SOLUTIONS TO DIFFERENTIATION OF FUNCTIONS USING THE QUOTIENT RULE



SOLUTION 12 : Differentiate tex2html_wrap_inline913 .

Then

tex2html_wrap_inline915

(Apply the product rule in the second part of the numerator. Recall that tex2html_wrap_inline917.)

tex2html_wrap_inline919

tex2html_wrap_inline921

tex2html_wrap_inline923

tex2html_wrap_inline925

(Begin to simplify the final answer by getting a common denominator in the numerator.)

tex2html_wrap_inline927

(Simplify the numerator and invert and multiply by the reciprocal of the denominator.)

tex2html_wrap_inline929

tex2html_wrap_inline931

tex2html_wrap_inline933

(Recall that tex2html_wrap_inline935 .)

tex2html_wrap_inline937

tex2html_wrap_inline939 .

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SOLUTION 13 : Differentiate tex2html_wrap_inline941 . First apply the chain rule. Then

tex2html_wrap_inline943

(Now apply the quotient rule and recall that tex2html_wrap_inline945 .)

tex2html_wrap_inline947

(Recall that tex2html_wrap_inline949 and tex2html_wrap_inline951 .)

tex2html_wrap_inline953

(Recall that tex2html_wrap_inline955 .)

tex2html_wrap_inline957

tex2html_wrap_inline959 .

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SOLUTION 14 : Differentiate tex2html_wrap_inline961 . Recall that tex2html_wrap_inline963 and apply the chain rule. Then

tex2html_wrap_inline965

(Now apply the quotient rule.)

tex2html_wrap_inline967

tex2html_wrap_inline969

tex2html_wrap_inline971

tex2html_wrap_inline973

tex2html_wrap_inline975 .

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SOLUTION 15 : Differentiate tex2html_wrap_inline110 . Apply the product rule first. Then

tex2html_wrap_inline112

(Recall that tex2html_wrap_inline114 and apply the quotient rule.)

tex2html_wrap_inline116

tex2html_wrap_inline118

tex2html_wrap_inline120

tex2html_wrap_inline122

tex2html_wrap_inline124

(Begin to simplify your answer by finding a common denominator.)

tex2html_wrap_inline126

tex2html_wrap_inline128

tex2html_wrap_inline130

(Factor x and tex2html_wrap_inline134 from the numerator.)

tex2html_wrap_inline136

tex2html_wrap_inline138

tex2html_wrap_inline140

tex2html_wrap_inline142 .

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SOLUTION 16 : Find an equation of the line tangent to the graph of tex2html_wrap_inline1011 at x=-1 . If x= -1 then tex2html_wrap_inline1017 so that the tangent line passes through the point (-1, 1 ) . The slope of the tangent line follows from the derivative of y . Then

tex2html_wrap_inline1023

tex2html_wrap_inline1025

tex2html_wrap_inline1027

tex2html_wrap_inline1029 .

The slope of the line tangent to the graph at x = -1 is

tex2html_wrap_inline1033 .

Thus, an equation of the tangent line is

y - 1 = -5 (x - (-1) ) or y = -5x - 4 .

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SOLUTION 17 : Find an equation of the line tangent to the graph of tex2html_wrap_inline1039 at tex2html_wrap_inline1041 . If tex2html_wrap_inline1041 then tex2html_wrap_inline1045 tex2html_wrap_inline1047 so that the tangent line passes through the point tex2html_wrap_inline1049 . The slope of the tangent line follows from the derivative of y . Then

tex2html_wrap_inline1053

tex2html_wrap_inline1055 .

There is no need to simpliy this derivative. Just let tex2html_wrap_inline1041 in the derivative equation. Thus, the slope of the line tangent to the graph at tex2html_wrap_inline1041 is

m = y'

tex2html_wrap_inline1063

tex2html_wrap_inline1065

tex2html_wrap_inline1067

tex2html_wrap_inline1069

tex2html_wrap_inline1071

tex2html_wrap_inline1073 .

Thus, an equation of the tangent line is

tex2html_wrap_inline1075 .

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SOLUTION 18 : Consider the function tex2html_wrap_inline1077 . Solve f'(x) = 0 for x . Solve f''(x) = 0 for x . The first derivative is

tex2html_wrap_inline1087

tex2html_wrap_inline1089

(Factor out the common terms of 2x and tex2html_wrap_inline1093 in the numerator.)

tex2html_wrap_inline1095

(Recall that tex2html_wrap_inline1097 and tex2html_wrap_inline1099 .)

tex2html_wrap_inline1101

tex2html_wrap_inline1103 .

(It is a fact that if tex2html_wrap_inline1105 then A = 0 .)

Thus,

2x (1-x) = 0 ,

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

so that

2x = 0 or 1-x = 0 ,

i.e.,

x = 0 or x = 1

are the only solutions to f'(x) = 0 .

The second derivative is

tex2html_wrap_inline1127

tex2html_wrap_inline1129

tex2html_wrap_inline1131

(Factor tex2html_wrap_inline1133 from the numerator.)

tex2html_wrap_inline1135

tex2html_wrap_inline1137

tex2html_wrap_inline1139 .

It follows that

tex2html_wrap_inline1141 ,

so that using the quadratic formula we get that

tex2html_wrap_inline1143

tex2html_wrap_inline1145

tex2html_wrap_inline1147

tex2html_wrap_inline1149 or tex2html_wrap_inline1151

are the only solutions to f''(x) = 0 .

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SOLUTION 19 : Find all points (x, y) on the graph of tex2html_wrap_inline1157 where tangent lines are perpendicular to the line 8x+2y = 1 . Since

8x+2y = 1 iff 2y = 1 - 8x iff tex2html_wrap_inline1165 ,

the slope of this given line is -4. Thus, a tangent line to the graph of f which is perpendicular to 8x+2y = 1 will have slope

tex2html_wrap_inline1171 .

The slope of a tangent line is also given by the derivative

tex2html_wrap_inline1173

tex2html_wrap_inline1175

tex2html_wrap_inline1177

tex2html_wrap_inline1179 .

Thus, equating slopes, it follows that

tex2html_wrap_inline1181

iff

tex2html_wrap_inline1183

iff

tex2html_wrap_inline1185

(Recall that tex2html_wrap_inline1187 .)

iff

| 2-x | = 2

iff

tex2html_wrap_inline1191

iff

x = 0 or x = 4 .

It follows that there are two distinct tangent lines which are perpendicular to the line 8x+2y = 1 . If x = 0 , then

tex2html_wrap_inline1201 ,

and if x = 4 , then

tex2html_wrap_inline1205 ,

so that the two distinct points of tangency are

( 0, 1/ 2 ) and ( 4, -3/ 2 ) .

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Duane Kouba
Wed Jul 30 20:10:53 PDT 1997