.
Then
(Apply the product rule in the second part of the numerator. Recall that 
.)
(Begin to simplify the final answer by getting a common denominator in the numerator.)
(Simplify the numerator and invert and multiply by the reciprocal of the denominator.)
(Recall that 
.)
.
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SOLUTION 13 : Differentiate 
. First apply the chain rule. Then
(Now apply the quotient rule and recall that
. Please not that there is a TYPO in the remaining steps of this solution. Every subsequent term (3x+1) should be (3x+2).)
(Recall that
and 
.)
(Recall that
.)
(Please note that there is a TYPO in the final answer. The term (3x+1) should be (3x+2).)
.
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SOLUTION 14 : Differentiate 
.
Recall that 
and apply the chain rule. Then
(Now apply the quotient rule.)
.
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SOLUTION 15 : Differentiate 
. Apply the product rule first. Then
(Recall that 
and apply the quotient rule.)
(Begin to simplify your answer by finding a common denominator.)
(Factor x and 
from the numerator.)
.
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SOLUTION 16 : Find an equation of the line tangent to the graph of 
at x=-1 . If x= -1 then
so that the tangent line passes through the point
(-1, 1 ) . The slope of the tangent line follows from the derivative of y . Then
.
The slope of the line tangent to the graph at x = -1 is
.
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 
at 
. If then
so that the tangent line passes through the point
. The slope of the tangent line follows from the derivative of y . Then
.
There is no need to simpliy this derivative. Just let in the derivative equation. Thus,
the slope of the line tangent to the graph at is
m = y'
.
Thus, an equation of the tangent line is
.
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SOLUTION 18 : Consider the function 
.
Solve f'(x) = 0 for x . Solve f''(x) = 0 for x . The first derivative is
(Factor out the common terms of 2x and 
in the numerator.)
(Recall that 
and 
.)
.
(It is a fact that if 
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
(Factor 
from the numerator.)
.
It follows that
,
so that using the quadratic formula we get that
or 
are the only solutions to f''(x) = 0 .
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SOLUTION 19 : Find all points (x, y) on the graph of 
where tangent lines are perpendicular to the line 8x+2y = 1 . Since
8x+2y = 1 iff 2y = 1 - 8x iff 
,
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
.
The slope of a tangent line is also given by the derivative
.
Thus, equating slopes, it follows that
iff
iff
(Recall that 
.)
iff
| 2-x | = 2
iff
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
,
and if x = 4 , then
,
so that the two distinct points of tangency are
( 0, -1/ 2 ) and ( 4, -3/ 2 ) .
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