html_wrap_inline694

i.e.,

tex2html_wrap_inline696 .

Thus, c = 0 and arctex2html_wrap_inline702 for all admissable values of x . We conclude that

arctex2html_wrap_inline708 .

Note that this final conclusion follows even more simply and directly from the definitions of these two inverse trigonometric functions.

SOLUTION 12: Differentiate tex2html_wrap_inline710 . Begin by applying the product rule to the first summand and the chain rule to the second summand. Then

tex2html_wrap_inline712

tex2html_wrap_inline714

tex2html_wrap_inline716

tex2html_wrap_inline718 .

SOLUTION 13: Find an equation of the line tangent to the graph of tex2html_wrap_inline720 at x=2 . If x = 2 , then tex2html_wrap_inline726, so that the line passes through the point tex2html_wrap_inline728. The slope of the tangent line follows from the derivative

tex2html_wrap_inline730

tex2html_wrap_inline732

tex2html_wrap_inline734

tex2html_wrap_inline736

tex2html_wrap_inline738

tex2html_wrap_inline740

(Recall that when dividing by a fraction, one must invert and multiply by the reciprocal. That is tex2html_wrap_inline742 .)

tex2html_wrap_inline744

tex2html_wrap_inline746 .

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

tex2html_wrap_inline750 .

Thus, an equation of the tangent line is

tex2html_wrap_inline752

or

tex2html_wrap_inline754

or

tex2html_wrap_inline756 .

SOLUTION 14: Evaluate tex2html_wrap_inline758 . Since tex2html_wrap_inline760 and tex2html_wrap_inline762, it follows that tex2html_wrap_inline758 takes the indeterminate form . Thus, we can apply L'Htex2html_wrap_inline768pital's Rule. Begin by differentiating the numerator and denominator separately. DO NOT apply the quotient rule ! Then

tex2html_wrap_inline758 = tex2html_wrap_inline772

= tex2html_wrap_inline774

(Recall that when dividing by a fraction, one must invert and multiply by the reciprocal. That is tex2html_wrap_inline776 .)

= tex2html_wrap_inline778

= tex2html_wrap_inline780

= tex2html_wrap_inline782 .

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 introducing variables x and tex2html_wrap_inline786. (See the diagram below.)

From trigonometry it follows that

tex2html_wrap_inline788,

so that

tex2html_wrap_inline790 .

In addition,

tex2html_wrap_inline792

so that

tex2html_wrap_inline794 .

It follows that

tex2html_wrap_inline796

tex2html_wrap_inline798,

that is, angle tex2html_wrap_inline800 is explicitly a function of distance x . Now find the value of x which maximizes the value of function tex2html_wrap_inline800. Begin by differentiating function tex2html_wrap_inline800 and setting the derivative equal to zero. Then

tex2html_wrap_inline810

tex2html_wrap_inline812

tex2html_wrap_inline814.

tex2html_wrap_inline816.

Now solve this equation for x . Then

tex2html_wrap_inline820

iff

tex2html_wrap_inline822

iff

tex2html_wrap_inline824

iff

tex2html_wrap_inline826

iff

tex2html_wrap_inline828

iff

tex2html_wrap_inline830 feet .

Use the first or second derivative test (The first derivative test is easier.) to verify that this value of x determines a maximum value for tex2html_wrap_inline800.




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Duane Kouba
Tue Sep 16 16:10:59 PDT 1997